In the 1780s, Luigi Galvani, a professor of anatomy at Bologna, noticed something he could not explain. A dissected frog’s leg lying on his laboratory bench twitched sharply whenever it was touched with metal instruments. Galvani concluded that the muscle itself stored a special kind of “animal electricity.”
Alessandro Volta, a physicist at Pavia, was not convinced. He noticed that the twitch appeared when two different metals touched the moist tissue. To Volta, the frog was only a detector; the real source of the electricity was the contact between two unlike metals through a moist conducting layer. To prove it, he removed the animal altogether. Around 1800 he stacked alternating discs of zinc and silver, separating each pair with cardboard soaked in brine, and built a tall pile. When he connected the two ends with a wire, current flowed steadily — the first device that could deliver electricity continuously rather than in a single spark.
The voltaic pile changed chemistry almost overnight. Within months, researchers in England connected such a pile to water and watched it separate into hydrogen and oxygen gas. Chemical change had produced electricity; now electricity was producing chemical change. That two-way traffic between chemical energy and electrical energy is what this chapter is about.
Q1. A 5% solution of common salt makes a bulb glow brightly, but a 5% sugar solution leaves it dark. Both are clear liquids that look identical — so what exactly is missing in one of them? Q2. When you dilute an electrolyte solution, its conductivity falls but its molar conductivity rises. How can both statements be true about the same solution at the same time? Q3. A torch cell pushes electrons out on its own, while an electroplating bath needs a battery to force a reaction along. What decides which of these two directions a cell will run in? Q4. Chemists say it is impossible to measure the potential of a single electrode. If that is so, how are the electrode potential values printed in every textbook table obtained? Q5. A car battery can be recharged thousands of times, but an ordinary dry cell cannot be recharged even once. What must be different about what happens at the electrodes in the two cases?
Keep these questions with you as you read. The chapter will give you everything you need to reason out each answer for yourself.
5.1 Introduction
- What is a redox reaction?
- Which form of energy is converted into electrical energy in dry cells?
- How is NaOH manufactured from NaCl?
A dry cell powers a torch, a wall clock, a TV remote and dozens of other small electrical and electronic devices. It does this because it generates electricity. But where does that electricity come from? There is no fuel burning inside a dry cell and no wire coming out of a power station.
The answer is that a chemical reaction takes place inside the cell. As the chemicals in the cell react, energy stored in their bonds is released, and the cell delivers that energy as an electric current. In a dry cell, therefore, chemical energy is converted into electrical energy.
The Reverse Journey
You have already met a process that runs the other way round — electrolysis. When electricity is passed through a solution containing ions, the ionic compound present is broken down. Electrolysis is the breaking down of an ionic compound by the passage of electricity.
Notice what that means. The breaking down of an electrolyte during electrolysis is a chemical reaction, and it takes place only because electricity is being passed through. Here electrical energy is converted into chemical energy — the exact reverse of what happens in a dry cell.
These two situations, taken together, define an entire branch of the subject.
Electrochemistry is the area of chemistry that deals with the interconversion of chemical energy and electrical energy.
An electrolyte is a substance that furnishes ions and conducts electricity when it is melted or dissolved in water; electrolysis is the decomposition of such a compound by passing electricity through it.
The two-way traffic between the two forms of energy, and the processes that belong to each direction, are set out in FC 5.1.
What Else Electrochemistry Covers
Electrochemistry is not only about producing current. It also deals with the resistance and the conductance of aqueous electrolytic solutions — that is, with how easily a solution lets electricity pass through it.
This turns out to be far more useful than it first sounds. When an electrolyte such as acetic acid dissolves in water, only a fraction of its molecules split into ions. Measuring how well the solution conducts tells us how large that fraction is. The determination of conductivities of aqueous electrolytic solutions therefore provides information on the extent of ionization of electrolytes in water, a topic you studied in Chapter 3.
Redox Reactions Are the Foundation
In Standard XI you learnt redox reactions — reactions in which electrons are transferred from one species to another. The species that loses electrons is said to be oxidised, and the species that gains them is said to be reduced. Because a free electron cannot simply float away in solution, the two events always occur together.
That electron transfer is exactly what an electric current is made of. Redox reactions therefore form the basis both for the generation of electricity by chemical reactions and for chemical reactions brought about by means of electricity. Such processes are carried out in a device called an electrochemical cell, and electrochemistry deals with the design and operation of these cells.
Why It Matters Outside the Classroom
The study of electrochemical cells is important in science and technology because it makes possible the manufacture of essential chemicals. You have learnt the preparation of sodium hydroxide (NaOH) — used in the manufacture of soaps, detergents and paper — by the electrolysis of sodium chloride (NaCl). Electrolysis is possibly the only means of producing fluorine. Processes such as electro-refining, used for the purification of metals, and electroplating, used for coating one metal on the surface of another, are also electrochemical processes.
Soon after the first battery made a steady current available, Humphry Davy in London turned that current on substances that no chemist had ever managed to break apart. In 1807 he passed electricity through molten potash and molten soda and obtained two shining, extremely reactive metals never seen before — potassium and sodium. Compounds that had resisted every furnace and every chemical reagent gave way to an electric current, and within a short period the same method yielded calcium, magnesium, barium and strontium as well. Electrolysis had arrived as a tool for making new elements, not merely for studying old ones.
The industrial electrolysis of brine (concentrated NaCl solution) is one of the largest chemical operations in the world. A single plant delivers three products at once: sodium hydroxide for soaps, detergents and paper, chlorine for water treatment and PVC, and hydrogen as a fuel. Fluorine, the most reactive of all elements, cannot be prised out of its compounds by any ordinary chemical reagent — electrolysis is essentially the only route to it. The same principle, run gently, gives the thin silver coating on cutlery and the chromium layer on a bicycle handlebar. A handling note that matters in any school laboratory: electroplating and refining baths contain concentrated solutions of metal salts, many of them toxic, and spent bath liquid must be collected for disposal rather than washed down a sink.
Where the Field Is Heading
Research in electrochemistry today is focused on the design of fuel cells. A fuel cell is a device in which a fuel such as hydrogen is supplied continuously and its combustion energy is drawn off directly as electricity, instead of being used to boil water and turn a turbine. Because they are compact and can run as long as fuel is fed to them, fuel cells are being explored as a convenient source of electricity. You will study how they work later in this chapter.
Classify each of the following as a conversion of chemical energy into electrical energy, or of electrical energy into chemical energy, and give one reason for each choice: (i) a mobile phone battery being charged from a wall socket, (ii) a torch cell lighting a bulb, (iii) silver being deposited on a spoon in a plating bath, (iv) a hydrogen–oxygen fuel cell powering a spacecraft.
5.2 Electric Conduction
We know that an electric current represents a charge transfer. Something carrying charge has to move from one place to another; if nothing moves, there is no current.
A charge transfer, or flow of electricity, occurs through substances called conductors. Not every conductor carries charge in the same way, however. There are two types of conductors, and they give rise to two types of conduction of electricity.
The difference lies in what actually moves inside the conductor — and, as you will see, that single difference decides whether the conductor is chemically changed by the current or left untouched.
5.2.2 Electrolytic or Ionic Conduction
Electrolytic conduction involves the conduction of electric current by the movement of ions of the electrolytes. In this type of conduction the charge transfer occurs in the form of movement of ions through molten electrolytes or through the aqueous solutions of electrolytes.
An ion is an atom or group of atoms carrying a net positive or negative charge. Positive ions (cations) drift towards the negative electrode and negative ions (anions) drift towards the positive electrode, so both kinds of ions contribute to the same current.
Substances such as ionic salts, and strong or weak acids and bases, are the electrolytes. These dissociate into ions when dissolved in polar solvents such as water. Water molecules, being polar, pull the oppositely charged ions of a crystal apart and surround them, keeping them separated in solution.
Dissolving is not the only route. Ionic solids dissociate into ions in the molten state as well — melting supplies enough energy to break the lattice apart, and the free ions can then move.
Electrolytic (ionic) conduction is the conduction of electric current by the movement of ions through a molten electrolyte or through an aqueous solution of an electrolyte.
Conduction through electrolytic conductors involves the transfer of matter from one part of the conductor to the other. Ions physically travel across the liquid and are discharged when they arrive at an electrode.
This means that the current flowing through an electrolytic conductor is accompanied by a chemical change. Something is always converted into something else at the electrodes — which is precisely why electrolysis works at all.
Metallic conduction vs Electrolytic conduction
| Charge carrier |
Electrons |
Cations and anions |
| Conductor |
Metals, graphite |
Molten electrolytes, aqueous solutions of electrolytes |
| Movement of matter |
None; only electrons drift |
Matter is transferred from one part of the conductor to the other |
| Chemical change |
No chemical change occurs |
Current is always accompanied by chemical change at the electrodes |
| Effect of temperature |
Conductivity generally decreases on heating |
Conductivity generally increases on heating (ions move more freely) |
That electrolytes break up into ions in solution before any current is applied was proposed in the 1880s by the young Swedish chemist Svante Arrhenius. His examiners at Uppsala were unimpressed — the idea that stable salt could fall apart into charged fragments in ordinary water struck them as absurd, and his thesis was passed with the lowest acceptable grade. Conductivity measurements steadily proved him right, and in 1903 the same work brought him the Nobel Prize in Chemistry.
5.3 Electrical Conductance of Solution
We have seen that an electrolytic solution carries current by the movement of its ions. The next question is a quantitative one: how well does a given solution carry that current? To answer it we borrow the language of ordinary electrical circuits.
According to Ohm’s law, the electrical resistance \(R\) of a conductor is equal to the electric potential difference \(V\) divided by the electric current \(I\):
\[R = \frac{V}{I} \qquad \text{...(5.1)}\]
Ohm’s law: for a given conductor at constant temperature, the electrical resistance is the ratio of the potential difference applied across it to the current flowing through it, \(R = V/I\).
The SI unit of potential is the volt (V) and that of current is the ampere (A). The unit of electrical resistance is the ohm, denoted by the symbol \(\Omega\) (omega). Thus \(\Omega = \text{VA}^{-1}\).
Conductance: Resistance Turned Around
Resistance measures opposition to current. When we are interested in how easily current passes, it is more natural to use the reciprocal quantity. The electrical conductance, \(G\), of a solution is the reciprocal of resistance:
\[G = \frac{1}{R} \qquad \text{...(5.2)}\]
A solution with a small resistance has a large conductance, and vice versa. The SI unit of \(G\) is the siemens, denoted by S, which is equal to \(\Omega^{-1}\). Therefore, we write
\[\text{S} = \Omega^{-1} = \text{AV}^{-1} = \text{CV}^{-1}\text{s}^{-1}\]
where C represents the coulomb, the unit of quantity of electricity, related to current strength in ampere and time in seconds as \(\text{C} = \text{A s}\). In words: one coulomb is the charge carried past a point when a current of one ampere flows for one second.
electrical conductance G (SI unit siemens, S, equal to \(\Omega^{-1}\); \(G = 1/R\), so a large \(G\) means an easily conducting solution. Conductance is always positive; there is no sign convention to worry about here.)
Two of the units in this section carry the names of their investigators. The ohm honours Georg Simon Ohm, the German schoolteacher who established the proportionality between potential difference and current in the 1820s and whose work was ignored for years afterwards. The siemens honours Werner von Siemens, the German engineer and industrialist of the same century; it is the reciprocal of the ohm, and for a long while chemists cheerfully wrote it as the “mho” — ohm spelt backwards — before the SI name was adopted.
Resistance Depends on the Shape of the Conductor
Resistance is not a property of a substance alone; it also depends on the size and shape of the piece being tested. A long thin wire resists more than a short thick one of the same metal.
The electrical resistance of a conductor is proportional to its length \(l\) and inversely proportional to its cross sectional area \(a\). Thus,
\[R \propto \frac{l}{a} \quad \text{or} \quad R = \rho\,\frac{l}{a} \qquad \text{...(5.3)}\]
where \(\rho\), the proportionality constant, is called the resistivity of the conductor. Setting \(l = 1\) and \(a = 1\) in Eq. (5.3) gives \(R = \rho\), so resistivity is the resistance of a conductor of unit length and unit cross sectional area. It is a property of the material itself, free of the accidents of shape.
What is the SI unit of resistivity?
5.3.1 Conductivity (\(k\))
We have seen that \(G = 1/R\), and that \(R\) is directly proportional to length and inversely proportional to its cross sectional area. It, therefore, follows that \(G\) is directly proportional to \(a\) and inversely proportional to the length \(l\). Thus
\[G \propto \frac{a}{l} \quad \text{or} \quad G = k\,\frac{a}{l} \qquad \text{...(5.4)}\]
The proportionality constant \(k\) is called conductivity. Putting \(l = 1\) and \(a = 1\) in Eq. (5.4) gives \(G = k\); that is, \(G = k\) if the length and cross sectional area of the conductor are unity.
Thus, conductivity is the electrical conductance of a conductor of unit length and unit area of cross section. In other words, the conductivity is the electrical conductance of a unit cube of the material.
Conductivity is the electrical conductance of a conductor of unit length and unit area of cross section — that is, the conductance of a unit cube of the material.
For solutions we give this quantity a more specific name. Conductivity of a solution of an electrolyte is called electrolytic conductivity, which refers to the electrical conductance of unit volume (\(1\ \text{m}^3\) or \(1\ \text{cm}^3\)) of solution.
Linking Conductivity to Resistivity
From Eq. (5.2) and Eq. (5.4), we write
\[k = G\,\frac{l}{a} = \frac{1}{R}\,\frac{l}{a} \qquad \text{...(5.5)}\]
Combination of Eq. (5.3) and Eq. (5.5) shows that \(k = 1/\rho\). This is worth pausing on: conductivity and resistivity are reciprocals of each other, exactly as conductance and resistance are. Everything said about resistivity can therefore be restated in terms of conductivity by turning it upside down.
Table 5.3a: Units of electrolytic conductivity
| Length |
m |
cm |
| Area |
m² |
cm² |
| Resistance |
\(\Omega\) |
\(\Omega\) |
| Conductivity |
\(\Omega^{-1}\ \text{m}^{-1}\) or \(\text{S m}^{-1}\) |
\(\Omega^{-1}\ \text{cm}^{-1}\) |
Note carefully that the common unit is based on the centimetre while the SI unit is based on the metre. Numerical problems in this chapter move between the two, so the unit written beside a value is not decoration — it is part of the answer.
5.3.2 Molar Conductivity (\(\Lambda\))
The electrolytic conductivity is not suitable for comparing the conductivities of different solutions. Why not?
The conductivity of a solution depends on the number of ions present in unit volume of solution. The solution of higher concentration contains more ions and exhibits higher conductivity than the solution of lower concentration. So a high conductivity may mean that the electrolyte is a good conductor — or merely that the solution is concentrated. To compare conductivities of different solutions, they must have the same concentration.
The way out is to divide the conductivity by the concentration, so that every electrolyte is judged on the same basis: one mole of it. In 1880, the German physicist F. W. G. Kohlrausch introduced the term molar conductivity, denoted by \(\Lambda\) (lambda).
The molar conductivity of an electrolytic solution is the electrolytic conductivity, \(k\), divided by its molar concentration \(c\).
\[\Lambda = \frac{k}{c} \qquad \text{...(5.6)}\]
Molar conductivity is the electrolytic conductivity of a solution divided by its molar concentration.
SI units of \(k\) are \(\text{S m}^{-1}\) and that of \(c\) are \(\text{mol m}^{-3}\). Hence SI units of \(\Lambda\) are \(\text{S m}^2\text{mol}^{-1}\). Common units employed for molar conductivity are \(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
molar conductivity \(\Lambda\) (SI unit \(\text{S m}^2\text{mol}^{-1}\); common unit \(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\). Since \(1\ \text{m} = 100\ \text{cm}\), \(1\ \text{S m}^2\text{mol}^{-1} = 10^4\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) — a factor examiners test often, so derive it rather than memorise it.)
Significance of Molar Conductivity
To understand the significance of \(\Lambda\), consider the volume of a solution containing 1 mole of dissolved electrolyte. Suppose the solution is placed between two parallel electrodes 1 cm apart and large enough to accommodate it.
The electrical conductance exhibited by this solution is the molar conductivity. In other words, the molar conductivity is the electrical conductance generated by all the ions obtained from 1 mole of the electrolyte.
Picture it as follows. For conductivity we fix the volume at one cubic centimetre and accept whatever number of ions happens to be in it. For molar conductivity we fix the amount at one mole and let the volume be whatever it needs to be. That single change of viewpoint is what makes comparison between different electrolytes fair.
Conductivity is the electrical conductance due to all the ions present in \(1\ \text{cm}^3\) of a given solution; molar conductivity is the electrical conductance due to the ions obtained from 1 mole of an electrolyte in a given volume of solution.
Conductivity (\(k\)) vs Molar conductivity (\(\Lambda\))
| What is held fixed |
Volume of solution (\(1\ \text{cm}^3\)) |
Amount of electrolyte (1 mol) |
| Definition |
Conductance of a unit cube of solution |
Conductance of all ions from 1 mol of electrolyte |
| Relation |
\(k = 1/\rho\) |
\(\Lambda = k/c\) |
| SI unit |
\(\text{S m}^{-1}\) |
\(\text{S m}^2\text{mol}^{-1}\) |
| Common unit |
\(\Omega^{-1}\text{cm}^{-1}\) |
\(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\) |
| Effect of dilution |
Decreases |
Increases |
| Use |
Tells how well this particular solution conducts |
Allows fair comparison between different electrolytes |
5.3.3 Relation Between \(k\) and \(\Lambda\)
Conductivity \(k\) is the electrical conductance of \(1\ \text{cm}^3\) of solution. If \(V\) is the volume of solution in \(\text{cm}^3\) containing 1 mole of dissolved electrolyte, its electrical conductance is \(\Lambda\).
Now think of that volume \(V\) as being made up of \(V\) separate cubes, each of \(1\ \text{cm}^3\). Each \(1\ \text{cm}^3\) portion in the volume \(V\) has conductance \(k\). Hence the total conductance of \(V\ \text{cm}^3\) is \(kV\), which is the molar conductivity.
Thus, we have
\[\Lambda = k\,V \qquad \text{...(5.7)}\]
To make this usable we must express \(V\) in terms of concentration. Concentration of solution
\[= c\ \text{mol L}^{-1} = \frac{c\ \text{mol L}^{-1}}{1000\ \text{cm}^3\text{L}^{-1}} = \frac{c}{1000}\ \text{mol cm}^{-3}\]
The step above is only a unit conversion: one litre is \(1000\ \text{cm}^3\), so dividing by 1000 rewrites the concentration per cubic centimetre instead of per litre.
Volume, \(V\), of solution in \(\text{cm}^3\) containing 1 mole of an electrolyte is the reciprocal of concentration. Therefore,
\[V = \frac{1}{\text{concentration}} = \frac{1000}{c}\ \text{cm}^3\text{mol}^{-1} \qquad \text{...(5.8)}\]
Substitution for \(V\) in Eq. (5.7) yields
\[\Lambda = \frac{1000\,k}{c} \qquad \text{...(5.9)}\]
This is the working equation of the whole section. Given any two of \(\Lambda\), \(k\) and \(c\) in the common units, you can find the third.
What must be the concentration of a solution of silver nitrate to have the molar conductivity of \(121.4\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) and the conductivity of \(2.428\times10^{-3}\ \Omega^{-1}\text{cm}^{-1}\) at \(25\ ^\circ\text{C}\)?
The molar conductivity of \(0.05\ \text{M}\) \(\mathrm{BaCl_2}\) solution at \(25\ ^\circ\text{C}\) is \(223\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\). What is its conductivity?
We know the molar conductivity and the concentration, and we are asked for the conductivity, so Eq. (5.9) is rearranged to make \(k\) the subject.
\[\Lambda = \frac{1000k}{c} \quad \text{or} \quad k = \frac{\Lambda\,c}{1000}\]
\[\Lambda = 223\ \Omega^{-1}\text{cm}^2\text{mol}^{-1},\qquad c = 0.05\ \text{mol L}^{-1}\]
Hence
\[k = \frac{223\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} \times 0.05\ \text{mol L}^{-1}}{1000\ \text{cm}^3\text{L}^{-1}} = 0.01115\ \Omega^{-1}\text{cm}^{-1}\]
Notice how the units settle the matter on their own: \(\text{mol}\) cancels against \(\text{mol}^{-1}\), \(\text{L}^{-1}\) against \(\text{L}\), and \(\text{cm}^2\) against \(\text{cm}^3\) to leave \(\Omega^{-1}\text{cm}^{-1}\), which is exactly the common unit of conductivity.
The conductivity of a \(0.20\ \text{M}\) solution of a strong electrolyte at \(298\ \text{K}\) is \(0.0248\ \Omega^{-1}\text{cm}^{-1}\). Calculate its molar conductivity in \(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\) and express the same answer in the SI unit \(\text{S m}^2\text{mol}^{-1}\).
Step 1 — molar conductivity in common units, from Eq. (5.9):
\[\Lambda = \frac{1000\,k}{c} = \frac{1000\ \text{cm}^3\text{L}^{-1} \times 0.0248\ \Omega^{-1}\text{cm}^{-1}}{0.20\ \text{mol L}^{-1}} = 124\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
Step 2 — conversion to SI units. Since \(1\ \text{cm} = 10^{-2}\ \text{m}\), we have \(1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\), and \(\Omega^{-1}\) is the same as S. Therefore
\[124\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} = 124 \times 10^{-4}\ \text{S m}^2\text{mol}^{-1} = 1.24\times10^{-2}\ \text{S m}^2\text{mol}^{-1}\]
The physical quantity has not changed at all; only the ruler used to report it has.
5.3.4 Variation of Conductivity With Concentration
What happens to these two quantities when we add water to a solution? They behave in opposite ways, and understanding why is a favourite examination point.
The electrolytic conductivity is the electrical conductance of unit volume (\(1\ \text{cm}^3\)) of solution. It depends on the number of current carrying ions present in unit volume of solution.
On dilution the total number of ions increases, as a result of the increased degree of dissociation. Adding water pushes a weak electrolyte further towards dissociation, so more ions exist in the beaker than before.
However, an increase in the total number of ions is not in proportion to the dilution. Therefore, the number of ions per unit volume of solution decreases. This results in a decrease of conductivity with decrease in concentration of solution.
The arithmetic below, given in the textbook, makes the point concrete.
Suppose \(100\ \text{cm}^3\) of solution of an electrolyte contains \(8\times10^{20}\) ions. The number of ions per \(\text{cm}^3\) is \(8\times10^{18}\).
If the solution is diluted to \(1000\ \text{cm}^3\), the total number of ions will increase, but not by a factor of 10. Assume that the number of ions increases from \(8\times10^{20}\) to \(64\times10^{20}\) on dilution. After dilution the number of ions per \(\text{cm}^3\) is \(6.8\times10^{18}\).
It is evident that the number of ions per \(\text{cm}^3\) decreases from \(8\times10^{18}\) to \(6.8\times10^{18}\) on dilution from \(100\ \text{cm}^3\) to \(1000\ \text{cm}^3\), and in turn, the conductivity decreases.
So dilution wins the race on total ions but loses it on ions per unit volume — and it is ions per unit volume that conductivity responds to.
5.3.5 Variation of Molar Conductivity With Concentration
Molar conductivity is measured on a different basis, and so it moves the other way.
The molar conductivity is the electrical conductance of 1 mole of an electrolyte in a given volume of solution. The amount of electrolyte is now fixed at one mole, however much water surrounds it.
The increasing number of ions produced in solution by 1 mole of the electrolyte leads to increased molar conductivity. Since dilution increases the degree of dissociation, that same one mole yields more ions than before, and every extra ion adds to the conductance being counted.
Q2. When you dilute an electrolyte solution, its conductivity falls but its molar conductivity rises. How can both statements be true about the same solution at the same time?
A. Because the two quantities count different things. Conductivity asks how many ions are packed into \(1\ \text{cm}^3\); dilution spreads the ions through much more water, so the count per cubic centimetre goes down and \(k\) falls. Molar conductivity asks how much conductance the ions from one whole mole of electrolyte deliver; dilution increases the degree of dissociation, so that one mole releases more ions than before and \(\Lambda\) rises. Same solution, two different bases of comparison — hence two opposite trends, with no contradiction.
A student dilutes \(100\ \text{cm}^3\) of \(0.1\ \text{M}\) acetic acid to \(1000\ \text{cm}^3\) and measures both \(k\) and \(\Lambda\) before and after. State how each of the two readings changes, and justify each answer in one sentence by referring to the number of ions per unit volume and the number of ions per mole.
5.3.6 Variation of Molar Conductivity With Concentration for Strong and Weak Electrolytes
The variation of molar conductivity with concentration in the case of strong and weak electrolytes is qualitatively different. Both rise on dilution, but they rise in different shapes — and that difference turns out to be extremely useful.
i. Strong Electrolytes
The molar conductivity of a solution of a strong electrolyte increases rapidly with dilution. It approaches the limiting value for \(0.001\ \text{M}\) or \(0.0001\ \text{M}\) solution. The dilution has no effect on molar conductivity thereafter — the curve flattens out.
The maximum limiting value of molar conductivity is the molar conductivity at zero concentration or at infinite dilution. It is denoted by \(\Lambda_0\). The zero concentration or infinite dilution means the solution is so dilute that further dilution does not increase the molar conductivity.
Why should there be a ceiling at all? A strong electrolyte was almost completely dissociated to begin with, so dilution cannot produce many more ions. What it does is separate the existing ions so widely that they stop interfering with one another’s motion; once that interference has been removed as far as it can be, nothing further is gained.
During the nineteenth century F. Kohlrausch, with repeated experiments, showed that the molar conductivity of strong electrolytes varies linearly with the square root of concentration as:
\[\Lambda = \Lambda_0 - a\sqrt{c} \qquad \text{...(5.10)}\]
where \(a\) is a constant. For strong electrolytes a plot of \(\Lambda\) versus \(\sqrt{c}\) is linear, as shown in Fig. 5.1.
Compare Eq. (5.10) with the straight-line form \(y = mx + c\): plotting \(\Lambda\) on the \(y\)-axis against \(\sqrt{c}\) on the \(x\)-axis gives a line of slope \(-a\) whose intercept on the \(\Lambda\) axis is \(\Lambda_0\).
Molar conductivity of strong electrolytes at zero concentration can be determined by extrapolation of the linear part of the \(\Lambda\) versus \(\sqrt{c}\) curve, as shown in Fig. 5.1. Extrapolation simply means extending the straight line backwards until it meets the axis, since we cannot actually prepare a solution of zero concentration and measure it.
ii. Weak Electrolytes
The molar conductivity of weak electrolytes increases rapidly on dilution. For concentrations of \(0.001\ \text{M}\) or \(0.0001\ \text{M}\), the \(\Lambda\) value is lower than \(\Lambda_0\), the molar conductivity at zero concentration. There is no flattening out; the curve is still climbing steeply when the solution is already very dilute.
For weak electrolytes the variation of \(\Lambda\) with \(\sqrt{c}\) shown in Fig. 5.1 is not linear. The reason lies in the chemistry: for a weak electrolyte, dilution keeps producing genuinely new ions by pushing the dissociation equilibrium forward, and this effect does not fade away as concentration approaches zero.
This method (extrapolation) cannot be used for weak electrolytes, since the \(\Lambda\) versus \(\sqrt{c}\) curve does not approach linearity. Kohlrausch’s law is useful for calculating \(\Lambda_0\) of weak electrolytes — which is exactly the subject of the next subsection.
Behaviour of strong and weak electrolytes on dilution
| Dissociation at ordinary concentration |
Almost complete |
Very small |
| Shape of \(\Lambda\) vs \(\sqrt{c}\) plot |
Linear |
Not linear |
| Behaviour near \(0.001\ \text{M}\) |
\(\Lambda\) has practically reached \(\Lambda_0\) |
\(\Lambda\) is still well below \(\Lambda_0\) |
| Main reason \(\Lambda\) rises on dilution |
Reduced interference between ions already present |
Increase in degree of dissociation |
| How \(\Lambda_0\) is obtained |
Extrapolation of the linear plot |
Kohlrausch’s law, using \(\Lambda_0\) values of strong electrolytes |
the \(\Lambda\) versus \(\sqrt{c}\) plot of Fig. 5.1 for a strong and a weak electrolyte (Read Fig. 5.1 and answer: (a) Which curve belongs to the strong electrolyte, and what feature of the plot tells you so? (b) On which curve can \(\Lambda_0\) be read directly off the intercept, and why can it not be read the same way on the other? (c) At the far right of the plot, which electrolyte has the higher molar conductivity, and does the gap between the two curves widen or narrow as the solution is diluted?)
5.3.7 Kohlrausch Law of Independent Migration of Ions
If a strong electrolyte’s \(\Lambda_0\) can be obtained by extrapolation but a weak electrolyte’s cannot, we need a way of building the weak electrolyte’s value out of quantities we can measure. Kohlrausch’s law provides exactly that, and it rests on a simple physical idea: at infinite dilution the ions are so far apart that each one behaves as though the others were not there.
The law states that at infinite dilution each ion migrates independently of its co-ion and contributes to the total molar conductivity of an electrolyte irrespective of the nature of the other ion to which it is associated.
Kohlrausch’s law of independent migration of ions: at infinite dilution, every ion migrates independently of the co-ion present and makes its own fixed contribution to the total molar conductivity of the electrolyte, whatever the other ion associated with it may be.
Both cation and anion contribute to the molar conductivity of the electrolyte at zero concentration, and thus \(\Lambda_0\) is the sum of the molar conductivity of the cation and that of the anion at zero concentration. Thus,
\[\Lambda_0 = n_{\oplus}\,\lambda^0_{\oplus} + n_{\ominus}\,\lambda^0_{\ominus} \qquad \text{...(5.11)}\]
where \(\lambda_{\oplus}\) and \(\lambda_{\ominus}\) are the molar conductivities of the cation and anion, respectively, and \(n_{\oplus}\) and \(n_{\ominus}\) are the number of moles of cation and anion specified in the chemical formula of the electrolyte.
The factors \(n_{\oplus}\) and \(n_{\ominus}\) matter. One mole of \(\mathrm{BaCl_2}\) furnishes one mole of \(\mathrm{Ba^{2+}}\) but two moles of \(\mathrm{Cl^{-}}\), so the chloride contribution must be counted twice.
Applications of Kohlrausch Theory
1. Calculating \(\Lambda_0\) of an electrolyte from ionic values. The theory can be used to calculate the molar conductivity of an electrolyte at zero concentration. For example,
\[\Lambda_0 (\mathrm{KCl}) = \lambda^0_{\mathrm{K^{\oplus}}} + \lambda^0_{\mathrm{Cl^{\ominus}}}\]
\[\Lambda_0 [\mathrm{Ba(OH)_2}] = \lambda^0_{\mathrm{Ba^{2\oplus}}} + 2\,\lambda^0_{\mathrm{OH^{\ominus}}}\]
Knowing the molar conductivities of ions at infinite dilution, \(\Lambda_0\) values of any electrolyte can be obtained.
2. Getting \(\Lambda_0\) of a weak electrolyte from strong ones. The theory is particularly useful in calculating \(\Lambda_0\) values of weak electrolytes from those of strong electrolytes. For example, \(\Lambda_0\) of acetic acid can be calculated by knowing those of HCl, NaCl and \(\mathrm{CH_3COONa}\), as described below:
\[\Lambda_0(\mathrm{HCl}) + \Lambda_0(\mathrm{CH_3COONa}) - \Lambda_0(\mathrm{NaCl})\]
\[= \lambda^0_{\mathrm{H^{\oplus}}} + \lambda^0_{\mathrm{Cl^{\ominus}}} + \lambda^0_{\mathrm{CH_3COO^{\ominus}}} + \lambda^0_{\mathrm{Na^{\oplus}}} - \lambda^0_{\mathrm{Na^{\oplus}}} - \lambda^0_{\mathrm{Cl^{\ominus}}}\]
\[= \lambda^0_{\mathrm{H^{\oplus}}} + \lambda^0_{\mathrm{CH_3COO^{\ominus}}} = \Lambda_0(\mathrm{CH_3COOH})\]
Thus,
\[\Lambda_0(\mathrm{CH_3COOH}) = \Lambda_0(\mathrm{HCl}) + \Lambda_0(\mathrm{CH_3COONa}) - \Lambda_0(\mathrm{NaCl})\]
Because \(\Lambda_0\) values of the strong electrolytes HCl, \(\mathrm{CH_3COONa}\) and NaCl can be determined by the extrapolation method, the \(\Lambda_0\) of acetic acid can be obtained.
The trick is to choose three strong electrolytes whose ions add up, with the unwanted ones (\(\mathrm{Na^{\oplus}}\) and \(\mathrm{Cl^{\ominus}}\) here) cancelling in pairs and leaving behind exactly the cation and anion of the weak electrolyte.
Calculate the molar conductivity of AgI at zero concentration if the molar conductivities of NaI, \(\mathrm{AgNO_3}\) and \(\mathrm{NaNO_3}\) at zero concentration are respectively \(126.9\), \(133.4\) and \(121.5\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
According to Kohlrausch’s law, write each given electrolyte as the sum of its two ionic contributions:
\(\Lambda_0(\mathrm{NaI}) = \lambda^0_{\mathrm{Na^{\oplus}}} + \lambda^0_{\mathrm{I^{\ominus}}} = 126.9\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
\(\Lambda_0(\mathrm{AgNO_3}) = \lambda^0_{\mathrm{Ag^{\oplus}}} + \lambda^0_{\mathrm{NO_3^{\ominus}}} = 133.4\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
\(\Lambda_0(\mathrm{NaNO_3}) = \lambda^0_{\mathrm{Na^{\oplus}}} + \lambda^0_{\mathrm{NO_3^{\ominus}}} = 121.5\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
Eq. (i) + eq. (ii) − eq. (iii) gives
\[\Lambda_0(\mathrm{NaI}) + \Lambda_0(\mathrm{AgNO_3}) - \Lambda_0(\mathrm{NaNO_3})\]
\[= \lambda^0_{\mathrm{Na^{\oplus}}} + \lambda^0_{\mathrm{I^{\ominus}}} + \lambda^0_{\mathrm{Ag^{\oplus}}} + \lambda^0_{\mathrm{NO_3^{\ominus}}} - \lambda^0_{\mathrm{Na^{\oplus}}} - \lambda^0_{\mathrm{NO_3^{\ominus}}}\]
\[= \lambda^0_{\mathrm{Ag^{\oplus}}} + \lambda^0_{\mathrm{I^{\ominus}}} = \Lambda_0(\mathrm{AgI})\]
\[= 126.9\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} + 133.4\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} - 121.5\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
\[= 138.8\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
The sodium and nitrate contributions cancelled out, leaving precisely the two ions of silver iodide — a salt so sparingly soluble that measuring its solution directly would be hopeless.
Calculate molar conductivities at zero concentration for \(\mathrm{CaCl_2}\) and \(\mathrm{Na_2SO_4}\). Given: molar ionic conductivities of \(\mathrm{Ca^{2\oplus}}\), \(\mathrm{Cl^{\ominus}}\), \(\mathrm{Na^{\oplus}}\) and \(\mathrm{SO_4^{2\ominus}}\) ions are respectively \(104\), \(76.4\), \(50.1\) and \(159.6\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
According to Kohlrausch’s law, each ion is counted as many times as it appears in the formula.
- \(\Lambda_0(\mathrm{CaCl_2}) = \lambda^0_{\mathrm{Ca^{2\oplus}}} + 2\lambda^0_{\mathrm{Cl^{\ominus}}}\)
\[= 104\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} + 2 \times 76.4\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
\[= 256.8\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
- \(\Lambda_0(\mathrm{Na_2SO_4}) = 2\lambda^0_{\mathrm{Na^{\oplus}}} + \lambda^0_{\mathrm{SO_4^{2\ominus}}}\)
\[= 2 \times 50.1\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} + 159.6\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
\[= 259.8\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
Calculate \(\Lambda_0\) (\(\mathrm{CH_2ClCOOH}\)) if \(\Lambda_0\) values for HCl, KCl and \(\mathrm{CH_2ClCOOK}\) are respectively \(4.261\), \(1.499\) and \(1.132\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
Write the Kohlrausch expression for \(\Lambda_0\) of each of the following in terms of ionic molar conductivities, taking care with the number of moles of each ion: (i) \(\mathrm{K_2SO_4}\), (ii) \(\mathrm{Al_2(SO_4)_3}\), (iii) \(\mathrm{Mg(NO_3)_2}\), (iv) \(\mathrm{NH_4OH}\) expressed using the values for \(\mathrm{NH_4Cl}\), NaOH and NaCl.
5.3.8 Molar Conductivity and Degree of Dissociation of Weak Electrolytes
Once \(\Lambda_0\) of a weak electrolyte is known, a genuinely useful quantity comes within reach. The degree of dissociation, \(\alpha\), is the fraction of the dissolved electrolyte that has actually broken up into ions.
The argument is direct. At infinite dilution the electrolyte is completely dissociated and gives the conductance \(\Lambda_0\). At the working concentration \(c\) only a fraction of it has dissociated, and it gives the smaller conductance \(\Lambda_c\). The ratio of the two therefore measures that fraction.
The degree of dissociation (\(\alpha\)) of a weak electrolyte is related to its molar conductivity at a given concentration \(c\) by the equation
\[\alpha = \frac{\Lambda_c}{\Lambda_0} \qquad \text{...(5.12)}\]
where \(\Lambda_c\) is the molar conductivity of the weak electrolyte at concentration \(c\), and \(\Lambda_0\) is the molar conductivity at zero concentration.
The degree of dissociation of a weak electrolyte at a given concentration is the ratio of its molar conductivity at that concentration to its molar conductivity at zero concentration.
Since \(\Lambda_c\) is always smaller than \(\Lambda_0\), the value of \(\alpha\) always lies between 0 and 1, and it is often quoted as a percentage.
Obtain the expression for dissociation constant in terms of \(\Lambda_c\) and \(\Lambda_0\) using Ostwald’s dilution law.
The molar conductivity of \(0.01\ \text{M}\) acetic acid at \(25\ ^\circ\text{C}\) is \(16.5\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\). Calculate its degree of dissociation in \(0.01\ \text{M}\) solution and dissociation constant if molar conductivity of acetic acid at zero concentration is \(390.7\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
The degree of dissociation,
\[\alpha = \frac{\Lambda_c}{\Lambda_0} = \frac{16.6\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}}{390.7\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}} = 0.0422\]
So only about 4% of the acetic acid has dissociated — which is precisely what “weak electrolyte” means in numbers.
\[K_a = \frac{\alpha^2 c}{1-\alpha} = \frac{(0.0422)^2 \times 0.01}{(1 - 0.0422)} = 1.85\times10^{-5}\]
The expression used for \(K_a\) is Ostwald’s dilution law from Chapter 3: for a weak acid dissociating as \(\mathrm{HA} \rightleftharpoons \mathrm{H^{\oplus}} + \mathrm{A^{\ominus}}\), the equilibrium concentrations of the two ions are each \(\alpha c\) while that of the undissociated acid is \((1-\alpha)c\), which on substitution into the equilibrium expression gives \(\alpha^2c/(1-\alpha)\).
The molar conductivity of a \(0.001\ \text{M}\) solution of a weak monobasic acid is \(48.15\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) and its molar conductivity at zero concentration is \(390.5\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\). Calculate the degree of dissociation of the acid at this concentration and its dissociation constant.
5.3.9 Measurement of Conductivity
Everything above assumed that \(k\) could be measured. In practice it is not measured directly at all — what the experiment yields is a resistance, from which conductivity is calculated. The conductivity of a solution can be determined from resistance measurements by the Wheatstone bridge.
Conductivity Cell
The conductivity cell consists of a glass tube with two platinum plates coated with a thin layer of finely divided platinum black. This coating is achieved by the electrolysis of a solution of chloroplatinic acid.
Why roughen the electrodes in this way? The finely divided black deposit gives an enormously larger surface area than a smooth plate, which helps the electrode exchange charge freely with the solution and keeps the measured resistance a property of the solution rather than of the electrode surface. The cell is dipped in a solution whose resistance is to be measured, as shown in Fig. 5.2.
Cell Constant
The conductivity of an electrolytic solution is given by Eq. (5.5),
\[k = \frac{1}{R}\,\frac{l}{a}\]
For a given cell, the ratio of the separation (\(l\)) between the two electrodes divided by the area of cross section (\(a\)) of the electrode is called the cell constant. Thus,
\[\text{Cell constant} = \frac{l}{a} \qquad \text{...(5.13)}\]
The cell constant of a conductivity cell is the ratio of the distance between its electrodes to their area of cross section.
The SI unit of cell constant is \(\text{m}^{-1}\), which is conveniently expressed in \(\text{cm}^{-1}\). The Eq. (5.5) then becomes
\[k = \frac{\text{cell constant}}{R} \qquad \text{...(5.14)}\]
This is the whole point of the cell constant: it is a fixed number for a given cell, so once it is known, a single resistance reading converts straight into a conductivity. It is not measured with a scale and ruler, because the effective area of a platinised plate is not the area you can see; it is obtained by calibration, as described next.
The determination of molar conductivity consists of three steps, summarised in FC 5.2.
1. Determination of Cell Constant
The cell constant is determined using the \(1\ \text{M}\), \(0.1\ \text{M}\) or \(0.01\ \text{M}\) KCl solutions. The conductivity of KCl solution is well tabulated at various temperatures, so it can be treated as known. The resistance of KCl solution is measured by the Wheatstone bridge (refer to Standard XII Physics Textbook, Chapter 9).
standard potassium chloride solution (A strong electrolyte whose conductivity has been measured to high accuracy at every common temperature and concentration → needs only to be made up accurately to \(1\ \text{M}\), \(0.1\ \text{M}\) or \(0.01\ \text{M}\) and kept at the stated temperature → used as the calibrating standard that converts a conductivity cell into a measuring instrument, by supplying the one solution whose \(k\) we already know.)
In Fig. 5.3, AB is the uniform wire. \(R_x\) is the variable known resistance placed in one arm of the Wheatstone bridge.
The conductivity cell containing KCl solution of unknown resistance is placed in the other arm of the Wheatstone bridge. D is a current detector. F is the sliding contact that moves along AB. A.C. represents the source of alternating current.
Alternating current is used rather than direct current for a chemical reason: a steady direct current would electrolyse the solution and change its composition during the measurement, whereas a rapidly alternating current reverses before any appreciable electrolysis can occur.
The sliding contact is moved along AB until no current flows. The detector D shows no deflection. The null point is, thus, obtained at C.
According to the Wheatstone bridge principle,
\[\frac{R_{\text{solution}}}{l\,(\mathrm{AC})} = \frac{R_x}{l\,(\mathrm{BC})}\]
Hence,
\[R_{\text{solution}} = \frac{l(\mathrm{AC})}{l(\mathrm{BC})} \times R_x \qquad \text{...(5.15)}\]
By measuring lengths AC and BC and knowing \(R_x\), the resistance of KCl solution can be calculated. The cell constant is given by Eq. (5.13), and in practice is obtained as
\[\text{Cell constant} = k_{\mathrm{KCl}} \times R_{\text{solution}}\]
The conductivity of KCl solution is known. The cell constant, thus, can be calculated. This expression is nothing but Eq. (5.14) rearranged, applied to the one solution whose conductivity we already know.
2. Determination of Conductivity of the Given Solution
KCl solution in the conductivity cell in step (1) is replaced by the given solution whose conductivity is to be measured. Its resistance is measured by the process described in step (1). The conductivity of the given solution is then calculated as:
\[k = \frac{\text{Cell constant}}{R_{\text{solution}}}\]
Because the same cell is used, the cell constant found in step 1 still applies — which is why the calibration is done first and only once.
3. Calculation of Molar Conductivity
The molar conductivity of the given solution is then calculated using Eq. (5.9).
\[\Lambda = \frac{1000\,k}{c}\]
The same three-step logic runs inside every hand-held EC (electrical conductivity) meter. Farmers dip one into irrigation water and into soil suspensions to judge salinity, because dissolved salts raise conductivity long before any crop damage is visible; aquarists and dairy plants use the identical instrument to monitor water quality, and pharmaceutical units use it to certify that their purified water carries almost no dissolved ions. The meter contains a small conductivity cell of known cell constant, and its “calibrate” button does exactly what step 1 above does. A laboratory caution belongs with this section too: chloroplatinic acid, used to deposit platinum black on the electrodes, is corrosive and expensive, so platinised electrodes are never scrubbed clean — they are rinsed gently and kept immersed in distilled water.
A conductivity cell containing \(0.01\ \text{M}\) KCl gives at \(25\ ^\circ\text{C}\) the resistance of \(604\) ohms. The same cell containing \(0.001\ \text{M}\) \(\mathrm{AgNO_3}\) gives resistance of \(6530\) ohms. Calculate the molar conductivity of \(0.001\ \text{M}\) \(\mathrm{AgNO_3}\). [Conductivity of \(0.01\ \text{M}\) KCl at \(25\ ^\circ\text{C}\) is \(0.00141\ \Omega^{-1}\text{cm}^{-1}\)]
This is the three-step procedure of FC 5.2, worked through with numbers.
- Calculation of cell constant
\[\text{Cell constant} = k_{\mathrm{KCl}} \times R_{\mathrm{KCl}}\]
\[= 0.00141\ \Omega^{-1}\text{cm}^{-1} \times 604\ \Omega = 0.852\ \text{cm}^{-1}\]
- Calculation of conductivity of \(\mathrm{AgNO_3}\)
\[k = \frac{\text{Cell constant}}{R} \quad \text{where } R = 6530\ \Omega\]
\[= \frac{0.852\ \text{cm}^{-1}}{6530\ \Omega} = 1.3\times10^{-4}\ \Omega^{-1}\text{cm}^{-1}\]
- Calculation of molar conductivity of \(\mathrm{AgNO_3}\)
\[\Lambda = \frac{1000k}{c} \quad \text{where } c = 0.001\ \text{M}\]
\[= \frac{1000\ \text{cm}^3\text{L}^{-1} \times 1.3\times10^{-4}\ \Omega^{-1}\text{cm}^{-1}}{0.001\ \text{mol L}^{-1}} = 130\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\]
Note that the cell constant found in step (i) from the KCl reading is carried straight into step (ii) — the same cell was used for both solutions, and that is what makes the whole method work.
A conductivity cell has a cell constant of \(1.25\ \text{cm}^{-1}\). When filled with a \(0.05\ \text{M}\) solution of an electrolyte it shows a resistance of \(250\ \Omega\) at \(298\ \text{K}\). Calculate (i) the conductivity of the solution and (ii) its molar conductivity in \(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
5.4 Electrochemical Cells
We now leave measurement behind and turn to the devices themselves. An electrochemical cell consists of two metal plates or carbon (graphite) rods. These electronic conductors are dipped into an electrolytic or ionic conductor.
Look closely at that sentence, because it contains the whole design idea. Both types of conduction met in Section 5.2 are present in a single device: electrons travel through the solid plates and the outside wire, while ions travel through the solution. Where the two meet — at the surface of each plate — the charge has to change hands, and that hand-over is a chemical reaction.
An electrochemical cell is a device in which two electronic conductors (metal plates or graphite rods) are dipped into an electrolytic conductor, so that a redox reaction and an electric current are coupled to each other.
5.4.1 Electrochemical Reactions
What is the reaction involving transfer of electrons from one chemical species to another called?
The chemical reaction occurring in electrochemical cells involves transfer of electrons from one species to the other. It is a redox reaction, as we learnt in Standard XI, Chapter 6.
Electrochemical reactions are made of oxidation and reduction half reactions. Writing a redox change as two halves is not merely a bookkeeping habit here; in a cell the two halves are physically separated.
The oxidation half reaction occurs at one electrode and the reduction half reaction occurs at the other electrode. The species being oxidised hands its electrons to one plate; those electrons travel through the external wire; and at the other plate they are handed to the species being reduced. That journey through the wire is the current we use.
The net cell reaction is the sum of these half reactions. When the two halves are added, the electrons must cancel completely — no free electrons may appear in the overall equation, because every electron released at one electrode is consumed at the other.
5.4.2 Electrodes
Electrodes are the surfaces on which oxidation and reduction half reactions take place. An electrode is therefore not just a wire terminal — it is a reaction site.
Electrodes may or may not participate in the reactions. A zinc plate dipping in zinc ion solution is itself oxidised, so it takes part. A platinum or graphite rod, on the other hand, may only supply a surface where electrons are exchanged, while the substances reacting come from the solution or from a gas bubbled over it. The electrodes which do not take part in reactions are inert electrodes.
An inert electrode is one that provides a surface for electron transfer without itself being oxidised or reduced during the cell reaction.
The two electrodes are named by what happens on them, not by their charge or their position.
Cathode: It is an electrode at which the reduction takes place. At this electrode the species undergoing reduction gains electrons.
Anode: It is an electrode at which oxidation takes place. At this electrode, the species undergoing oxidation loses electrons.
Cathode is the electrode at which reduction occurs; anode is the electrode at which oxidation occurs.
Hold on to these two definitions exactly as stated. They stay true in every cell you will meet in this chapter, whereas the signs attached to the electrodes do not, as the next subsection shows.
In a cell, a strip of zinc dipping in zinc sulphate solution is connected through a wire to a copper strip dipping in copper sulphate solution. Zinc goes into solution as \(\mathrm{Zn^{2\oplus}}\) ions and copper is deposited on the copper strip. Write the two half reactions, identify the anode and the cathode, and write the net cell reaction.
Zinc is going into solution, which means each zinc atom is losing electrons — an oxidation.
\[\mathrm{Zn}\text{(s)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation)}\]
Copper is being deposited, which means each \(\mathrm{Cu^{2\oplus}}\) ion in solution is gaining electrons — a reduction.
\[\mathrm{Cu^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\text{(s)} \qquad \text{(reduction)}\]
By the definitions above, the zinc strip, where oxidation occurs, is the anode; the copper strip, where reduction occurs, is the cathode.
Adding the two half reactions, the two electrons on each side cancel and no free electrons remain:
\[\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad \text{(net cell reaction)}\]
5.4.3 Types of Electrochemical Cells
There are two types of electrochemical cells, and they correspond exactly to the two directions of energy conversion introduced in Section 5.1.
1. Electrolytic cell: In this type of cell, a nonspontaneous reaction, known as electrolysis, is forced to occur by passing a direct current from an external source into the solution. Left to itself the reaction would never happen; the external battery supplies the energy that drives it. In such cells electrical energy is converted into chemical energy. The anode of an electrolytic cell is positive and the cathode is negative.
2. Galvanic or voltaic cell: In a galvanic (voltaic) cell a spontaneous chemical reaction produces electricity. Here the reaction is one that wants to happen anyway, and the cell simply collects the energy it releases. In these cells chemical energy is converted into electrical energy. The anode of a galvanic cell is negative and the cathode is positive.
Why the Signs Reverse
Students lose marks on this point every year, so it is worth reasoning out rather than memorising.
In a galvanic cell nobody imposes a charge from outside. The anode becomes negative simply because oxidation is piling up electrons on it, and those electrons are then pushed out into the wire. In an electrolytic cell the external battery decides everything: the plate connected to the battery’s positive terminal is pulled electron-poor, so it becomes the site where the solution is forced to give up electrons — that is, oxidation, which by definition makes it the anode, and it carries a positive sign.
The rule that never changes is the definition: anode = oxidation, cathode = reduction. The sign follows from the situation.
Electrolytic cell vs Galvanic (voltaic) cell
| Nature of the cell reaction |
Nonspontaneous; forced to occur |
Spontaneous; occurs on its own |
| Source of current |
Direct current supplied from an external source |
The cell itself generates the current |
| Energy conversion |
Electrical energy → chemical energy |
Chemical energy → electrical energy |
| Sign of anode |
Positive |
Negative |
| Sign of cathode |
Negative |
Positive |
| Reaction at anode |
Oxidation |
Oxidation |
| Reaction at cathode |
Reduction |
Reduction |
| Everyday example |
Electroplating bath, electro-refining of a metal |
Torch cell, car battery on discharge |
Q3. A torch cell pushes electrons out on its own, while an electroplating bath needs a battery to force a reaction along. What decides which of these two directions a cell will run in?
A. Whether the redox reaction inside is spontaneous. If the reaction releases energy of its own accord, the cell runs as a galvanic cell and delivers that energy as electricity — the torch cell. If the reaction would never occur by itself, only an external supply of electrical energy can force it, and the arrangement works as an electrolytic cell — the plating bath. Later in this chapter you will be able to decide which case applies before setting up the cell at all, by comparing the standard potentials of the two electrodes.
The whole classification, and the single question that decides it, is summarised in FC 5.3.
Distinguish between electrolytic and galvanic cells.
For each arrangement below, state whether it works as a galvanic cell or an electrolytic cell, name the electrode at which oxidation occurs, and give the sign of that electrode: (i) a silver spoon being coated with silver from a silver salt solution using a battery, (ii) a dry cell lighting a bulb, (iii) molten sodium chloride being decomposed by a direct current, (iv) a car battery while the engine is running and the battery is being charged.
5.5 Electrolytic Cell
An electrolytic cell consists of a container in which the electrolyte is placed. Two electrodes are immersed in the electrolyte and connected to a source of direct current.
The external source is what makes this cell different from every other cell in the chapter. It does not wait for a reaction to happen; it drives one.
At the anode (+) a species oxidises with the removal of electrons. These electrons are pulled from the anode and pushed to the cathode through an external circuit. The electrons are supplied to species at the cathode, which are reduced.
Notice the direction of traffic. Electrons are withdrawn at one plate and delivered at the other, and it is the battery that does the withdrawing and delivering. The chemistry follows wherever the battery leads.
Electrolysis is the process of breaking down of an ionic compound, in the molten state or in aqueous solution, by the passage of electricity.
The chemist promised in the previous section was Michael Faraday, who coined those Greek-rooted names and who was the first person to explain electrolysis, nearly two hundred years ago. He came to the subject by an unlikely route: apprenticed to a London bookbinder, with almost no formal schooling, he read the science volumes that passed through the workshop and talked his way into a post as assistant at the Royal Institution. In the 1830s his patient measurements established that the amount of a substance liberated at an electrode is proportional to the quantity of electricity passed through the cell — the quantitative result on which the whole of Section 5.5.3 rests.
5.5.1 Electrolysis of Molten NaCl
Construction of Cell
The electrolytic cell consists of a container in which fused NaCl is placed. Two graphite electrodes are immersed in it. They are connected by metallic wires to a source of direct current, that is, a battery. This is shown in Fig. 5.4.
The carbon electrode connected to the positive terminal of the battery is the anode, and that connected to the negative terminal of the battery is the cathode.
Note that graphite is chosen deliberately. In electrolysis the electrodes are usually inert — platinum or graphite — because we want them to supply a surface for electron transfer without joining in the reaction themselves and contaminating the products.
Reactions Occurring in the Cell
Fused NaCl contains \(\mathrm{Na^{\oplus}}\) and \(\mathrm{Cl^{\ominus}}\) ions which are freely mobile. Melting has destroyed the crystal lattice, so the ions are no longer locked in place.
When potential is applied, the cathode attracts \(\mathrm{Na^{\oplus}}\) ions and the anode attracts \(\mathrm{Cl^{\ominus}}\) ions. As these are charged particles, their migration results in an electric current. When these ions reach the respective electrodes, they are discharged according to the following reactions.
Oxidation Half Reaction at Anode
\(\mathrm{Cl^{\ominus}}\) ions migrate to the anode. Each \(\mathrm{Cl^{\ominus}}\) ion that reaches the anode gives one electron to the anode. It oxidises to a neutral Cl atom in the primary process. Two Cl atoms then combine to form chlorine gas in the secondary process.
\[2\,\mathrm{Cl^{\ominus}}\text{(l)} \rightarrow \mathrm{Cl}\text{(g)} + \mathrm{Cl}\text{(g)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(primary process)}\]
\[\mathrm{Cl}\text{(g)} + \mathrm{Cl}\text{(g)} \rightarrow \mathrm{Cl_2}\text{(g)} \qquad \text{(secondary process)}\]
\[2\,\mathrm{Cl^{\ominus}}\text{(l)} \rightarrow \mathrm{Cl_2}\text{(g)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(overall oxidation)}\]
Splitting the change into a primary and a secondary process is worth noticing. Only the primary step involves electrons; the secondary step is ordinary chemistry — two reactive atoms pairing up into a stable molecule.
The battery sucks electrons so produced at the anode and pushes them to the cathode through a wire in the external circuit. The battery thus serves as an electron pump. The electrons from the battery enter into the solution through the cathode and leave the solution through the anode.
Reduction Half Reaction at Cathode
The electrons supplied by the battery are used in cathodic reduction. Each \(\mathrm{Na^{\oplus}}\) ion that reaches the cathode accepts an electron from the cathode and is reduced to metallic sodium.
\[\mathrm{Na^{\oplus}}\text{(l)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Na}\text{(l)}\]
The state symbol (l) is not a slip: the cell runs above the melting point of both the salt and the sodium, so the metal formed collects as a liquid.
Net Cell Reaction
The net cell reaction is the sum of the two electrode reactions. Before adding, the reduction half reaction is multiplied by 2 so that the electrons cancel exactly.
\[2\,\mathrm{Cl^{\ominus}}\text{(l)} \rightarrow \mathrm{Cl_2}\text{(g)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation half reaction)}\]
\[2\,\mathrm{Na^{\oplus}}\text{(l)} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{Na}\text{(l)} \qquad \text{(reduction half reaction)}\]
\[2\,\mathrm{Na^{\oplus}}\text{(l)} + 2\,\mathrm{Cl^{\ominus}}\text{(l)} \rightarrow 2\,\mathrm{Na}\text{(l)} + \mathrm{Cl_2}\text{(g)} \qquad \text{(overall cell reaction)}\]
The whole sequence, from the moment the salt melts to the moment the products are collected, is traced in FC 5.4.
Results of Electrolysis of Molten NaCl
A pale green \(\mathrm{Cl_2}\) gas is released at the anode.
A molten silvery-white sodium is formed at the cathode.
Decomposition of NaCl into metallic sodium and \(\mathrm{Cl_2}\)(g) is nonspontaneous. Common salt is an extremely stable compound; it will not fall apart on its own however long we wait. The electrical energy supplied by the battery forces the reaction to occur — which is precisely the definition of an electrolytic cell given in Section 5.4.3.
A general pattern emerges here that holds for every molten ionic compound: when a molten ionic compound is electrolysed, a metal is formed at the negative electrode and a nonmetal at the positive electrode. The metal cation is attracted to the negative plate and picks up electrons there; the nonmetal anion goes to the positive plate and gives electrons up.
Molten magnesium chloride, \(\mathrm{MgCl_2}\), is electrolysed between inert electrodes. Write the oxidation half reaction, the reduction half reaction and the balanced overall cell reaction, name the product formed at each electrode, and state the sign of each electrode.
5.5.2 Electrolysis of Aqueous NaCl
Electrolysis of an aqueous NaCl can be carried out in the cell used for the electrolysis of molten NaCl using inert electrodes shown in Fig. 5.4. The fused NaCl is replaced by a moderately concentrated aqueous solution of NaCl.
Now, however, there is a third substance in the container that was absent before — water. The water involved in electrolysis of aqueous NaCl leads to electrode reactions that differ from electrolysis of molten NaCl. At each electrode there are now two candidates competing to react, and we have to work out which one wins.
To decide, we use the standard potential \(E^0\) of each competing half reaction. For the moment, treat it simply as a measure of how strongly a species tends to undergo that change: the higher (more positive) the value for a reduction, the greater the tendency of that species to be reduced. Standard potentials are defined properly in Section 5.7.1.
Reduction Half Reaction at Cathode
At the cathode, two reduction reactions compete. One is the reduction of \(\mathrm{Na^{\oplus}}\) ions as in the case of molten NaCl.
\[\text{i.}\quad \mathrm{Na^{\oplus}}\text{(aq)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Na}\text{(s)}, \qquad E^0 = -2.71\ \text{V}\]
The other is the reduction of water to hydrogen gas.
\[\text{ii.}\quad 2\,\mathrm{H_2O}\text{(l)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}\text{(g)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)}, \qquad E^0 = -0.83\ \text{V}\]
The standard potential (Section 5.7.1) for the reduction of water is higher than that for the reduction of \(\mathrm{Na^{\oplus}}\). Comparing the two numbers, \(-0.83\ \text{V}\) is the less negative and therefore the higher value. This implies that water has a much greater tendency to get reduced than the \(\mathrm{Na^{\oplus}}\) ion. Hence reaction (ii), that is, reduction of water, is the cathode reaction when the aqueous NaCl is electrolysed.
Oxidation Half Reaction at Anode
At the anode there will be competition between the oxidation of \(\mathrm{Cl^{\ominus}}\) ion to \(\mathrm{Cl_2}\) gas, as in the case of molten NaCl, and the oxidation of water to \(\mathrm{O_2}\) gas.
\[\text{i.}\quad 2\,\mathrm{Cl^{\ominus}}\text{(aq)} \rightarrow \mathrm{Cl_2}\text{(g)} + 2\,\mathrm{e^{\ominus}}, \qquad E^0_{oxi} = -1.36\ \text{V}\]
\[\text{ii.}\quad 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{O_2}\text{(g)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + 4\,\mathrm{e^{\ominus}}, \qquad E^0_{oxi} = -0.4\ \text{V}\]
The standard electrode potential for the oxidation of water is greater than that of \(\mathrm{Cl^{\ominus}}\) ion, or in other words water has the greater tendency to undergo oxidation. It is, therefore, expected that the anode half reaction would be the oxidation of water.
The experiments have shown, however, that the gas produced at the anode is \(\mathrm{Cl_2}\) and not \(\mathrm{O_2}\). This suggests that the anode reaction is oxidation of \(\mathrm{Cl^{\ominus}}\) to \(\mathrm{Cl_2}\) gas. This is because of the overvoltage, discussion of which is beyond the scope of the present book.
This is an important lesson in itself: a prediction from standard potentials tells us what is thermodynamically favourable, but the experiment decides. Here the oxygen-forming reaction, though favoured on paper, is so sluggish at the electrode surface that chlorine is released instead.
It has been found experimentally that the actual voltage required for electrolysis is greater than that calculated using standard potentials. This additional voltage required is the overpotential.
Overpotential is the extra voltage, over and above the value calculated from standard potentials, that must actually be applied to make an electrolysis proceed at a measurable rate.
Overall Cell Reaction
It is the sum of the electrode reactions.
\[2\,\mathrm{Cl^{\ominus}}\text{(aq)} \rightarrow \mathrm{Cl_2}\text{(g)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation at anode)}\]
\[2\,\mathrm{H_2O}\text{(l)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}\text{(g)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \qquad \text{(reduction at cathode)}\]
\[2\,\mathrm{Cl^{\ominus}}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{Cl_2}\text{(g)} + \mathrm{H_2}\text{(g)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \qquad \text{(overall cell reaction)}\]
Results of Electrolysis of Aqueous NaCl
\(\mathrm{H_2}\) gas is liberated at the cathode.
\(\mathrm{Cl_2}\) gas is released at the anode.
Because \(\mathrm{Na^{\oplus}}\) ions remain unreacted and \(\mathrm{OH^{\ominus}}\) ions are formed at the cathode, NaCl solution is converted to NaOH solution.
Result (iii) answers the recall question with which this chapter opened. The sodium ions never leave the solution; the chloride ions do. What is left behind in the container is \(\mathrm{Na^{\oplus}}\) and \(\mathrm{OH^{\ominus}}\) — sodium hydroxide.
Electrolysis of molten NaCl vs Electrolysis of aqueous NaCl
| Species present |
\(\mathrm{Na^{\oplus}}\), \(\mathrm{Cl^{\ominus}}\) only |
\(\mathrm{Na^{\oplus}}\), \(\mathrm{Cl^{\ominus}}\) and \(\mathrm{H_2O}\) |
| Reaction at cathode |
\(\mathrm{Na^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Na}\) |
\(2\,\mathrm{H_2O} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2} + 2\,\mathrm{OH^{\ominus}}\) |
| Reaction at anode |
\(2\,\mathrm{Cl^{\ominus}} \rightarrow \mathrm{Cl_2} + 2\,\mathrm{e^{\ominus}}\) |
\(2\,\mathrm{Cl^{\ominus}} \rightarrow \mathrm{Cl_2} + 2\,\mathrm{e^{\ominus}}\) |
| Product at cathode |
Molten sodium metal |
Hydrogen gas |
| Product at anode |
Chlorine gas |
Chlorine gas |
| Left in the container |
Salt is consumed |
NaCl solution becomes NaOH solution |
Electrolysis is not confined to chemical works. Sea water is the source of 300000 tonnes of magnesium produced every year by electrolysis, which is why the lightest structural metal in common use can be made almost anywhere there is a coastline and a power supply. A more decorative application is electrochemical art: aluminium, chromium and tin can be coloured by an electrochemical process called anodizing. In this process the metal anode oxidises to give a metal oxide coat; when an organic dye is added to the electrolyte, the dye molecules soak into the spongy surface of the coating and become trapped as the metal oxide surface hardens. The bright anodised finish on a water bottle or a laptop shell is exactly this. A green point worth noting is that the chlor-alkali process discussed above yields hydrogen as a by-product at the cathode, and modern plants capture it as a fuel rather than venting it.
A dilute aqueous solution of sodium sulphate is electrolysed between inert platinum electrodes. Predict the gas liberated at each electrode, write the half reaction responsible in each case, and explain why the sulphate ion is not discharged at the anode.
5.5.3 Quantitative Aspects of Electrolysis
So far we have asked what is formed. We now ask how much — and this is where electrolysis becomes a calculation.
a. The mass of reactant consumed or the mass of product formed at an electrode during electrolysis can be calculated by knowing the stoichiometry of the half reaction at the electrode.
The logic is a chain with four links: current and time give charge; charge gives moles of electrons; the half reaction converts moles of electrons into moles of product; and molar mass converts moles of product into grams. FC 5.5 sets out the chain.
i. Calculation of Quantity of Electricity Passed
To calculate the quantity of electricity (\(Q\)) passed during electrolysis, the amount of current, \(I\), passed through the cell is measured. The time for which the current is passed is noted.
\[Q\,\text{(C)} = I\,\text{(A)} \times t\,\text{(s)} \qquad \text{...(5.16)}\]
The time must be in seconds. A problem that states minutes or hours is asking you to convert before substituting.
ii. Calculation of Moles of Electrons Passed
The total charge passed is \(Q\)(C). The charge of one mole of electrons is 96500 coulombs (C). It is referred to as one faraday (1 F). Hence,
\[\text{Moles of electrons actually passed} = \frac{Q\,\text{(C)}}{96500\ \text{(C/mol e}^{\ominus})} \qquad \text{...(5.17)}\]
One faraday is the quantity of electric charge carried by one mole of electrons, equal to 96500 coulombs.
quantity of electricity \(Q\) and the faraday (SI unit of charge is the coulomb, C, where \(1\ \text{C} = 1\ \text{A}\times1\ \text{s}\); the faraday is not a separate SI unit but a convenient package of charge, \(1\ \text{F} = 96500\ \text{C mol}^{-1}\). Always put current in amperes and time in seconds before using Eq. (5.16), or the coulombs will be wrong by a factor of 60 or 3600.)
iv. Calculation of Mass of Product
Mass of product
\[W = \text{moles of product} \times \text{molar mass of product}\]
\[= \frac{I\,\text{(A)} \times t\,\text{(s)}}{96500\ \text{(C/mol e}^{\ominus})} \times \text{mole ratio} \times \text{molar mass of product} \qquad \text{...(5.20)}\]
What is the mass of Cu metal produced at the cathode during the passage of 5 ampere current through \(\mathrm{CuSO_4}\) solution for 100 minutes? Molar mass of Cu is \(63.5\ \text{g mol}^{-1}\).
- Stoichiometry for the formation of Cu is
\[\mathrm{Cu^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} = \mathrm{Cu}\text{(s)}\]
Hence,
\[\text{mole ratio} = \frac{1\ \text{mol}}{2\ \text{mol e}^{\ominus}}\]
- Mass of Cu formed,
\[W = \frac{I\,\text{(A)} \times t\,\text{(s)}}{96500\ \text{(C/mol e}^{\ominus})} \times \text{mole ratio} \times \text{molar mass of Cu}\]
\[= \frac{5\ \text{A} \times 100 \times 60\ \text{s}}{96500\ \text{(C/mol e}^{\ominus})} \times \frac{1\ \text{mol}}{2\ \text{mol e}^{\ominus}} \times 63.5\ \text{g mol}^{-1}\]
\[= 9.87\ \text{g}\]
The 100 minutes had to become \(100 \times 60\) seconds before it could be multiplied by amperes to give coulombs.
How long will it take to produce \(2.415\ \text{g}\) of Ag metal from its salt solution by passing a current of 3 ampere? Molar mass of Ag is \(107.9\ \text{g mol}^{-1}\).
- Stoichiometry:
\[\mathrm{Ag^{\oplus}}\text{(aq)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ag}\text{(s)}\]
\[\text{mole ratio} = \frac{1\ \text{mol}}{1\ \text{mol e}^{\ominus}}\]
- Here the mass is known and the time is the unknown, so Eq. (5.20) is used in reverse.
\[W = \frac{I\,\text{(A)} \times t\,\text{(s)}}{96500\ \text{(C/mol e}^{\ominus})} \times \text{mole ratio} \times \text{molar mass of Ag}\]
\[2.415\ \text{g} = \frac{3\ \text{A} \times t}{96500\ \text{(C/mol e}^{\ominus})} \times \frac{1\ \text{mol}}{1\ \text{mol e}^{\ominus}} \times 107.9\ \text{g mol}^{-1}\]
\[t = \frac{2.415 \times 96500\ (\text{C} = \text{As})}{3\ \text{A} \times 107.9}\]
\[= 720\ \text{s} = 12\ \text{min}\]
How many moles of electrons are required for reduction of 3 moles of \(\mathrm{Zn^{2\oplus}}\) to Zn? How many Faradays of electricity will be required?
- The balanced equation for the reduction of \(\mathrm{Zn^{2\oplus}}\) to Zn is
\[\mathrm{Zn^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Zn}\text{(s)}\]
The equation shows that 1 mole of \(\mathrm{Zn^{2\oplus}}\) is reduced to Zn by 2 moles of electrons. For the reduction of 3 moles of \(\mathrm{Zn^{2\oplus}}\), 6 moles of electrons will be required.
Faraday (96500 coulombs) is the amount of charge on one mole of electrons. Therefore, for 6 moles of electrons, 6 F of electricity will be required.
Observe that no current or time was given, and none was needed — this problem is pure stoichiometry of the half reaction.
What current strength, in amperes, must be passed through a solution of copper sulphate for 30 minutes to deposit \(1.27\ \text{g}\) of copper at the cathode? Molar mass of Cu \(= 63.5\ \text{g mol}^{-1}\).
Step 1 — moles of copper required:
\[n(\mathrm{Cu}) = \frac{1.27\ \text{g}}{63.5\ \text{g mol}^{-1}} = 0.0200\ \text{mol}\]
Step 2 — moles of electrons. The half reaction is \(\mathrm{Cu^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\text{(s)}\), so each mole of copper needs 2 moles of electrons:
\[n(\mathrm{e^{\ominus}}) = 2 \times 0.0200\ \text{mol} = 0.0400\ \text{mol}\]
Step 3 — charge required, from Eq. (5.17) used in reverse:
\[Q = 0.0400\ \text{mol} \times 96500\ \text{C mol}^{-1} = 3860\ \text{C}\]
Step 4 — current, from Eq. (5.16), remembering that \(30\ \text{min} = 1800\ \text{s}\):
\[I = \frac{Q}{t} = \frac{3860\ \text{C}}{1800\ \text{s}} = 2.14\ \text{A}\]
The chain of FC 5.5 has simply been walked backwards, from the mass at the end to the current at the start.
b. Two Cells Connected in Series
Suppose two cells containing different electrolytes are connected in series. The same quantity of electricity is passed through them, because in a series circuit there is only one path and every electron that goes through the first cell must also go through the second. The masses of the substances liberated at the electrodes of the two cells are related as given below.
The mass of the substance produced at the electrode of the first cell is given by
\[W_1 = \frac{Q\,\text{(C)}}{96500\ \text{(C/mol e}^{\ominus})} \times (\text{mole ratio})_1 \times M_1\]
Hence,
\[\frac{Q\,\text{(C)}}{96500\ \text{(C/mol e}^{\ominus})} = \frac{W_1}{(\text{mole ratio})_1 \times M_1}\]
Similarly, the mass of substance liberated at the electrode of the second cell is \(W_2\) in the equation,
\[\frac{Q\,\text{(C)}}{96500\ \text{(C/mol e}^{\ominus})} = \frac{W_2}{(\text{mole ratio})_2 \times M_2}\]
\(M_1\) and \(M_2\) are the molar masses of the substances produced at the electrodes of cells 1 and 2.
Because \(\dfrac{Q\,\text{(C)}}{96500\ \text{(C/mol e}^{\ominus})}\) is the same for both, we have
\[\frac{W_1}{(\text{mole ratio})_1 \times M_1} = \frac{W_2}{(\text{mole ratio})_2 \times M_2} \qquad \text{...(5.21)}\]
The beauty of Eq. (5.21) is that the current and the time have vanished from it. If one mass is known, the other follows without knowing how long the experiment ran.
In a certain electrolysis experiment \(4.36\ \text{g}\) of Zn are deposited in one cell containing \(\mathrm{ZnSO_4}\) solution. Calculate the mass of Al deposited in another cell containing \(\mathrm{AlCl_3}\) solution connected in series with the \(\mathrm{ZnSO_4}\) cell. Molar masses of Zn and Al are \(65.4\ \text{g mol}^{-1}\) and \(27\ \text{g mol}^{-1}\), respectively.
Cell 1:
\[\mathrm{Zn^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Zn}\text{(s)}\]
\[(\text{mole ratio})_1 = \frac{1\ \text{mol}}{2\ \text{mol e}^{\ominus}}\]
Cell 2:
\[\mathrm{Al^{3\oplus}}\text{(aq)} + 3\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Al}\text{(s)}\]
\[(\text{mole ratio})_2 = \frac{1\ \text{mol}}{3\ \text{mol e}^{\ominus}}\]
\[\frac{W_1}{(\text{mole ratio})_1 \times M_1} = \frac{W_2}{(\text{mole ratio})_2 \times M_2}\]
\[W_1 = 4.36\ \text{g},\quad M_1 = 65.4\ \text{g mol}^{-1},\quad M_2 = 27\ \text{g mol}^{-1}\]
Substitution of the quantities gives
\[\frac{4.36\ \text{g}}{1\ \text{mol}/2\ \text{mol e}^{\ominus} \times 65.4\ \text{g mol}^{-1}} = \frac{W_2}{1\ \text{mol}/3\ \text{mol e}^{\ominus} \times 27\ \text{g mol}^{-1}}\]
or
\[\frac{4.36\ \text{g} \times 2}{65.4} = \frac{W_2 \times 3}{27}\]
Hence,
\[W_2 = \frac{4.36\ \text{g} \times 2 \times 27}{65.4 \times 3} = 1.2\ \text{g}\]
Aluminium needs three electrons per atom and has a smaller molar mass than zinc, so the same charge deposits far less of it by mass.
A current of \(2\ \text{A}\) is passed for 50 minutes through two cells connected in series, one containing \(\mathrm{AgNO_3}\) solution and the other containing \(\mathrm{CuSO_4}\) solution. Calculate the mass of silver and the mass of copper deposited. (Molar masses: Ag \(= 108\ \text{g mol}^{-1}\), Cu \(= 63.5\ \text{g mol}^{-1}\).)
5.6 Galvanic or Voltaic Cell
In galvanic or voltaic cells, electricity is generated through the use of spontaneous chemical reactions. The reaction is one that would happen anyway; the cell’s job is to make it happen in such a way that the electrons are forced to travel through a wire on their journey, where we can put them to work.
A galvanic (or voltaic) cell is made of two half cells. Both names, as you saw at the start of this chapter, honour the two Italian scientists whose disagreement over twitching frog legs led to the first battery. Each half cell consists of a metal strip immersed in the solution of its own ions of known concentration. For example, a strip of zinc metal immersed in \(1\ \text{M}\) aqueous solution of zinc ions forms a half cell.
A half cell is a single electrode in contact with a solution of its own ions, at which either the oxidation half reaction or the reduction half reaction of the cell takes place.
It follows that two metal plates and the solutions of their ions with known concentrations are required for the construction of a galvanic (voltaic) cell. Two half cells are constructed by immersing the two metal plates in solutions of their respective ions placed in separate containers.
The word separate is doing important work in that sentence. If the two solutions were mixed, the oxidising and reducing species would meet directly and simply react in the beaker, releasing their energy as heat. Keeping them apart forces the electrons to take the long way round.
Completing the Circuit
The two half cells so constructed are combined together to form the galvanic cell. The metal plates, called electrodes, are connected through a voltmeter by a conducting wire for transfer of electrons between them. The arrangement is shown in Fig. 5.6a.
To complete the circuit, the two solutions are connected by a conducting medium through which cations and anions move from one compartment to the other. This requirement is fulfilled by a salt bridge.
Why is the wire alone not enough? Because a circuit must be closed. Electrons leaving the anode would very quickly leave that solution with a surplus of positive charge, while the cathode compartment would build up a surplus of negative charge, and this charge separation would stop the reaction almost the instant it began.
The cell you met in Standard XI as the Daniel cell was devised in 1836 by John Frederic Daniell, a professor at King’s College London. The batteries then available faded within minutes because bubbles of hydrogen collected on one plate and choked the reaction. Daniell’s answer was to give each electrode its own solution and separate the two, so that no gas was produced at all. The result was the first battery that could deliver a steady current for hours, and it became the working power supply of the early electric telegraph — the first practical electrical technology in history.
5.6.1 Salt Bridge
In a galvanic cell, the two solutions are connected by a salt bridge. It is a U tube containing a saturated solution of an inert electrolyte such as KCl or \(\mathrm{NH_4NO_3}\), and 5% agar solution.
The word “inert” here is the key to the choice. The ions of the electrolyte do not react with the ions of the electrode solutions or with the electrodes. Potassium, chloride, ammonium and nitrate ions will neither precipitate the metal ions nor be oxidised or reduced themselves, so the bridge carries charge without joining the chemistry.
A salt bridge is a U tube filled with a gel of an inert electrolyte, which connects the two half cells of a galvanic cell so that ions can pass between them without the two solutions mixing.
Preparing and Storing a Salt Bridge
Salt bridge is prepared by filling a U tube with hot saturated solution of the salt and agar agar solution, allowing it to cool. The cooled solution sets into a gel which does not come out on inverting the tube.
The agar is not there to conduct anything; it is there to hold the salt solution in place. The ions remain free to move through the gel, but the liquid cannot run out of the tube into the beakers. The salt bridge is kept dipped in distilled water when not in use, as shown in Fig. 5.5, so that the gel does not dry out.
A salt bridge does not have to be a U tube. Take a long strip of good quality filter paper, laminate it, cut the two ends open, and dip those ends in a saturated solution of KCl for 24 hours; the strip can then be used as a salt bridge simply by dipping its two ends into the two solutions. Try this before reading on, and predict whether the voltmeter reading will differ from the one obtained with a gel-filled U tube. The same idea, scaled down and toughened, sits inside every commercial cell you own: a torch cell or a mobile phone battery contains no U tube, but it does contain a porous separator soaked in electrolyte, doing exactly the job described here — letting ions through while keeping the two reacting materials from touching each other.
Functions of Salt Bridge
The salt bridge serves the following functions:
It provides an electrical contact between the two solutions and thereby completes the electrical circuit.
It prevents mixing of the two solutions.
It maintains electrical neutrality in both the solutions by transfer of ions.
Function (iii) deserves a sentence of explanation, because it is the one students most often state without understanding. As metal ions are formed at the anode, that solution tends to become positively charged; as metal ions are deposited at the cathode, that solution tends to become negatively charged. Anions from the bridge migrate towards the anode compartment and cations towards the cathode compartment, exactly cancelling both build-ups as fast as they arise.
5.6.3 Writing of Cell Reaction
The cell reaction corresponding to the cell notation is written on the assumption that the right hand side electrode is the cathode (+) and the left hand side electrode is the anode (−).
As mentioned in Section 5.4.2, the oxidation half reaction occurs at the anode and the reduction half reaction at the cathode. It, therefore, follows that in a galvanic cell the oxidation half reaction takes place on the left hand side electrode and the reduction half reaction on the right hand side electrode.
The following steps are followed to write the cell reaction.
Write the oxidation half reaction for the left hand side electrode (anode) and the reduction half reaction for the right hand side electrode (cathode).
Add the two electrode half reactions to get the overall cell reaction. While adding, the electrons must be balanced. For this purpose, it may be necessary to multiply one or both of the half reactions by a suitable numerical factor(s). No electrons should appear in the overall reaction.
It is important to note that the individual half reactions may be written with one or more electrons. For example, half reactions for \(\mathrm{H_2}\) gas, whether written as
\[2\,\mathrm{H^{\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}\text{(g)} \quad \text{or} \quad \mathrm{H^{\oplus}}\text{(aq)} + \mathrm{e^{\ominus}} \rightarrow \tfrac{1}{2}\,\mathrm{H_2}\text{(g)}\]
makes no difference. In writing the overall cell reaction, the electrons must be balanced.
A Worked Illustration
Consider the cell,
\[\mathrm{Al}\text{(s)} \,|\, \mathrm{Al^{3\oplus}}\text{(1M)} \,\|\, \mathrm{Ni^{2\oplus}}\text{(1M)} \,|\, \mathrm{Ni}\text{(s)}\]
with electrons flowing through the external circuit from the aluminium electrode to the nickel electrode.
The oxidation at the anode is
\[\mathrm{Al}\text{(s)} \rightarrow \mathrm{Al^{3\oplus}}\text{(1M)} + 3\,\mathrm{e^{\ominus}}\]
The reduction half reaction at the cathode is
\[\mathrm{Ni^{2\oplus}}\text{(1M)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Ni}\text{(s)}\]
To balance the electrons, the oxidation reaction is multiplied by 2 and the reduction reaction by 3. Both halves then involve six electrons, six being the lowest common multiple of 3 and 2. The two half reactions so obtained, when added, give the overall cell reaction. Thus,
\[2\,\mathrm{Al}\text{(s)} \rightarrow 2\,\mathrm{Al^{3\oplus}}\text{(1M)} + 6\,\mathrm{e^{\ominus}} \qquad \text{(oxidation half reaction)}\]
\[3\,\mathrm{Ni^{2\oplus}}\text{(1M)} + 6\,\mathrm{e^{\ominus}} \rightarrow 3\,\mathrm{Ni}\text{(s)} \qquad \text{(reduction half reaction)}\]
\[2\,\mathrm{Al}\text{(s)} + 3\,\mathrm{Ni^{2\oplus}}\text{(1M)} \rightarrow 2\,\mathrm{Al^{3\oplus}}\text{(1M)} + 3\,\mathrm{Ni}\text{(s)} \qquad \text{(overall cell reaction)}\]
No electrons remain in the final equation — always check this before writing your answer, because a stray electron is a sure sign that the multiplying factors were wrong.
Write the electrode reactions and the overall cell reaction for the cell
\[\mathrm{Cr}\text{(s)} \,|\, \mathrm{Cr^{3\oplus}}\text{(1M)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(1M)} \,|\, \mathrm{Cu}\text{(s)}\]
and state at which electrode electrons enter the external circuit.
By convention the left hand electrode is the anode, where oxidation occurs:
\[\mathrm{Cr}\text{(s)} \rightarrow \mathrm{Cr^{3\oplus}}\text{(1M)} + 3\,\mathrm{e^{\ominus}} \qquad \text{(oxidation at anode)}\]
The right hand electrode is the cathode, where reduction occurs:
\[\mathrm{Cu^{2\oplus}}\text{(1M)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\text{(s)} \qquad \text{(reduction at cathode)}\]
The halves involve 3 and 2 electrons, so multiply the first by 2 and the second by 3 to bring both to six electrons:
\[2\,\mathrm{Cr}\text{(s)} \rightarrow 2\,\mathrm{Cr^{3\oplus}}\text{(1M)} + 6\,\mathrm{e^{\ominus}}\]
\[3\,\mathrm{Cu^{2\oplus}}\text{(1M)} + 6\,\mathrm{e^{\ominus}} \rightarrow 3\,\mathrm{Cu}\text{(s)}\]
Adding, and cancelling the six electrons on each side:
\[2\,\mathrm{Cr}\text{(s)} + 3\,\mathrm{Cu^{2\oplus}}\text{(1M)} \rightarrow 2\,\mathrm{Cr^{3\oplus}}\text{(1M)} + 3\,\mathrm{Cu}\text{(s)} \qquad \text{(overall cell reaction)}\]
Electrons enter the external circuit at the chromium electrode, the anode, since that is where oxidation releases them.
Write the electrode reactions and the overall cell reaction for the Daniel cell you learnt in Standard XI.
For the cell \(\mathrm{Mg}\text{(s)} \,|\, \mathrm{Mg^{2\oplus}}\text{(1M)} \,\|\, \mathrm{Al^{3\oplus}}\text{(1M)} \,|\, \mathrm{Al}\text{(s)}\), write the oxidation half reaction, the reduction half reaction and the balanced overall cell reaction, and state the direction in which cations from the salt bridge migrate while the cell is working.
5.7 Electrode Potential and Cell Potential
We know that a galvanic cell drives electrons through a wire. The next question is how hard it pushes them — and where that push comes from.
A galvanic cell is composed of two half cells, each consisting of an electronic conductor (metal plate) and an electrolytic conductor (solution of ions) in contact. At the surface of separation of the solid metal and the solution, there exists a difference of electrical potential.
Why should there be one? At that surface a half reaction is going on: metal atoms tend to enter the solution as ions leaving electrons behind, while metal ions from the solution tend to deposit on the plate. These two tendencies do not cancel exactly, so the metal ends up carrying a charge slightly different from that of the solution touching it, and a potential difference builds up across the boundary, as sketched in Fig. 5.7a.
This potential difference, established due to the electrode half reaction occurring at the electrode surface, is the electrode potential.
Electrode potential is the potential difference set up between a metal and the solution in contact with it, as a result of the half reaction occurring at that electrode surface.
Oxidation Potential and Reduction Potential
The potential is associated with each of the half reactions, be it oxidation or reduction. The potential associated with an oxidation reaction is the oxidation potential, while that associated with reduction gives the reduction potential.
For the same electrode these two are the same measurement viewed from opposite directions, so they carry the same number with opposite signs.
Oxidation potential vs Reduction potential
| Half reaction described |
\(\mathrm{M} \rightarrow \mathrm{M^{n\oplus}} + n\,\mathrm{e^{\ominus}}\) |
\(\mathrm{M^{n\oplus}} + n\,\mathrm{e^{\ominus}} \rightarrow \mathrm{M}\) |
| Electrode of a galvanic cell it belongs to |
Anode (−) |
Cathode (+) |
| Relation between the two |
Numerically equal, opposite in sign |
Numerically equal, opposite in sign |
| Which one IUPAC adopts as “the” electrode potential |
— |
This one |
The overall cell potential, also called the electromotive force (emf), is made of the contributions from each of the electrodes. In other words, the cell potential is the algebraic sum of the electrode potentials,
\[E_{cell} = E_{oxi}\,\text{(anode)} + E_{red}\,\text{(cathode)} \qquad \text{...(5.22)}\]
where \(E_{oxi}\) is the oxidation potential of the anode (−) and \(E_{red}\) is the reduction potential of the cathode (+).
When a galvanic cell operates, electrons are generated at the anode. These electrons move through the external circuit to the cathode. The cell potential is the force that pushes electrons away from the anode (−) and pulls them toward the cathode, where they are consumed.
Think of it as the electrical equivalent of a height difference: the larger the potential difference, the harder the electrons are driven around the circuit.
electrode potential \(E\) and cell potential \(E_{cell}\) (SI unit volt, V. Sign convention: by IUPAC the electrode potential quoted is always the reduction potential, so reversing a half reaction reverses the sign of its potential. A galvanic cell has \(E_{cell}\) positive; a negative value means the reaction as written is not spontaneous and the cell would have to be driven the other way.)
5.7.1 Standard Potentials
The electrode potential and the cell potential depend on the concentrations of solutions, the pressures of gases and the temperature. Two chemists measuring the same electrode under different conditions would get different numbers, and their results could not be compared.
To facilitate comparison of different galvanic cells, it is therefore necessary to measure the cell voltage under a given set of standard conditions of concentration and temperature.
The standard conditions chosen are \(1\ \text{M}\) concentration of solution, \(1\ \text{atm}\) pressure for gases, solids and liquids in pure form, and \(25\ ^\circ\text{C}\). The voltage measured under these conditions is called the standard potential, designated as \(E^0\).
Standard potential \(E^0\) is the potential measured when all solutions are \(1\ \text{M}\), all gases are at \(1\ \text{atm}\) pressure, all solids and liquids are in the pure form, and the temperature is \(25\ ^\circ\text{C}\).
The standard cell potential is the algebraic sum of the standard electrode potentials, similar to Eq. (5.22).
\[E^0_{cell} = E^0_{oxi}\,\text{(anode)} + E^0_{red}\,\text{(cathode)} \qquad \text{...(5.23)}\]
Here \(E^0_{oxi}\) is the standard oxidation potential and \(E^0_{red}\) is the standard reduction potential.
The IUPAC Convention
Quoting two different potentials for every electrode would be needlessly confusing, so one of them has been adopted as standard.
According to IUPAC convention, the standard potential of an electrode is taken as its standard reduction potential.
It must be realised that the standard oxidation potential of any electrode is numerically equal to its standard reduction potential with the reversal of sign. For example, the standard oxidation potential of the \(\mathrm{Zn^{2\oplus}}\text{(1M)} \,|\, \mathrm{Zn}\) electrode is \(0.76\ \text{V}\). Its standard reduction potential will be \(-0.76\ \text{V}\).
Hereafter the standard reduction potential will be called the standard potential — the voltage associated with a reduction reaction. Every value you will meet in Table 5.1 later in this chapter is of this kind.
Writing the Cell Potential in Terms of Standard Potentials
It follows that the standard cell potential (emf) is written in terms of the standard potentials of the electrodes. In Eq. (5.23), \(E^0_{oxi}\text{(anode)}\) is replaced by \(-E^0_{red}\text{(anode)}\), since reversing the half reaction reverses the sign. We then write,
\[E^0_{cell} = -E^0_{red}\,\text{(anode)} + E^0_{red}\,\text{(cathode)}\]
Omitting the subscript \(red\), we have
\[E^0_{cell} = E^0\,\text{(cathode, +)} - E^0\,\text{(anode, −)} \qquad \text{...(5.24)}\]
This compact result — cathode minus anode — is the one used in every numerical problem in this chapter. Both values substituted into it must be standard reduction potentials, taken straight from the table without changing any sign.
While constructing a galvanic cell from two electrodes, the electrode with the higher standard potential is the cathode (+) and that with the lower standard potential is the anode (−). The difference in electrical potential between anode and cathode is the cell voltage.
Reason it out rather than memorising it: the electrode with the higher standard potential has the stronger tendency to be reduced, so it takes the electrons — and taking electrons is what a cathode does. Assigning the electrodes this way also guarantees that \(E^0_{cell}\) from Eq. (5.24) comes out positive, as it must for a spontaneous cell.
Standard potentials are \(E^0 = -0.44\ \text{V}\) for the \(\mathrm{Fe^{2\oplus}}\,|\,\mathrm{Fe}\) electrode and \(E^0 = +0.34\ \text{V}\) for the \(\mathrm{Cu^{2\oplus}}\,|\,\mathrm{Cu}\) electrode. Identify the anode and the cathode of the galvanic cell built from these two electrodes, calculate its standard cell potential, and write the cell notation.
Step 1 — assign the electrodes. Copper has the higher standard potential (\(+0.34\ \text{V}\) against \(-0.44\ \text{V}\)), so the copper electrode is the cathode (+) and the iron electrode is the anode (−).
Step 2 — apply Eq. (5.24):
\[E^0_{cell} = E^0\text{(cathode)} - E^0\text{(anode)} = 0.34\ \text{V} - (-0.44\ \text{V}) = 0.78\ \text{V}\]
The value is positive, confirming that the cell reaction is spontaneous as assigned.
Step 3 — write the notation with the anode on the left and the cathode on the right, following the conventions of Section 5.6.2:
\[\mathrm{Fe}\text{(s)} \,|\, \mathrm{Fe^{2\oplus}}\text{(1M)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(1M)} \,|\, \mathrm{Cu}\text{(s)}\]
The standard potentials are \(E^0 = -2.37\ \text{V}\) for \(\mathrm{Mg^{2\oplus}}\,|\,\mathrm{Mg}\) and \(E^0 = -0.14\ \text{V}\) for \(\mathrm{Sn^{2\oplus}}\,|\,\mathrm{Sn}\). Decide which electrode acts as the anode, calculate \(E^0_{cell}\), write the cell notation and write the overall cell reaction.
5.7.2 Dependence of Cell Potential on Concentration (Nernst Equation)
The standard cell potential tells us whether or not the reactants in their standard states form the products in their standard states spontaneously. But laboratories and batteries rarely operate at exactly \(1\ \text{M}\), and a working cell’s concentrations change continuously as it discharges.
To predict the spontaneity of reactions for anything other than standard concentration conditions, we need to know how the voltage of a galvanic cell varies with concentration. The dependence of cell voltage on concentrations is given by the Nernst equation.
For any general reaction,
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\]
the cell voltage is given by
\[E_{cell} = E^0_{cell} - \frac{RT}{nF}\ln\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\]
\[= E^0_{cell} - \frac{2.303\,RT}{nF}\log_{10}\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \qquad \text{...(5.25)}\]
where \(n\) = moles of electrons used in the reaction, \(F\) = Faraday = \(96500\ \text{C}\), \(T\) = temperature in kelvin, and \(R\) = gas constant = \(8.314\ \text{J K}^{-1}\text{mol}^{-1}\).
Two small pieces of mathematics are worth naming here. The switch from \(\ln\) to \(\log_{10}\) brings in the factor \(2.303\), because \(\ln x = 2.303\log_{10}x\); and a logarithm appears at all because the potential depends on the ratio of product to reactant concentrations, so that a ten-fold change in concentration alters the voltage by a fixed small amount rather than by a factor of ten.
At \(25\ ^\circ\text{C}\),
\[\frac{2.303\,RT}{F} = 0.0592\ \text{V}\]
Therefore at \(25\ ^\circ\text{C}\), Eq. (5.25) becomes
\[E_{cell} = E^0_{cell} - \frac{0.0592\ \text{V}}{n}\log_{10}\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \qquad \text{...(5.26)}\]
The Eq. (5.25) or Eq. (5.26) is the Nernst equation.
Nernst equation: at a given temperature the cell potential is the standard cell potential reduced by a term \(\dfrac{2.303\,RT}{nF}\log_{10}Q\), where the logarithmic term contains the concentrations of products over reactants, each raised to its coefficient in the balanced cell reaction, and \(n\) is the number of moles of electrons transferred.
Reading the Two Terms
The first term on the right hand side of the Nernst equation represents standard state electrochemical conditions. The second term is the correction for non standard state conditions.
The cell potential equals the standard potential if the concentrations of reactants and products are \(1\ \text{M}\) each. Thus, if \([\mathrm{A}] = [\mathrm{B}] = [\mathrm{C}] = [\mathrm{D}] = 1\ \text{M}\), then the ratio inside the logarithm becomes 1, \(\log_{10}1 = 0\), the whole correction term vanishes, and
\[E_{cell} = E^0_{cell}\]
If a gaseous substance is present in the cell reaction, its concentration term is replaced by the partial pressure of the gas. Concentrations of solids and pure liquids are taken to be unity and so never appear in the ratio at all — a pure solid has no concentration that can be varied.
The equation is named for Walther Nernst, the German physical chemist who worked it out in the 1880s while still in his twenties. His interests ranged far beyond electrochemistry — he formulated the heat theorem that became the third law of thermodynamics, for which he received the Nobel Prize in Chemistry in 1920 — but it is this compact voltage-versus-concentration relation that carries his name into every school laboratory.
The Nernst equation can be used to calculate cell potential and electrode potential.
i. Calculation of Cell Potential
Consider the cell
\[\mathrm{Cd}\text{(s)} \,|\, \mathrm{Cd^{2\oplus}}\text{(aq)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(aq)} \,|\, \mathrm{Cu}\]
Let us first write the cell reaction.
\[\mathrm{Cd}\text{(s)} \rightarrow \mathrm{Cd^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation at anode)}\]
\[\mathrm{Cu^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\text{(s)} \qquad \text{(reduction at cathode)}\]
\[\mathrm{Cd}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Cd^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad \text{(overall cell reaction)}\]
Here \(n = 2\).
The potential of the cell is given by the Nernst equation,
\[E_{cell} = E^0_{cell} - \frac{0.0592}{2}\log_{10}\frac{[\mathrm{Cd^{2\oplus}}]}{[\mathrm{Cu^{2\oplus}}]} \qquad \text{at } 25\ ^\circ\text{C}\]
(Concentrations of solids and pure liquids are taken to be unity — which is why neither \(\mathrm{Cd}\)(s) nor \(\mathrm{Cu}\)(s) appears in the ratio.)
ii. Calculation of Electrode Potential
The same equation works for a single electrode, using its own half reaction. Consider \(\mathrm{Zn^{2\oplus}}\text{(aq)} \,|\, \mathrm{Zn}\text{(s)}\).
The reduction reaction for the electrode is
\[\mathrm{Zn^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Zn}\text{(s)}\]
Applying the Nernst equation, the electrode potential is given by
\[E_{Zn} = E^0_{Zn} - \frac{0.0592}{2}\log_{10}\frac{1}{[\mathrm{Zn^{2\oplus}}]}\]
\[= E^0_{Zn} + \frac{0.0592}{2}\log_{10}[\mathrm{Zn^{2\oplus}}] \qquad \text{at } 25\ ^\circ\text{C}\]
The sign changed because the reciprocal was turned the right way up inside the logarithm: \(\log_{10}(1/x) = -\log_{10}x\). Solid zinc, being a pure solid, again contributes 1 to the numerator.
Calculate the voltage of the cell, \(\mathrm{Sn}\text{(s)} \,|\, \mathrm{Sn^{2\oplus}}\text{(0.02M)} \,\|\, \mathrm{Ag^{\oplus}}\text{(0.01M)} \,|\, \mathrm{Ag}\text{(s)}\) at \(25\ ^\circ\text{C}\). \(E^0_{Sn} = -0.136\ \text{V}\), \(E^0_{Ag} = 0.800\ \text{V}\).
First we write the cell reaction. The left hand electrode is the anode and the right hand electrode the cathode, and the silver half reaction is multiplied by 2 to balance the electrons.
\[\mathrm{Sn}\text{(s)} \rightarrow \mathrm{Sn^{2\oplus}}\text{(0.02M)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation at anode)}\]
\[[\mathrm{Ag^{\oplus}}\text{(0.01M)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ag}\text{(s)}] \times 2 \qquad \text{(reduction at cathode)}\]
\[\mathrm{Sn}\text{(s)} + 2\,\mathrm{Ag^{\oplus}}\text{(0.01M)} \rightarrow \mathrm{Sn^{2\oplus}}\text{(0.02 M)} + 2\,\mathrm{Ag}\text{(s)} \qquad \text{(overall cell reaction)}\]
The cell potential is given by
\[E_{cell} = E^0_{cell} - \frac{0.0592\ \text{V}}{2}\log_{10}\frac{[\mathrm{Sn^{2\oplus}}]}{[\mathrm{Ag^{\oplus}}]^2}\]
Note that \([\mathrm{Ag^{\oplus}}]\) is squared because 2 moles of \(\mathrm{Ag^{\oplus}}\) appear in the balanced cell reaction, and that \(n = 2\) because two electrons are transferred.
\[E^0_{cell} = E^0_{Ag} - E^0_{Sn} = 0.8\ \text{V} + 0.136\ \text{V} = 0.936\ \text{V}\]
Hence,
\[E_{cell} = 0.936\ \text{V} - \frac{0.0592\ \text{V}}{2}\log_{10}\frac{0.02}{(0.01)^2}\]
\[= 0.936\ \text{V} - \frac{0.0592\ \text{V}}{2}\log_{10}200\]
\[= 0.936\ \text{V} - \frac{0.0592\ \text{V}}{2} \times 2.301\]
\[= 0.936\ \text{V} - 0.0681\ \text{V} = 0.8679\ \text{V}\]
The working voltage is a little below the standard value, because the concentrations used are not the standard \(1\ \text{M}\).
The standard potential of the electrode, \(\mathrm{Zn^{2\oplus}}\text{(0.02 M)} \,|\, \mathrm{Zn}\text{(s)}\) is \(-0.76\ \text{V}\). Calculate its potential.
Electrode reaction:
\[\mathrm{Zn^{2\oplus}}\text{(0.02M)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Zn}\text{(s)}\]
\[E_{Zn} = E^0_{Zn} - \frac{0.0592\ \text{V}}{n}\log_{10}\frac{1}{[\mathrm{Zn^{2\oplus}}]}\]
\[= -0.76\ \text{V} + \frac{0.0592\ \text{V}}{2}\log_{10}(0.02)\]
\[= -0.76\ \text{V} + \frac{0.0592\ \text{V}}{2} \times (-1.6990)\]
\[= -0.76\ \text{V} - 0.0503\ \text{V} = -0.81\ \text{V}\]
Since \(0.02\) is less than 1, its logarithm is negative, and the correction makes the electrode potential slightly more negative than the standard value.
The Nernst equation is not confined to problem sets — it is the working principle of an entire family of measuring instruments. A pH meter contains an electrode whose potential varies with hydrogen ion concentration exactly as the equation predicts, so a voltage reading can be displayed directly as pH. The same idea, with electrodes made selective for one particular ion, gives the ion-selective electrodes used in hospital analysers to report sodium and potassium levels in a few drops of blood, and in environmental laboratories to measure fluoride or nitrate in water samples. In each case the instrument measures a voltage and the Nernst equation converts it into a concentration.
For the cell \(\mathrm{Zn}\text{(s)} \,|\, \mathrm{Zn^{2\oplus}}\text{(0.1M)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(0.01M)} \,|\, \mathrm{Cu}\text{(s)}\) at \(25\ ^\circ\text{C}\), given \(E^0_{Zn} = -0.76\ \text{V}\) and \(E^0_{Cu} = 0.34\ \text{V}\): (i) write the overall cell reaction, (ii) state the value of \(n\), (iii) calculate \(E^0_{cell}\), and (iv) calculate \(E_{cell}\).
5.8 Thermodynamics of Galvanic Cells
A galvanic cell is a redox reaction with wires attached. Since every chemical reaction obeys the laws of thermodynamics, the voltage a cell delivers must be connected to the thermodynamic quantities you met in an earlier chapter — and it is. This section builds that bridge.
5.8.1 Gibbs Energy of Cell Reactions and Cell Potential
The electrical work done in a galvanic cell is the electricity (charge) passed multiplied by the cell potential.
\[\text{Electrical work} = \text{amount of charge passed} \times \text{cell potential}\]
This is the ordinary definition of electrical work: pushing a larger charge, or pushing it across a larger potential difference, means doing more work.
The charge of one mole of electrons is \(F\) coulombs. For a cell reaction involving \(n\) moles of electrons,
\[\text{charge passed} = nF \text{ coulombs}\]
Hence,
\[\text{electrical work} = nFE_{cell}\]
W. Gibbs in 1878 concluded that the electrical work done in a galvanic cell is equal to the decrease in Gibbs energy, \(-\Delta G\), of the cell reaction. It then follows that
\[\text{Electrical work} = -\Delta G\]
and thus
\[-\Delta G = nFE_{cell}\]
or
\[\Delta G = -nFE_{cell} \qquad \text{...(5.27)}\]
Under standard state conditions, we write
\[\Delta G^0 = -nFE^0_{cell} \qquad \text{...(5.28)}\]
The Gibbs energy change of a cell reaction is related to the cell potential by \(\Delta G = -nFE_{cell}\), where \(n\) is the number of moles of electrons transferred in the balanced cell reaction and \(F\) is the Faraday constant; under standard state conditions this becomes \(\Delta G^0 = -nFE^0_{cell}\).
Gibbs energy change \(\Delta G\) of a cell reaction (SI unit \(\text{J mol}^{-1}\), commonly reported in \(\text{kJ}\); \(E_{cell}\) in volts and \(F\) in \(\text{C mol}^{-1}\) give \(\Delta G\) directly in joules, since \(1\ \text{V} \times 1\ \text{C} = 1\ \text{J}\). Sign convention: \(\Delta G\) negative means the reaction is spontaneous — and because of the minus sign in Eq. (5.27), that corresponds to \(E_{cell}\) being positive.)
Josiah Willard Gibbs was a professor at Yale who published his monumental work on chemical thermodynamics between 1876 and 1878 in the proceedings of a small local scientific society, in a mathematical style so dense that almost nobody in America read it. It was European scientists — Maxwell in Britain among the first — who recognised what had been written and carried the ideas back into mainstream chemistry. The quantity that carries his name today is the one that decides, for every reaction in this book, whether it will go.
Spontaneity, Read Off a Voltmeter
For a chemical reaction to be spontaneous, \(\Delta G\) must be negative. Because \(\Delta G = -nFE_{cell}\), \(E_{cell}\) must be positive for a cell reaction if it is spontaneous.
This is a remarkably convenient result. Deciding spontaneity normally calls for enthalpy and entropy data; for a redox reaction that can be set up as a cell, the same decision can be made by looking at whether the voltmeter reading is positive.
Why \(E^0_{cell}\) Is an Intensive Property
The Eq. (5.28) explains why \(E^0_{cell}\) is an intensive property — one that does not depend on how much material is taken.
We know that \(\Delta G^0\) is an extensive property, since its value depends on the amount of substance. If the stoichiometric equation of the redox reaction is multiplied by 2 — that is, the amounts of substances oxidised and reduced are doubled — \(\Delta G^0\) doubles. The moles of electrons transferred also doubles.
The ratio,
\[E^0_{cell} = -\frac{\Delta G^0}{nF}\]
then becomes
\[E^0_{cell} = -\frac{2\Delta G^0}{2\,nF} = -\frac{\Delta G^0}{nF}\]
Thus, \(E^0_{cell}\) remains the same on multiplying the redox reaction by 2. It means \(E^0_{cell}\) is independent of the amount of substance and is an intensive property.
Both the numerator and the denominator were doubled, so the ratio was untouched — which is exactly why this argument works.
For the reaction \(\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)}\), \(E^0_{cell} = 1.10\ \text{V}\). State, with a reason in each case, what happens to (i) \(E^0_{cell}\), (ii) \(\Delta G^0\) and (iii) \(n\) when the whole equation is multiplied by 3.
5.8.2 Standard Cell Potential and Equilibrium Constant
A cell does not go on working for ever. As it discharges, reactants are used up and products accumulate until the cell reaction reaches equilibrium and the voltage falls to zero. It should therefore be possible to connect the standard cell potential with the equilibrium constant of the cell reaction — and Gibbs energy is the link between them.
The relation between the standard Gibbs energy change of a cell reaction and the standard cell potential is given by Eq. (5.28).
\[-\Delta G^0 = nFE^0_{cell}\]
The relation between the standard Gibbs energy change of a chemical reaction and its equilibrium constant, as given in thermodynamics, is:
\[\Delta G^0 = -RT\ln K \qquad \text{...(5.29)}\]
Combining Eq. (5.28) and Eq. (5.29), we have
\[-nFE^0_{cell} = -RT\ln K\]
or
\[E^0_{cell} = \frac{RT}{nF}\ln K\]
\[= \frac{2.303\,RT}{nF}\log_{10}K\]
\[= \frac{0.0592}{n}\log_{10}K \qquad \text{at } 25\ ^\circ\text{C} \qquad \text{...(5.30)}\]
The last step uses the same numerical value, \(2.303RT/F = 0.0592\ \text{V}\) at \(25\ ^\circ\text{C}\), that appeared in the Nernst equation in Section 5.7.2.
The standard cell potential is related to the equilibrium constant of the cell reaction by \(E^0_{cell} = \dfrac{2.303\,RT}{nF}\log_{10}K\), which at \(25\ ^\circ\text{C}\) becomes \(E^0_{cell} = \dfrac{0.0592}{n}\log_{10}K\).
Signs of \(E^0_{cell}\), \(\Delta G^0\) and the value of \(K\)
| Positive |
Negative |
Greater than 1 |
Spontaneous; the cell delivers current |
| Zero |
Zero |
Equal to 1 |
At equilibrium; no current flows |
| Negative |
Positive |
Less than 1 |
Non-spontaneous; the reverse reaction is spontaneous |
Write expressions to calculate the equilibrium constant from (i) concentration data, (ii) thermochemical data, (iii) electrochemical data.
Calculate the standard Gibbs energy change and the equilibrium constant at \(25\ ^\circ\text{C}\) for the cell reaction,
\[\mathrm{Cd}\text{(s)} + \mathrm{Sn^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Cd^{2\oplus}}\text{(aq)} + \mathrm{Sn}\text{(s)}\]
Given: \(E^0_{Cd} = -0.403\ \text{V}\) and \(E^0_{Sn} = -0.136\ \text{V}\). Write the formula of the cell.
- The cell is made of two electrodes, \(\mathrm{Cd^{2\oplus}}\text{(aq)} \,|\, \mathrm{Cd}\text{(s)}\) and \(\mathrm{Sn^{2\oplus}}\text{(aq)} \,|\, \mathrm{Sn}\text{(s)}\). The \(E^0\) value for the \(\mathrm{Sn^{2\oplus}}\text{(aq)} \,|\, \mathrm{Sn}\text{(s)}\) electrode is higher than that of the \(\mathrm{Cd^{2\oplus}}\text{(aq)} \,|\, \mathrm{Cd}\text{(s)}\) electrode. Hence, the \(\mathrm{Sn^{2\oplus}}\text{(aq)} \,|\, \mathrm{Sn}\text{(s)}\) electrode is the cathode and \(\mathrm{Cd^{2\oplus}}\text{(aq)} \,|\, \mathrm{Cd}\text{(s)}\) the anode. The cell formula is
\[\mathrm{Cd}\text{(s)} \,|\, \mathrm{Cd^{2\oplus}}\text{(aq)} \,\|\, \mathrm{Sn^{2\oplus}}\text{(aq)} \,|\, \mathrm{Sn}\text{(s)}\]
- \(\Delta G^0 = -nF\,E^0_{cell}\)
\[E^0_{cell} = E^0_{Sn} - E^0_{Cd} = -0.136\ \text{V} - (-0.403\ \text{V}) = 0.267\ \text{V}\]
\[n = 2\ \text{mol e}^{\ominus}\]
\[\Delta G^0 = -2\ \text{mol e}^{\ominus} \times 96500\ \text{C/mol e}^{\ominus} \times 0.267\ \text{V}\]
\[= -51531\ \text{V C} = -51531\ \text{J} = -51.53\ \text{kJ}\]
The value \(n = 2\) was read from the cell reaction: cadmium goes to \(\mathrm{Cd^{2\oplus}}\), so two electrons are transferred. The unit conversion at the end uses \(1\ \text{V C} = 1\ \text{J}\).
- \(E^0_{cell} = \dfrac{0.0592\ \text{V}}{2}\log_{10}K\)
\[0.267\ \text{V} = \frac{0.0592\ \text{V}}{2}\log_{10}K\]
\[\log_{10}K = \frac{0.267 \times 2}{0.0592} = 9.0203\]
\[K = \text{antilog } 9.0203 = 1.05\times10^{9}\]
A modest-looking voltage of about a quarter of a volt corresponds to an equilibrium constant of a thousand million — a reminder of how sensitively \(K\) responds to \(E^0_{cell}\), since the potential appears in the exponent.
For the Daniell cell, \(\mathrm{Zn}\text{(s)} \,|\, \mathrm{Zn^{2\oplus}}\text{(1M)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(1M)} \,|\, \mathrm{Cu}\text{(s)}\), the standard cell potential is \(1.10\ \text{V}\) at \(25\ ^\circ\text{C}\). Calculate the standard Gibbs energy change and the equilibrium constant for the cell reaction.
The cell reaction is
\[\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)}\]
so two electrons are transferred and \(n = 2\).
Step 1 — standard Gibbs energy change, from Eq. (5.28):
\[\Delta G^0 = -nFE^0_{cell} = -2 \times 96500\ \text{C mol}^{-1} \times 1.10\ \text{V}\]
\[= -212300\ \text{J mol}^{-1} = -212.3\ \text{kJ mol}^{-1}\]
The large negative value tells us at once that the reaction is strongly spontaneous.
Step 2 — equilibrium constant, from Eq. (5.30):
\[\log_{10}K = \frac{n\,E^0_{cell}}{0.0592} = \frac{2 \times 1.10}{0.0592} = 37.16\]
\[K = \text{antilog } 37.16 = 1.45\times10^{37}\]
An equilibrium constant of this size means the reaction goes essentially to completion: at equilibrium, practically no \(\mathrm{Cu^{2\oplus}}\) would be left in the solution.
Two very different uses grow out of Eq. (5.28) and Eq. (5.30). The first belongs to battery design: since \(\Delta G^0 = -nFE^0_{cell}\), the energy a cell can supply per mole of reaction is set by its voltage, so designers hunt for electrode pairs whose standard potentials lie as far apart as possible. That is why a lithium-based cell, built around a metal with a strongly negative standard potential, delivers around \(3.6\ \text{V}\) where a zinc-based dry cell manages about \(1.5\ \text{V}\), and why the lithium cell stores far more energy for the same mass. The second use belongs to the laboratory: an equilibrium constant of \(10^{9}\), let alone \(10^{37}\), could never be found by measuring concentrations at equilibrium, because the leftover reactant would be far too dilute to detect. A voltmeter reading of a fraction of a volt gives it in one measurement.
For a certain cell reaction involving the transfer of 2 moles of electrons, the equilibrium constant at \(25\ ^\circ\text{C}\) is \(4.0\times10^{10}\). Calculate (i) \(E^0_{cell}\) and (ii) \(\Delta G^0\) for the reaction, and state whether the reaction as written is spontaneous.
5.9 Reference Electrodes
Every oxidation needs to be accompanied by reduction. The occurrence of only oxidation or only reduction is not possible, as you learnt in the Std. XI Chemistry Textbook, Chapter 6.
In a galvanic cell, oxidation and reduction occur simultaneously. The potential associated with the redox reaction can be experimentally measured. For the measurement of potential, two electrodes need to be combined together, so that the redox reaction occurs.
This is a genuine difficulty, not a technicality. A voltmeter has two terminals and reports a difference; there is no way to connect it to one electrode alone. Whatever number we obtain always belongs to a pair.
A Way Around the Difficulty
What would happen if the potential of one of the electrodes in a galvanic cell were zero? Can we measure the potential of such a galvanic cell?
There are two electrodes combined together and a redox reaction results. The measured cell potential is the algebraic sum of two electrode potentials. One electrode potential is zero. Therefore, the measured cell potential is equal to the potential of the other electrode.
From the foregoing arguments, it follows that it is necessary to choose an arbitrary standard electrode as a reference point.
The chemists have chosen the hydrogen gas electrode, consisting of \(\mathrm{H_2}\) gas at \(1\ \text{atm}\) pressure in contact with \(1\ \text{M}\) \(\mathrm{H^{\oplus}}\) ion solution, as a primary reference electrode. The potential of this electrode has arbitrarily been taken as zero. The electrode is called the standard hydrogen electrode (SHE).
Note the word arbitrarily. Nobody has proved that this electrode’s potential is truly zero; it has been agreed upon as the zero of the scale, in the same way that sea level is agreed upon as the zero for measuring the height of a mountain. Every standard potential in Table 5.1 is therefore a height measured from this one agreed sea level.
Primary and Secondary Reference Electrodes
We will see that the SHE is not the most convenient electrode to work with. Several other electrodes, namely calomel, silver–silver chloride and glass electrodes, with known potentials, are used as secondary reference electrodes. The potentials of these electrodes are determined using the SHE.
A secondary reference electrode is therefore not a rival zero; it is an electrode whose potential has already been measured against the SHE once and for all, so that everyday measurements can be made against it instead.
A reference electrode is an electrode whose potential is arbitrarily taken as zero or is exactly known.
Primary vs Secondary reference electrode
| Example |
Standard hydrogen electrode (SHE) |
Calomel, silver–silver chloride, glass electrodes |
| Potential |
Arbitrarily taken as zero |
Exactly known, having been measured against SHE |
| Ease of use |
Difficult to set up and maintain |
Compact, robust and convenient |
| Role |
Defines the zero of the whole potential scale |
Used for routine day-to-day measurements |
5.9.1 Standard Hydrogen Electrode (SHE)
Construction
The SHE consists of a platinum plate, coated with platinum black, used as the electrode. This plate is connected to the external circuit through a sealed narrow glass tube containing mercury. It is surrounded by an outer jacket.
The platinum electrode is immersed in \(1\ \text{M}\) \(\mathrm{H^{\oplus}}\) ion solution. The solution is kept saturated with dissolved \(\mathrm{H_2}\) by bubbling hydrogen gas under \(1\ \text{atm}\) pressure through the side tube of the jacket, as shown in Fig. 5.6.
Platinum does not take part in the electrode reaction. It is an inert electrode and serves as the site for electron transfer. The three requirements of the standard state — \(1\ \text{M}\) acid, \(1\ \text{atm}\) hydrogen and \(25\ ^\circ\text{C}\) — must all be met, because the electrode’s potential is only zero under exactly these conditions.
Electrode Reaction
The platinum black, capable of adsorbing large quantities of \(\mathrm{H_2}\) gas, allows the change from gaseous to ionic form and the reverse process to occur. The rough black surface holds a supply of hydrogen right at the point where the reaction happens, so the electrode responds quickly and reproducibly.
The reduction half reaction at the electrode is
\[2\,\mathrm{H^{\oplus}}\text{(1M)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}\text{(g, 1atm)}, \qquad E^0_{H_2} = 0.000\ \text{V}\]
Application of SHE
The SHE is used as a primary reference electrode to determine the standard potentials of other electrodes. The procedure is always the same, and is set out in FC 5.8.
To determine the standard potential of \(\mathrm{Zn^{2\oplus}}\text{(1M)} \,|\, \mathrm{Zn}\text{(s)}\), it is combined with the SHE to form the cell,
\[\mathrm{Zn} \,|\, \mathrm{Zn^{2\oplus}}\text{(1M)} \,\|\, \mathrm{H^{\oplus}}\text{(1M)} \,|\, \mathrm{H_2}\text{(g, 1atm)} \,|\, \mathrm{Pt}\]
This is shown in Fig. 5.7.
The standard cell potential, \(E^0_{cell}\), is measured.
\[E^0_{cell} = E^0_{H_2} - E^0_{Zn} = -E^0_{Zn}, \qquad \text{because } E^0_{H_2} \text{ is zero}\]
Thus, the measured emf of the cell is equal to the standard potential of the \(\mathrm{Zn^{2\oplus}}\text{(1M)} \,|\, \mathrm{Zn}\text{(s)}\) electrode.
Q4. Chemists say it is impossible to measure the potential of a single electrode. If that is so, how are the electrode potential values printed in every textbook table obtained?
A. By measuring pairs and fixing one member of the pair by agreement. A voltmeter can only report a difference between two electrodes, so no single electrode potential can ever be measured on its own. Chemists therefore chose one electrode — the standard hydrogen electrode — and defined its potential as exactly zero. Any other electrode is then coupled with the SHE, and since one of the two contributions is zero by definition, the whole measured cell potential belongs to the other electrode. Every value in Table 5.1 was obtained in this way, which is why they are all relative to hydrogen rather than absolute.
A cell is set up by combining a copper electrode dipping in \(1\ \text{M}\) \(\mathrm{Cu^{2\oplus}}\) solution with a standard hydrogen electrode at \(25\ ^\circ\text{C}\). The copper electrode is found to be the positive electrode, and the measured standard cell potential is \(0.34\ \text{V}\). Determine the standard potential of the \(\mathrm{Cu^{2\oplus}}\,|\,\mathrm{Cu}\) electrode and write the cell notation.
Step 1 — assign the electrodes. The copper electrode is positive, so it is the cathode; the SHE is therefore the anode.
Step 2 — apply Eq. (5.24):
\[E^0_{cell} = E^0\text{(cathode)} - E^0\text{(anode)} = E^0_{Cu} - E^0_{H_2}\]
Step 3 — substitute, remembering that \(E^0_{H_2} = 0.000\ \text{V}\) by definition:
\[0.34\ \text{V} = E^0_{Cu} - 0 \qquad \text{so} \qquad E^0_{Cu} = +0.34\ \text{V}\]
Step 4 — write the notation with the anode on the left:
\[\mathrm{Pt} \,|\, \mathrm{H_2}\text{(g, 1atm)} \,|\, \mathrm{H^{\oplus}}\text{(1M)} \,\|\, \mathrm{Cu^{2\oplus}}\text{(1M)} \,|\, \mathrm{Cu}\text{(s)}\]
The positive sign of \(E^0_{Cu}\) carries a chemical meaning: \(\mathrm{Cu^{2\oplus}}\) is easier to reduce than \(\mathrm{H^{\oplus}}\), which is why copper took the electrons and acted as the cathode. Compare this with zinc, whose \(E^0\) came out negative for the opposite reason.
A nickel electrode in \(1\ \text{M}\) \(\mathrm{Ni^{2\oplus}}\) solution is combined with a standard hydrogen electrode at \(25\ ^\circ\text{C}\). The nickel electrode is found to be the negative electrode and the cell potential is \(0.25\ \text{V}\). Write the cell notation, state which electrode is the cathode, and calculate the standard potential of the \(\mathrm{Ni^{2\oplus}}\,|\,\mathrm{Ni}\) electrode.
Difficulties in Setting Up SHE
For all its importance, the SHE is an awkward electrode to work with in practice.
It is difficult to obtain pure and dry hydrogen gas. Traces of impurity settle on the platinum black surface and stop it working properly.
The pressure of hydrogen gas cannot be maintained exactly at \(1\ \text{atm}\) throughout the measurement.
The concentration of \(\mathrm{H^{\oplus}}\) ion solution cannot be exactly maintained at \(1\ \text{M}\). Due to the bubbling of gas into the solution, evaporation of water may take place. This results in changing the concentration of the solution.
Each of these three difficulties attacks one of the three standard conditions the electrode depends on, and if any of them drifts, the electrode’s potential is no longer zero. This is precisely why the secondary reference electrodes listed earlier exist.
No school or industrial laboratory keeps a standard hydrogen electrode on the bench. Hydrogen is flammable and forms explosive mixtures with air, a cylinder of the pure dry gas is expensive, and the whole assembly must be watched constantly to hold its pressure and concentration steady. What laboratories use instead is a sealed, pocket-sized secondary reference — most often a silver–silver chloride or a calomel electrode — sitting inside the same plastic body as the sensing electrode of a pH meter or a water-testing probe. Its potential was established against the SHE once, in a standards laboratory, and it then needs nothing but occasional rinsing and storage in its filling solution. The SHE defines the scale; the secondary electrodes do the daily work.
Hydrogen Gas Electrode
The SHE is only one particular case — the case in which the acid is exactly \(1\ \text{M}\) and the gas exactly \(1\ \text{atm}\). What if they are not?
For the hydrogen gas electrode, \(\mathrm{H^{\oplus}}\text{(aq)} \,|\, \mathrm{H_2}(\text{g}, P_{H_2}) \,|\, \mathrm{Pt}\), the value of \([\mathrm{H^{\oplus}}]\) and the pressure of hydrogen gas differ from unity.
Electrode reaction:
\[2\,\mathrm{H^{\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}(\text{g},\ P_{H_2})\]
From the Nernst equation,
\[E_{H_2} = E^0_{H_2} - \frac{0.0592}{2}\log_{10}\frac{P_{H_2}}{[\mathrm{H^{\oplus}}]^2}\]
\[= -\frac{0.0592}{2}\log_{10}\frac{P_{H_2}}{[\mathrm{H^{\oplus}}]^2}\]
because \(E^0_{H_2} = 0\).
Two points from Section 5.7.2 are being used here. The gas appears as its partial pressure rather than a concentration, and \([\mathrm{H^{\oplus}}]\) is squared because two moles of \(\mathrm{H^{\oplus}}\) appear in the balanced half reaction. Since the standard term vanishes, the potential of a hydrogen gas electrode is nothing but the Nernst correction term itself — which is exactly what makes such an electrode a sensitive detector of hydrogen ion concentration.
A hydrogen gas electrode operates at \(25\ ^\circ\text{C}\) with hydrogen gas at \(1\ \text{atm}\) pressure and a solution in which \([\mathrm{H^{\oplus}}] = 0.01\ \text{M}\). Write the electrode reaction and calculate the potential of the electrode.
5.10 Galvanic Cells Useful in Day-to-Day Life
Everything in the last four sections has been building towards the cells you actually handle — the one in a torch, the one under a car bonnet, the one in a wristwatch. Voltaic (or galvanic) cells in common use can be classified as primary and secondary cells.
i. Primary voltaic cells: When a galvanic cell discharges during current generation, the chemicals are consumed. In a primary voltaic cell, once the chemicals are completely consumed, the cell reaction stops. The cell reaction cannot be reversed even after reversing the direction of current flow; that is, these cells cannot be recharged. The most familiar example is the dry cell.
ii. Secondary voltaic cells: In a secondary voltaic cell, the chemicals consumed during current generation can be regenerated. For this purpose an external potential slightly greater than the cell potential is applied across the electrodes. This results in reversal of the direction of current flow, causing the reversal of the cell reaction. This is recharging of the cell. The voltaic cells which can be recharged are called secondary voltaic cells.
Notice why the applied potential must be greater than the cell’s own potential: the cell is pushing back, and the external source has to overcome that push before it can drive the reaction the wrong way round.
It is remarkable to see that secondary cells are galvanic cells during discharge and electrolytic cells during recharging. The same box of chemicals runs in both of the modes described in Section 5.4.3, depending on which way the current is flowing. Examples of secondary cells are the lead storage battery, the mercury cell and the nickel-cadmium cell.
A primary cell is one whose reaction cannot be reversed, so that it is discarded once its chemicals are consumed; a secondary cell is one whose reaction can be reversed by applying an external potential slightly greater than the cell potential, so that it can be recharged and used again.
Primary vs Secondary voltaic cells
| Can it be recharged? |
No |
Yes |
| Fate of the chemicals |
Consumed permanently |
Regenerated on recharging |
| Behaviour while supplying current |
Galvanic cell |
Galvanic cell |
| Behaviour while being charged |
— |
Electrolytic cell |
| Examples |
Dry cell (Leclanché cell) |
Lead storage battery, NICAD cell, mercury cell |
5.10.1 Dry Cell (Leclanché Cell)
It is a cell without a liquid component, but the electrolyte is not completely dry. It is a viscous aqueous paste. The name is therefore only half accurate: what makes the cell “dry” is that nothing can spill out of it, not that water is absent.
Construction
The container of the cell is made of zinc, which serves as the anode (−). It is lined from inside with a porous paper to separate it from the other material of the cell. The porous lining lets ions through while keeping the zinc from touching the cathode mixture directly — the same job the salt bridge did in Section 5.6.1.
An inert graphite rod in the centre of the cell, immersed in the electrolyte paste, serves as the cathode. It is surrounded by a paste of manganese dioxide (\(\mathrm{MnO_2}\)) and carbon black. The graphite itself takes no part in the reaction; it only collects electrons.
The rest of the cell is filled with an electrolyte. It is a moist paste of ammonium chloride (\(\mathrm{NH_4Cl}\)) and zinc chloride (\(\mathrm{ZnCl_2}\)). Some starch is added to the paste to make it thick, so that it cannot leak out.
The cell is sealed at the top to prevent drying of the paste by evaporation of moisture. See Fig. 5.8.
The design is named after Georges Leclanché, a French engineer who devised it in the 1860s. His original version was a wet cell — a glass jar of ammonium chloride solution with a zinc rod and a porous pot packed with manganese dioxide — and it was adopted almost at once by telegraph companies, because unlike its rivals it could sit idle for long periods and still work when a signal came. The sealed paste version in your torch is a repackaging of Leclanché’s chemistry, unchanged in essentials for more than a century and a half.
Cell Reactions
i. Oxidation at anode: When the cell operates, the current is drawn from the cell and metallic zinc is oxidised to zinc ions. The container is literally consumed as the cell works.
\[\mathrm{Zn}\text{(s)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}}\]
ii. Reduction at cathode: The electrons liberated in the oxidation at the anode flow along the container and migrate to the cathode. At the cathode \(\mathrm{NH_4^{\oplus}}\) ions are reduced.
\[2\,\mathrm{NH_4^{\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{NH_3}\text{(aq)} + \mathrm{H_2}\text{(g)}\]
Hydrogen gas produced in the reduction reaction is oxidised by \(\mathrm{MnO_2}\), which prevents its collection on the cathode.
\[\mathrm{H_2}\text{(g)} + 2\,\mathrm{MnO_2}\text{(s)} \rightarrow \mathrm{Mn_2O_3}\text{(s)} + \mathrm{H_2O}\text{(l)}\]
The net reduction reaction at the cathode is the combination of these two reactions.
\[2\,\mathrm{NH_4^{\oplus}}\text{(aq)} + 2\,\mathrm{MnO_2}\text{(s)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Mn_2O_3}\text{(s)} + 2\,\mathrm{NH_3}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
manganese dioxide, \(\mathrm{MnO_2}\) (A solid oxidising agent packed around the cathode → needs only to be in contact with the graphite rod and the moist paste → its job is to oxidise the hydrogen gas produced at the cathode the instant it forms, converting it to water and preventing a layer of gas bubbles from insulating the electrode and stopping the cell. A reagent used in this role is called a depolariser.)
iii. Net cell reaction: The net cell reaction is the sum of the oxidation at the anode and the reduction at the cathode.
\[\mathrm{Zn}\text{(s)} + 2\,\mathrm{NH_4^{\oplus}}\text{(aq)} + 2\,\mathrm{MnO_2}\text{(s)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + \mathrm{Mn_2O_3}\text{(s)} + 2\,\mathrm{NH_3}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
The ammonia produced combines with \(\mathrm{Zn^{2\oplus}}\) to form a soluble compound containing a complex ion.
\[\mathrm{Zn^{2\oplus}}\text{(aq)} + 4\,\mathrm{NH_3}\text{(aq)} \rightarrow [\mathrm{Zn(NH_3)_4}]^{2\oplus}\text{(aq)}\]
This last step matters for a practical reason: it removes the ammonia, which would otherwise collect as a gas around the cathode and interrupt the cell just as hydrogen would.
Alkaline Dry Cell
The Leclanché dry cell works under acidic conditions, due to the presence of \(\mathrm{NH_4Cl}\). The difficulty with this dry cell is that the Zn anode corrodes due to its reaction with \(\mathrm{H^{\oplus}}\) ions coming from \(\mathrm{NH_4^{\oplus}}\) ions. Because the container is itself the anode, this corrosion eats away the cell even when it is not being used.
This results in shortening the life of the dry cell. To avoid this, a modified form of the dry cell called the alkaline dry cell has been proposed. In the alkaline dry cell, NaOH or KOH is used as the electrolyte in place of \(\mathrm{NH_4Cl}\).
The alkaline dry cell has a longer life than the acidic dry cell, since the Zn corrodes more slowly in the absence of the acidic ammonium ion.
Uses of Dry Cell
The dry cell is used as a source of power in flashlights, portable radios, tape recorders, clocks and so forth.
5.10.2 Lead Storage Battery (Lead Accumulator)
The lead accumulator stores electrical energy due to the regeneration of the original reactants during recharging. It functions as a galvanic cell and as an electrolytic cell as well — a secondary cell in the fullest sense.
Construction
A group of lead plates packed with spongy lead serves as the anode (−). Another group of lead plates bearing lead dioxide (\(\mathrm{PbO_2}\)) serves as the cathode (+).
The lead is described as “spongy” because a porous mass exposes far more surface to the acid than a smooth plate would, and the current a cell can deliver depends on how much surface is reacting. To provide a large reacting surface, the cell contains several plates of each type. The two types of plates are alternately arranged, as shown in Fig. 5.9.
The electrodes are immersed in an electrolytic aqueous solution of 38% (by mass) sulphuric acid of density \(1.2\ \text{g/mL}\).
Notation of the cell: The cell is formulated as
\[\mathrm{Pb}\text{(s)} \,|\, \mathrm{PbSO_4}\text{(s)} \,|\, 38\%\ \mathrm{H_2SO_4}\text{(aq)} \,|\, \mathrm{PbSO_4}\text{(s)} \,|\, \mathrm{PbO_2}\text{(s)} \,|\, \mathrm{Pb}\text{(s)}\]
The lead accumulator was built in 1859 by the French physicist Gaston Planté, who rolled two lead sheets separated by cloth into a spiral and immersed them in dilute sulphuric acid. It was the first rechargeable battery ever made, and it survives today because of an unusual virtue: it can deliver a very large current for a few seconds — exactly what is needed to turn a car’s starter motor — and then be refilled with energy by the vehicle’s own generator.
a. Cell Reactions During Discharge
i. Oxidation at anode (−): When the cell provides current, spongy lead is oxidised to \(\mathrm{Pb^{2\oplus}}\) ions and negative charge accumulates on the lead plates. The \(\mathrm{Pb^{2\oplus}}\) ions so formed combine with \(\mathrm{SO_4^{2\ominus}}\) ions from \(\mathrm{H_2SO_4}\) to form insoluble \(\mathrm{PbSO_4}\). The net oxidation is the sum of these two processes.
\[\mathrm{Pb}\text{(s)} \rightarrow \mathrm{Pb^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation)}\]
\[\mathrm{Pb^{2\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow \mathrm{PbSO_4}\text{(s)} \qquad \text{(precipitation)}\]
\[\mathrm{Pb}\text{(s)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow \mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{e^{\ominus}} \qquad \text{...(i) (overall oxidation)}\]
The insolubility of lead sulphate is the crucial feature. It does not wander off into the acid; it stays stuck to the plate that produced it, ready to be turned back into lead when the cell is recharged.
ii. Reduction at cathode (+): The electrons produced at the anode travel through the external circuit and re-enter the cell at the cathode. At the cathode \(\mathrm{PbO_2}\) is reduced to \(\mathrm{Pb^{2\oplus}}\) ions in the presence of \(\mathrm{H^{\oplus}}\) ions. Subsequently the \(\mathrm{Pb^{2\oplus}}\) ions so formed combine with \(\mathrm{SO_4^{2\ominus}}\) ions from \(\mathrm{H_2SO_4}\) to form insoluble \(\mathrm{PbSO_4}\) that gets coated on the electrode.
\[\mathrm{PbO_2}\text{(s)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pb^{2\oplus}}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \qquad \text{(reduction)}\]
\[\mathrm{Pb^{2\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow \mathrm{PbSO_4}\text{(s)} \qquad \text{(precipitation)}\]
\[\mathrm{PbO_2}\text{(s)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \qquad \text{...(ii) (overall reduction)}\]
So both plates, however different they look at the start, end up coated with the same white solid — lead sulphate.
iii. Net cell reaction during discharge: The net cell reaction is the sum of the overall oxidation at the anode and the overall reduction at the cathode.
\[\mathrm{Pb}\text{(s)} + \mathrm{PbO_2}\text{(s)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + 2\,\mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow 2\,\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)}\]
or
\[\mathrm{Pb}\text{(s)} + \mathrm{PbO_2}\text{(s)} + 2\,\mathrm{H_2SO_4}\text{(aq)} \rightarrow 2\,\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \qquad \text{...(iii)}\]
As the cell operates to generate current, \(\mathrm{H_2SO_4}\) is consumed. Its concentration (density) decreases and the cell potential is decreased. The cell potential thus depends on the sulphuric acid concentration (density).
That last sentence has a practical consequence you may have seen at a garage: the state of charge of a car battery can be judged by measuring the density of its acid with a hydrometer, without any electrical test at all. The link back to theory is the Nernst equation of Section 5.7.2 — as a reactant concentration falls, so does the cell potential.
b. Cell Reactions During Recharging
The potential of the lead accumulator is \(2\ \text{V}\). It must be recharged when the cell potential falls to \(1.8\ \text{V}\). To recharge the cell, an external potential slightly greater than \(2\ \text{V}\) needs to be applied across the electrodes.
During recharging the cell functions as an electrolytic cell. The anode and cathode are interchanged, with the \(\mathrm{PbO_2}\) electrode being the anode (+) and the lead electrode the cathode (−). The names swap because the reactions have swapped — oxidation now occurs at the \(\mathrm{PbO_2}\) plate — exactly as the definitions of Section 5.4.2 require.
iv. Oxidation at anode (+): It is the reverse of the reduction reaction (ii) at the cathode that occurs during discharge.
\[\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{PbO_2}\text{(s)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{...(iv)}\]
v. Reduction at cathode (−): It is the reverse of the oxidation reaction (i) at the anode that occurs during discharge.
\[\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pb}\text{(s)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \qquad \text{...(v)}\]
vi. Net cell reaction: It is the sum of reactions (iv) and (v), or the reverse of the net cell reaction (iii) that occurs during discharge.
\[2\,\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{Pb}\text{(s)} + \mathrm{PbO_2}\text{(s)} + 2\,\mathrm{H_2SO_4}\text{(aq)}\]
The above reaction shows that \(\mathrm{H_2SO_4}\) is regenerated. Its concentration (density) and, in turn, the cell potential increase.
The complete cycle is drawn in FC 5.9.
Q5. A car battery can be recharged thousands of times, but an ordinary dry cell cannot be recharged even once. What must be different about what happens at the electrodes in the two cases?
A. Where the products end up. In the lead accumulator both electrodes become coated with insoluble lead sulphate, which clings to the very plate that produced it; when an external potential greater than \(2\ \text{V}\) is applied, that solid is sitting exactly where it must be for the reaction to run backwards, and lead and lead dioxide are rebuilt on their own plates. In a dry cell nothing of the kind happens: the zinc container is eaten away and the ammonia produced dissolves off as a complex ion, so there is no product left in place to be converted back. The general rule, which the NICAD cell in the next subsection also illustrates, is that a cell can be recharged when its products are solids that adhere to the electrode surface.
Applications of Lead Accumulator
It is used as a source of direct current in the laboratory.
A \(12\ \text{V}\) lead storage battery, constructed by connecting six \(2\ \text{V}\) cells in series, is used in automobiles and inverters.
Note the word battery used strictly here: a battery is a set of cells joined together, and six two-volt cells in series give twelve volts because potentials in series add.
A single cell of a lead accumulator delivers a potential of \(2.0\ \text{V}\), and its discharge reaction transfers 2 moles of electrons as written in Eq. (iii). Calculate the maximum electrical work obtainable per mole of reaction and the corresponding value of \(\Delta G\). State what happens to this value when six such cells are joined in series to make a \(12\ \text{V}\) battery.
Step 1 — electrical work, from Section 5.8.1:
\[\text{Electrical work} = nFE_{cell} = 2 \times 96500\ \text{C mol}^{-1} \times 2.0\ \text{V} = 386000\ \text{J mol}^{-1} = 386\ \text{kJ mol}^{-1}\]
Step 2 — Gibbs energy change, from Eq. (5.27):
\[\Delta G = -nFE_{cell} = -386\ \text{kJ mol}^{-1}\]
The negative sign confirms that the discharge reaction is spontaneous, which is why the battery delivers current without any external help.
Step 3 — the series battery. Joining six cells in series raises the terminal voltage to \(12\ \text{V}\), but \(\Delta G\) per mole of reaction in each cell is unchanged at \(-386\ \text{kJ mol}^{-1}\), since cell potential is an intensive property. What the six cells give is six times as much reaction proceeding at any moment, and hence six times the total energy available.
Using the discharge reactions of the lead accumulator, answer the following: (i) name the substance formed at both electrodes during discharge; (ii) state whether the density of the electrolyte increases or decreases during recharging, and why; (iii) identify which electrode acts as the anode during discharge and which acts as the anode during recharging.
5.10.3 Nickel-Cadmium or NICAD Storage Cell
The nickel-cadmium cell is a secondary dry cell. In other words, it is a dry cell that can be recharged — it combines the convenience of a sealed paste cell with the reusability of an accumulator.
The anode of the NICAD storage cell is cadmium metal. The cathode is nickel(IV) oxide, \(\mathrm{NiO_2}\), supported on Ni. The electrolyte solution is basic.
The electrode reactions and overall cell reaction are as follows:
\[\mathrm{Cd}\text{(s)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \rightarrow \mathrm{Cd(OH)_2}\text{(s)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(anodic oxidation)}\]
\[\mathrm{NiO_2}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Ni(OH)_2}\text{(s)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \qquad \text{(cathodic reduction)}\]
\[\mathrm{Cd}\text{(s)} + \mathrm{NiO_2}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{Cd(OH)_2}\text{(s)} + \mathrm{Ni(OH)_2}\text{(s)} \qquad \text{(overall cell reaction)}\]
The reaction product at each electrode is a solid that adheres to the electrode surface. Therefore the cell can be recharged — the same principle that makes the lead accumulator rechargeable, applied to a different pair of metals.
The potential of the cell is about \(1.4\ \text{V}\). The cell has a longer life than other dry cells. It can be used in electronic watches, calculators, photographic equipment, and similar devices.
5.10.4 Mercury Battery
The mercury battery is a secondary dry cell and can be recharged. The mercury battery consists of a zinc anode, amalgamated with mercury. The cathode is a paste of Hg and carbon. The electrolyte is strongly alkaline and made of a paste of KOH and ZnO.
The electrode reactions and net cell reaction are:
\[\mathrm{Zn(Hg)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \rightarrow \mathrm{ZnO}\text{(s)} + \mathrm{H_2O}\text{(l)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(anode oxidation)}\]
\[\mathrm{HgO}\text{(s)} + \mathrm{H_2O}\text{(l)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Hg}\text{(l)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)} \qquad \text{(cathode reduction)}\]
\[\mathrm{Zn(Hg)} + \mathrm{HgO}\text{(s)} \rightarrow \mathrm{ZnO}\text{(s)} + \mathrm{Hg}\text{(l)} \qquad \text{(overall cell reaction)}\]
The overall reaction involves only solid substances. There is no change in electrolyte composition during operation.
This last point explains the battery’s great advantage. Because the hydroxide ions consumed at one electrode are produced again at the other, the electrolyte never becomes more dilute, so the Nernst correction term stays put and the voltage does not drift downwards as the cell is used. The mercury battery, therefore, provides a more constant voltage (\(1.35\ \text{V}\)) than the Leclanché dry cell. It also has a considerably higher capacity and longer life than the dry cell.
The mercury dry cell finds use in hearing aids, electric watches, pacemakers, and similar equipment, where a steady voltage over a long period matters more than the cost of the cell.
The four cells of this section
| Type |
Primary |
Secondary |
Secondary |
Secondary |
| Anode |
Zn container |
Spongy Pb plates |
Cd metal |
Zn amalgamated with Hg |
| Cathode |
Graphite rod with \(\mathrm{MnO_2}\) + carbon |
Pb plates bearing \(\mathrm{PbO_2}\) |
\(\mathrm{NiO_2}\) supported on Ni |
Paste of Hg and carbon |
| Electrolyte |
Moist paste of \(\mathrm{NH_4Cl}\) and \(\mathrm{ZnCl_2}\) |
38% \(\mathrm{H_2SO_4}\), density \(1.2\ \text{g/mL}\) |
Basic solution |
Alkaline paste of KOH and ZnO |
| Potential |
About \(1.5\ \text{V}\) in use |
\(2\ \text{V}\) per cell |
About \(1.4\ \text{V}\) |
\(1.35\ \text{V}\) |
| Typical uses |
Flashlights, radios, clocks |
Automobiles, inverters, laboratory DC supply |
Watches, calculators, photographic equipment |
Hearing aids, watches, pacemakers |
Every cell in the table above ends its life as a disposal problem, and the chemistry tells you why. Lead is a cumulative poison that damages the nervous system, particularly in children; cadmium is toxic to the kidneys; mercury is toxic in almost any form. This is why button cells and used batteries must never be thrown into household waste or burnt, and why many countries have restricted mercury cells to the few medical uses where nothing else will do. The encouraging side is that the lead accumulator is among the most successfully recycled manufactured objects in the world: the plates and the acid are both recovered and returned to new batteries, and a scrapped car battery is worth money precisely because of that. When you next replace a battery, the greenest step available is simply to hand the old one back to the dealer.
Compare the dry cell and the NICAD cell by answering the following: (i) write the anode reaction in each; (ii) explain, in terms of where the products go, why one can be recharged and the other cannot; (iii) state which of the two would be the better choice for a device that is used daily for several years, and give one reason.
5.11 Fuel Cells
Burning hydrogen in air releases a great deal of energy, but the energy comes out as heat. The functioning of fuel cells is based on the fact that combustion reactions are of redox type and can be used to generate electricity.
That is a striking idea. If a combustion reaction is really an electron transfer, then separating its two halves onto two electrodes — exactly as we did for the zinc–copper reaction in Section 5.4.1 — should let the electrons travel through a wire instead of releasing their energy as a flame.
Fuel cells differ from ordinary galvanic cells in that the reactants are not placed within the cell. They are continuously supplied to the electrodes from a reservoir. A dry cell carries its chemicals inside it and dies when they are used up; a fuel cell is more like an engine, working for as long as it is fed.
In these cells one of the reactants is a fuel, such as hydrogen gas or methanol. The other reactant, such as oxygen, is the oxidant.
A fuel cell is a galvanic cell in which the reactants — a fuel and an oxidant — are supplied continuously to the electrodes from outside, so that the cell converts the energy of a combustion reaction directly into electrical energy.
The simplest fuel cell is the hydrogen–oxygen fuel cell.
The idea is far older than the technology. In 1839 the Welsh judge and scientist William Grove, having watched electrolysis split water into hydrogen and oxygen, asked whether the process could be run backwards. He arranged platinum electrodes in tubes of hydrogen and oxygen standing in dilute acid, joined them, and obtained a current. He called it a gas voltaic battery. It produced too little power to be of any practical use in his lifetime, and more than a century passed before the same principle, much refined, was used to power spacecraft.
5.11.1 Hydrogen–Oxygen Fuel Cell
In the \(\mathrm{H_2}\)–\(\mathrm{O_2}\) fuel cell, the fuel is hydrogen gas. Oxygen gas is the oxidising agent. The energy of the combustion of hydrogen is converted into electrical energy.
Construction
The anode and cathode are porous carbon rods containing a small amount of finely divided platinum metal that acts as a catalyst. Two features of that sentence matter. The rods are porous so that gas can reach the electrolyte through the whole thickness of the electrode rather than only at its outer surface, and the platinum is finely divided so that a tiny quantity of an expensive metal presents a large catalytic area.
The electrolyte is a hot aqueous solution of KOH. The carbon rods immersed in the electrolyte are shown in Fig. 5.10.
Hydrogen gas is continuously bubbled through the anode, and oxygen gas through the cathode, into the electrolyte.
Cell Reactions
i. Oxidation at anode (−): At the anode hydrogen gas is oxidised to \(\mathrm{H_2O}\).
\[2\,\mathrm{H_2}\text{(g)} + 4\,\mathrm{OH^{\ominus}}\text{(aq)} \rightarrow 4\,\mathrm{H_2O}\text{(l)} + 4\,\mathrm{e^{\ominus}}\]
ii. Reduction at cathode (+): The electrons released at the anode travel through the external circuit to the cathode. Here \(\mathrm{O_2}\) is reduced to \(\mathrm{OH^{\ominus}}\).
\[\mathrm{O_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)} + 4\,\mathrm{e^{\ominus}} \rightarrow 4\,\mathrm{OH^{\ominus}}\text{(aq)}\]
Look at what the hydroxide ions are doing across these two equations. Four are consumed at the anode and four are produced at the cathode, so the electrolyte is not used up at all; the \(\mathrm{OH^{\ominus}}\) ions simply carry charge through the solution from one electrode to the other.
iii. Net cell reaction: The overall cell reaction is the sum of the electrode reactions (i) and (ii).
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}\]
The overall cell reaction is the combustion of \(\mathrm{H_2}\) to form liquid water. Interestingly, the fuel \(\mathrm{H_2}\) gas and the oxidant \(\mathrm{O_2}\) do not react directly — they never meet, being kept at separate electrodes, and their reaction is completed only through the external circuit.
The chemical energy released during the formation of the O–H bond is directly converted into electrical energy accompanying the above combustion reaction. The cell continues to operate as long as \(\mathrm{H_2}\) and \(\mathrm{O_2}\) gases are supplied to the electrodes.
The operation of the cell is summarised in FC 5.10.
Cell Potential
The cell potential is given by
\[E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.4\ \text{V} - (-0.83\ \text{V}) = 1.23\ \text{V}\]
This is Eq. (5.24) of Section 5.7.1 applied to the two half reactions above, and the positive value confirms that the cell reaction is spontaneous.
For the hydrogen–oxygen fuel cell, \(E^0_{cell} = 1.23\ \text{V}\) at \(25\ ^\circ\text{C}\). Using the net cell reaction, determine the number of moles of electrons transferred and calculate the standard Gibbs energy change for the reaction as written.
Step 1 — find \(n\). In the net reaction
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}\]
the half reactions show that 4 moles of electrons pass from the anode to the cathode, so \(n = 4\). This is the step most often got wrong: \(n\) comes from the balanced half reactions, not from the number of molecules in the overall equation.
Step 2 — apply Eq. (5.28):
\[\Delta G^0 = -nFE^0_{cell} = -4 \times 96500\ \text{C mol}^{-1} \times 1.23\ \text{V}\]
\[= -474780\ \text{J mol}^{-1} = -474.8\ \text{kJ mol}^{-1}\]
The large negative value is simply the energy of combustion of two moles of hydrogen, made available as electrical work rather than as heat.
Advantages of Fuel Cells
The reacting substances are continuously supplied to the electrodes. Unlike conventional galvanic cells, fuel cells do not have to be discarded on consumption of the chemicals.
They are nonpolluting, as the only reaction product is water.
Fuel cells provide electricity with an efficiency of about 70%, which is twice as large as the efficiency of thermal plants (only 40%).
Hydrogen fuel cell vs Thermal power plant
| Energy route |
Chemical energy → electrical energy, directly |
Chemical energy → heat → mechanical energy → electrical energy |
| Approximate efficiency |
About 70% |
About 40% |
| Product released |
Water only |
Combustion products of the fuel burnt |
| Working life |
Runs as long as fuel and oxidant are supplied |
Runs as long as fuel is supplied |
| Moving parts |
None in the cell itself |
Boiler, turbine, generator |
Drawbacks of Fuel Cell
\(\mathrm{H_2}\) gas is hazardous to handle and the cost of preparing \(\mathrm{H_2}\) is high. Hydrogen forms explosive mixtures with air over a wide range of compositions, and it has to be stored under pressure or at very low temperature, which is why fuel cell vehicles need specially engineered tanks.
In what ways are fuel cells and galvanic cells similar, and in what ways are they different?
Applications of Fuel Cells
The fuel cells are used on an experimental basis in automobiles.
The fuel cells are used for electrical power in the space programme.
In spacecraft the fuel cell is operated at such a high temperature that the water evaporates at the same rate as it is formed. The vapour is condensed and the pure water formed is used for drinking by the astronauts.
In future, fuel cells can possibly be explored as power generators in hospitals, hotels and homes.
The third application above is one of the neatest pieces of engineering in the space programme: a spacecraft cannot afford to carry both a power supply and a large water supply, and a fuel cell provides the two from the same tanks of hydrogen and oxygen. On the ground, the fuel cells that power electric vehicles incorporate a proton-conducting plastic membrane in place of a liquid electrolyte; these are called proton exchange membrane (PEM) fuel cells, and the solid membrane makes the stack compact enough to sit under a car floor. Smaller still is the direct methanol fuel cell (DMFC), proposed for cell phones and laptop computers, in which liquid methanol replaces hydrogen gas and removes the storage problem altogether — worth looking up if you have internet access. The environmental case is genuine but not unconditional: a hydrogen fuel cell emits nothing but water, yet it is only as clean as the electricity used to make the hydrogen in the first place.
For the hydrogen–oxygen fuel cell, answer the following: (i) write the oxidation and reduction half reactions and show how they add to give the net cell reaction; (ii) explain why the concentration of KOH in the electrolyte does not fall as the cell operates; (iii) state one reason why such a cell is described as nonpolluting and one practical difficulty that limits its use.
5.12 Electrochemical Series (Electromotive Series)
The standard potentials of a number of electrodes have been determined using the standard hydrogen electrode, by the method described in Section 5.9.1. These electrodes, with their half reactions, are arranged according to their decreasing standard potentials as shown in Table 5.1. This arrangement is called the electrochemical series.
The electrochemical series is the arrangement of electrodes, with their reduction half reactions, in the decreasing order of their standard potentials.
Once the potentials are laid out in a single ordered list, the table stops being a collection of numbers and becomes a predicting instrument. It can tell you which metal will displace which, which reagent will oxidise which, and whether a proposed reaction will go at all — before you touch a single chemical.
The older name for this arrangement, the electromotive series, preserves a piece of nineteenth-century thinking. When the effect was first studied, the quantity driving the current was imagined as a genuine force pushing electricity along a wire, and it was named the electromotive force. The name has stuck, but it is misleading: emf is not a force at all. It is a potential difference, measured in volts, and no force is being exerted on anything.
Table 5.1: The Standard Aqueous Electrode Potentials at 298 K (Electrochemical Series)
| \(\mathrm{F^{\ominus}}\,\vert\,\mathrm{F_2}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{F_2} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{F^{\ominus}}\) |
+2.870 |
| \(\mathrm{Au^{\oplus}}\,\vert\,\mathrm{Au}\) |
\(\mathrm{Au^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Au}\) |
+1.680 |
| \(\mathrm{Ce^{4\oplus}},\ \mathrm{Ce^{3\oplus}}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Ce^{4\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ce^{3\oplus}}\) |
+1.610 |
| \(\mathrm{Au^{3\oplus}}\,\vert\,\mathrm{Au}\) |
\(\mathrm{Au^{3\oplus}} + 3\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Au}\) |
+1.500 |
| \(\mathrm{Cl^{\ominus}}\,\vert\,\mathrm{Cl_2}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Cl_2} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{Cl^{\ominus}}\) |
+1.360 |
| \(\mathrm{Pt^{2\oplus}}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Pt^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pt}\) |
+1.200 |
| \(\mathrm{Br^{\ominus}}\,\vert\,\mathrm{Br_2}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Br_2} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{Br^{\ominus}}\) |
+1.080 |
| \(\mathrm{Hg^{2\oplus}}\,\vert\,\mathrm{Hg}\) |
\(\mathrm{Hg^{2\oplus}} + 2\,\mathrm{e^{-}} \rightarrow \mathrm{Hg}\) |
+0.854 |
| \(\mathrm{Ag^{\oplus}}\,\vert\,\mathrm{Ag}\) |
\(\mathrm{Ag^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ag}\) |
+0.799 |
| \(\mathrm{Hg_2^{2\oplus}}\,\vert\,\mathrm{Hg}\) |
\(\mathrm{Hg_2^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{Hg}\) |
+0.79 |
| \(\mathrm{Fe^{3\oplus}},\ \mathrm{Fe^{2\oplus}}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Fe^{3\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Fe^{2\oplus}}\) |
+0.771 |
| \(\mathrm{I^{\ominus}}\,\vert\,\mathrm{I_2(s)}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{I_2} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{I^{\ominus}}\) |
+0.535 |
| \(\mathrm{Cu^{2\oplus}}\,\vert\,\mathrm{Cu}\) |
\(\mathrm{Cu^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\) |
+0.337 |
| \(\mathrm{Ag}\,\vert\,\mathrm{AgCl(s)}\,\vert\,\mathrm{Cl^{\ominus}}\) |
\(\mathrm{AgCl}\text{(s)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ag} + \mathrm{Cl^{-}}\) |
+0.222 |
| \(\mathrm{Cu^{2\oplus}},\ \mathrm{Cu^{\oplus}}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Cu^{2\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Cu^{\oplus}}\) |
+0.153 |
| \(\mathrm{Sn^{4\oplus}},\ \mathrm{Sn^{2\oplus}}\,\vert\,\mathrm{Pt}\) |
\(\mathrm{Sn^{4\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Sn^{2\oplus}}\) |
+0.15 |
| \(\mathrm{H^{\oplus}}\,\vert\,\mathrm{H_2}\,\vert\,\mathrm{Pt}\) |
\(2\,\mathrm{H^{\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{H_2}\) |
0.00 |
| \(\mathrm{Pb^{2\oplus}}\,\vert\,\mathrm{Pb}\) |
\(\mathrm{Pb^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pb}\) |
−0.126 |
| \(\mathrm{Sn^{2\oplus}}\,\vert\,\mathrm{Sn}\) |
\(\mathrm{Sn^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Sn}\) |
−0.136 |
| \(\mathrm{Ni^{2\oplus}}\,\vert\,\mathrm{Ni}\) |
\(\mathrm{Ni^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Ni}\) |
−0.257 |
| \(\mathrm{Co^{2\oplus}}\,\vert\,\mathrm{Co}\) |
\(\mathrm{Co^{2\oplus}} + 2\,\mathrm{e^{-}} \rightarrow \mathrm{Co}\) |
−0.280 |
| \(\mathrm{Cd^{2\oplus}}\,\vert\,\mathrm{Cd}\) |
\(\mathrm{Cd^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cd}\) |
−0.403 |
| \(\mathrm{Fe^{2\oplus}}\,\vert\,\mathrm{Fe}\) |
\(\mathrm{Fe^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Fe}\) |
−0.440 |
| \(\mathrm{Cr^{3\oplus}}\,\vert\,\mathrm{Cr}\) |
\(\mathrm{Cr^{3\oplus}} + 3\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cr}\) |
−0.740 |
| \(\mathrm{Zn^{2\oplus}}\,\vert\,\mathrm{Zn}\) |
\(\mathrm{Zn^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Zn}\) |
−0.763 |
| \(\mathrm{Al^{3\oplus}}\,\vert\,\mathrm{Al}\) |
\(\mathrm{Al^{3\oplus}} + 3\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Al}\) |
−1.66 |
| \(\mathrm{Mg^{2\oplus}}\,\vert\,\mathrm{Mg}\) |
\(\mathrm{Mg^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Mg}\) |
−2.37 |
| \(\mathrm{Na^{\oplus}}\,\vert\,\mathrm{Na}\) |
\(\mathrm{Na^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Na}\) |
−2.714 |
| \(\mathrm{Ca^{2\oplus}}\,\vert\,\mathrm{Ca}\) |
\(\mathrm{Ca^{2\oplus}} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Ca}\) |
−2.866 |
| \(\mathrm{K^{\oplus}}\,\vert\,\mathrm{K}\) |
\(\mathrm{K^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{K}\) |
−2.925 |
| \(\mathrm{Li^{\oplus}}\,\vert\,\mathrm{Li}\) |
\(\mathrm{Li^{\oplus}} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Li}\) |
−3.045 |
Note: (i) all ions are at 1 M concentration in water. (ii) all gases are at 1 atm pressure.
Reading down the table, oxidising strength decreases; reading up it, reducing strength decreases. The two directions of that single ordering are drawn in Fig. 5.12a.
Key Points of Electrochemical Series
The half reactions are written as reductions. The oxidising agents and electrons appear on the left side of the half reactions, while the reducing agents are shown on the right side in the half reaction.
Below the hydrogen electrode the negative standard potential increases, and above the hydrogen electrode the positive standard potential increases.
\(E^0\) values apply to the reduction half reactions that occur in the forward direction as written. If you reverse a half reaction to use it as an oxidation, you must reverse the sign of its potential, as established in Section 5.7.1.
A higher (more positive) \(E^0\) value for a half reaction indicates its greater tendency to occur in the forward direction, and in turn a greater tendency for the substance to be reduced. Conversely, a low (more negative) \(E^0\) value of a half reaction corresponds to its greater tendency to occur in the reverse direction, or for the substance to be oxidised.
The half reactions are listed in order of their decreasing tendency in the forward direction. In other words, the table is a ranking: the reaction at the top happens most readily as written, and the one at the bottom least readily.
The left side of a half reaction has cations of metals or non-metallic molecules (oxidants). There are free metals or anions of non-metals on the right side (reductants).
standard potential and reactivity down the electrochemical series (from top ↓: \(E^0\) becomes less positive and then increasingly negative, because the species on the left holds electrons less and less strongly; so oxidising strength of the left-side species falls from top to bottom, while reducing strength of the right-side species rises from top to bottom. Landmark: hydrogen sits at exactly 0.00 V by definition, and every metal below it displaces hydrogen from acid while those above it do not.)
Applications of Electrochemical Series
i. Relative Strength of Oxidising Agents
The species on the left side of the half reactions are oxidising agents. The \(E^0\) value is a measure of the tendency of the species to accept electrons and get reduced.
In other words, the \(E^0\) value measures the strength of substances as oxidising agents. The larger the \(E^0\) value, the greater is the oxidising strength. The species at the top left side of the half reactions are strong oxidising agents. As we move down the table, the \(E^0\) value and the strength of oxidising agents decrease from top to bottom.
Identify the strongest and the weakest oxidising agents from the electrochemical series.
ii. Relative Strength of Reducing Agents
The species on the right side of the half reactions are reducing agents.
The half reactions at the bottom of the table, with large negative \(E^0\) values, have little or no tendency to occur in the forward direction as written. They tend to favour the reverse direction.
It follows that the species appearing at the bottom right side of half reactions, associated with large negative \(E^0\) values, are the effective electron donors. They serve as strong reducing agents. The strength of reducing agents increases from top to bottom as \(E^0\) values decrease.
Notice how neatly the two applications fit together. A single column of numbers ranks the left-hand species one way and the right-hand species the other way, because the two are the same tendency read in opposite directions.
Identify the strongest and the weakest reducing agents from the electrochemical series.
Top of the electrochemical series vs Bottom of the series
| \(E^0\) value |
Large and positive |
Large and negative |
| Tendency of the half reaction as written |
Occurs readily in the forward direction |
Tends to occur in the reverse direction |
| Left-side species |
Strong oxidising agents |
Weak oxidising agents |
| Right-side species |
Weak reducing agents |
Strong reducing agents |
| Example |
\(\mathrm{F_2}\) (\(+2.870\ \text{V}\)), the strongest oxidising agent listed |
\(\mathrm{Li}\) (\(-3.045\ \text{V}\)), the strongest reducing agent listed |
| Where such species are found in nature |
Metals near the top occur native (gold, platinum) |
Metals near the bottom occur only as compounds (Na, K, Ca) |
the ordered potentials of Table 5.1, as laid out on the scale of Fig. 5.12a (Using Fig. 5.12a and Table 5.1, answer: (a) Which species in the table is the strongest oxidising agent, and which is the strongest reducing agent? (b) Silver lies above hydrogen and zinc lies below it — what does each position predict about the reaction of that metal with dilute acid? (c) Two entries in the table involve only copper. Explain what makes them different from each other and why the metal appears in one of them but not in the other.)
iii. Spontaneity of Redox Reactions
A redox reaction in a galvanic cell is spontaneous only if the species with the higher \(E^0\) value is reduced (accepts electrons) and that with the lower \(E^0\) value is oxidised (donates electrons).
The standard cell potential must be positive for a cell reaction to be spontaneous under standard conditions — which is the criterion established in Section 5.8.1, where a positive \(E^0_{cell}\) was shown to correspond to a negative \(\Delta G^0\). A noteworthy application of the electromotive series is predicting the spontaneity of redox reactions from a knowledge of standard potentials.
The procedure is set out in FC 5.11 and then applied to a concrete question.
Suppose we ask a question: at standard conditions, would \(\mathrm{Ag^{\oplus}}\) ions oxidise metallic magnesium? To answer this question, we first write the oxidation of Mg by \(\mathrm{Ag^{\oplus}}\).
\[\mathrm{Mg}\text{(s)} \rightarrow \mathrm{Mg^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation)}\]
\[2\,\mathrm{Ag^{\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{Ag}\text{(s)} \qquad \text{(reduction)}\]
\[\mathrm{Mg}\text{(s)} + 2\,\mathrm{Ag^{\oplus}}\text{(aq)} \rightarrow \mathrm{Mg^{2\oplus}}\text{(aq)} + 2\,\mathrm{Ag}\text{(s)} \qquad \text{(overall reaction)}\]
From Table 5.1 of the electrochemical series we have \(E^0_{Mg} = -2.37\ \text{V}\) and \(E^0_{Ag} = 0.8\ \text{V}\). For the cell having Mg as anode and Ag as cathode,
\[E^0_{Cell} = E^0_{Ag} - E^0_{Mg} = 0.8\ \text{V} - (-2.37\ \text{V}) = 3.17\ \text{V}\]
The emf being positive, the cell reaction is spontaneous. Thus \(\mathrm{Ag^{\oplus}}\) ions do oxidise metallic Mg.
General Rules
An oxidising agent can oxidise any reducing agent that appears below it, and cannot oxidise the reducing agent appearing above it in the electrochemical series.
A reducing agent can reduce the oxidising agent located above it in the electrochemical series.
These two rules save a great deal of arithmetic. Once you can see that silver ions sit far above magnesium metal in the table, the answer to the question just worked out is available at a glance; the calculation only confirms it.
Using the standard potentials in Table 5.1, predict whether metallic iron will displace copper from a \(1\ \text{M}\) solution of \(\mathrm{Cu^{2\oplus}}\) ions at \(25\ ^\circ\text{C}\). Write the overall reaction and calculate \(E^0_{cell}\).
Step 1 — write the two half reactions in the direction proposed. Iron is to dissolve, so it is oxidised; copper ions are to be deposited, so they are reduced.
\[\mathrm{Fe}\text{(s)} \rightarrow \mathrm{Fe^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \qquad \text{(oxidation, anode)}\]
\[\mathrm{Cu^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Cu}\text{(s)} \qquad \text{(reduction, cathode)}\]
\[\mathrm{Fe}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Fe^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad \text{(overall reaction)}\]
Step 2 — read the standard potentials from Table 5.1: \(E^0_{Cu} = +0.337\ \text{V}\) and \(E^0_{Fe} = -0.440\ \text{V}\). Both are taken as reduction potentials, without altering any sign.
Step 3 — apply Eq. (5.24):
\[E^0_{cell} = E^0\text{(cathode)} - E^0\text{(anode)} = 0.337\ \text{V} - (-0.440\ \text{V}) = 0.777\ \text{V}\]
Step 4 — interpret. \(E^0_{cell}\) is positive, so \(\Delta G^0\) is negative and the reaction is spontaneous as written: iron will displace copper from the solution, and the iron surface becomes coated with reddish-brown copper. The same conclusion follows from the general rules, since \(\mathrm{Cu^{2\oplus}}\) lies above \(\mathrm{Fe}\) in the series.
From the \(E^0\) values given in Table 5.1, predict whether Sn can reduce \(\mathrm{I_2}\) or \(\mathrm{Ni^{2\oplus}}\).
The series explains, and is used to fight, the most expensive chemical process on earth — corrosion. Iron rusts because iron sits well below oxygen-bearing species in the table, so its oxidation is spontaneous wherever water and air can reach it. The defence is to give the iron something even more easily oxidised to sacrifice in its place. Galvanised sheets are coated with zinc (\(-0.763\ \text{V}\)), which corrodes in preference to the iron beneath (\(-0.440\ \text{V}\)) and goes on protecting it even where the coating is scratched. Ship hulls, underground pipelines and water tanks carry blocks of magnesium or zinc bolted to them for the same reason; the blocks are eaten away and periodically replaced, and are called sacrificial anodes. The same table also explains a familiar safety rule in reverse: metals near the bottom of the series, such as sodium and potassium, are stored under kerosene precisely because their oxidation is so strongly spontaneous that air and moisture are enough to set it off.
Using Table 5.1, predict in each case whether the reaction occurs spontaneously under standard conditions, giving \(E^0_{cell}\) with its sign for each: (i) \(\mathrm{Zn}\text{(s)} + \mathrm{Pb^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Zn^{2\oplus}}\text{(aq)} + \mathrm{Pb}\text{(s)}\); (ii) \(\mathrm{Cu}\text{(s)} + 2\,\mathrm{H^{\oplus}}\text{(aq)} \rightarrow \mathrm{Cu^{2\oplus}}\text{(aq)} + \mathrm{H_2}\text{(g)}\); (iii) \(2\,\mathrm{Br^{\ominus}}\text{(aq)} + \mathrm{Cl_2}\text{(g)} \rightarrow \mathrm{Br_2}\text{(l)} + 2\,\mathrm{Cl^{\ominus}}\text{(aq)}\).
Using only Table 5.1, arrange \(\mathrm{Cl_2}\), \(\mathrm{I_2}\), \(\mathrm{F_2}\) and \(\mathrm{Br_2}\) in increasing order of oxidising strength, and arrange Mg, Zn, Fe and Cu in increasing order of reducing strength. In each case state the feature of the table you used to decide the order.
5.13 Summary
Electrochemistry and the two directions of energy conversion
- Electrochemistry deals with the interconversion of chemical energy and electrical energy. A dry cell converts chemical energy into electrical energy; electrolysis converts electrical energy into chemical energy.
- Both directions rest on redox reactions, in which electrons are transferred from one species to another. These processes are carried out in an electrochemical cell.
- Electrolysis makes possible the manufacture of NaOH from NaCl, the production of fluorine, electro-refining of metals and electroplating.
Two kinds of conduction
- Metallic (electronic) conduction is carried by free electrons in the conduction band; no matter is transferred and no chemical change occurs.
- Electrolytic (ionic) conduction is carried by ions moving through a molten electrolyte or an aqueous solution. Matter is transferred, so the current is always accompanied by chemical change at the electrodes.
- Substances that do not dissociate in water (sucrose, urea) are nonelectrolytes. Those that dissociate are electrolytes: ionic salts and strong acids and bases are almost completely dissociated and are strong electrolytes; weak acids and weak bases dissociate very little and are weak electrolytes.
Conductance, conductivity and molar conductivity
- By Ohm’s law \(R = V/I\); conductance is its reciprocal, \(G = 1/R\), with SI unit siemens (\(\text{S} = \Omega^{-1}\)).
- \(R = \rho\,l/a\) and \(G = k\,a/l\), so conductivity \(k = 1/\rho\). Conductivity is the conductance of a unit cube of solution; its common unit is \(\Omega^{-1}\text{cm}^{-1}\).
- Molar conductivity is the conductance due to the ions from 1 mole of electrolyte:
\[\Lambda = \frac{k}{c}, \qquad \Lambda = \frac{1000\,k}{c}\]
with SI unit \(\text{S m}^2\text{mol}^{-1}\) and common unit \(\Omega^{-1}\text{cm}^2\text{mol}^{-1}\).
- On dilution, conductivity decreases (fewer ions per unit volume) while molar conductivity increases (more ions per mole of electrolyte).
- For strong electrolytes \(\Lambda = \Lambda_0 - a\sqrt{c}\), so a plot of \(\Lambda\) against \(\sqrt{c}\) is linear and \(\Lambda_0\) is obtained by extrapolation. For weak electrolytes the plot is not linear and this method fails.
Kohlrausch’s law and degree of dissociation
- Kohlrausch’s law of independent migration of ions: at infinite dilution each ion migrates independently of its co-ion and makes its own contribution, whatever the other ion may be.
\[\Lambda_0 = n_{\oplus}\lambda^0_{\oplus} + n_{\ominus}\lambda^0_{\ominus}\]
- The law gives \(\Lambda_0\) of a weak electrolyte from the values for strong electrolytes, for example \(\Lambda_0(\mathrm{CH_3COOH}) = \Lambda_0(\mathrm{HCl}) + \Lambda_0(\mathrm{CH_3COONa}) - \Lambda_0(\mathrm{NaCl})\).
- The degree of dissociation of a weak electrolyte is \(\alpha = \Lambda_c/\Lambda_0\), from which its dissociation constant follows by Ostwald’s dilution law.
Measurement of conductivity
- Resistance is measured with a conductivity cell and a Wheatstone bridge; the cell constant is \(l/a\), so that \(k = \text{cell constant}/R\).
- Molar conductivity is found in three steps: determine the cell constant using a standard KCl solution, measure the resistance of the given solution to obtain \(k\), then calculate \(\Lambda = 1000k/c\).
Electrochemical cells
- Every cell has two electrodes: oxidation occurs at the anode and reduction at the cathode, in both types of cell.
- In an electrolytic cell a non-spontaneous reaction is forced by an external direct current; the anode is positive and the cathode negative.
- In a galvanic (voltaic) cell a spontaneous reaction generates electricity; the anode is negative and the cathode positive.
Electrolysis
- Molten NaCl gives chlorine gas at the anode and molten sodium at the cathode:
\[2\,\mathrm{Na^{\oplus}}\text{(l)} + 2\,\mathrm{Cl^{\ominus}}\text{(l)} \rightarrow 2\,\mathrm{Na}\text{(l)} + \mathrm{Cl_2}\text{(g)}\]
- In aqueous NaCl, water competes at both electrodes. Water is reduced at the cathode to hydrogen, chloride is oxidised at the anode to chlorine because of overvoltage, and the solution left behind becomes NaOH:
\[2\,\mathrm{Cl^{\ominus}}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{Cl_2}\text{(g)} + \mathrm{H_2}\text{(g)} + 2\,\mathrm{OH^{\ominus}}\text{(aq)}\]
- The extra voltage needed in practice, above that calculated from standard potentials, is the overpotential.
Quantitative aspects of electrolysis
- \(Q = I \times t\); one faraday, \(96500\ \text{C}\), is the charge on one mole of electrons.
- The mole ratio comes from the balanced half reaction, and
\[W = \frac{I \times t}{96500} \times \text{mole ratio} \times \text{molar mass}\]
- For two cells in series the same charge passes through both, so
\[\frac{W_1}{(\text{mole ratio})_1 \times M_1} = \frac{W_2}{(\text{mole ratio})_2 \times M_2}\]
Galvanic cells: construction and notation
- A galvanic cell is built from two half cells, each a metal in a solution of its own ions, joined by a wire and a salt bridge.
- The salt bridge completes the circuit, prevents mixing of the solutions and maintains electrical neutrality in both.
- In the cell notation the anode is written at the extreme left and the cathode at the extreme right; a single vertical line marks a phase boundary and a double vertical line the salt bridge, as in \(\mathrm{Mg}\text{(s)} \,|\, \mathrm{Mg^{2\oplus}}\text{(1M)} \,\|\, \mathrm{Ag^{\oplus}}\text{(1M)} \,|\, \mathrm{Ag}\text{(s)}\).
- To write the cell reaction, take oxidation at the left electrode and reduction at the right, multiply as needed to balance electrons, and add so that no electrons remain.
Electrode potential, cell potential and the Nernst equation
- The potential difference set up at the metal–solution boundary is the electrode potential; the cell potential (emf) is the algebraic sum of the electrode potentials, \(E_{cell} = E_{oxi}\text{(anode)} + E_{red}\text{(cathode)}\).
- Standard conditions are \(1\ \text{M}\) solutions, \(1\ \text{atm}\) gases, pure solids and liquids, and \(25\ ^\circ\text{C}\). By IUPAC convention the standard potential of an electrode is its standard reduction potential, and
\[E^0_{cell} = E^0\text{(cathode)} - E^0\text{(anode)}\]
- The electrode with the higher standard potential is the cathode.
- The Nernst equation gives the potential at non-standard concentrations:
\[E_{cell} = E^0_{cell} - \frac{2.303\,RT}{nF}\log_{10}\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} = E^0_{cell} - \frac{0.0592\ \text{V}}{n}\log_{10}\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \quad \text{at } 25\ ^\circ\text{C}\]
Gases enter as partial pressures; solids and pure liquids are taken as unity.
Thermodynamics of galvanic cells
- Electrical work \(= nFE_{cell}\), and equals the decrease in Gibbs energy, so
\[\Delta G = -nFE_{cell}, \qquad \Delta G^0 = -nFE^0_{cell}\]
- \(\Delta G^0\) is extensive but \(E^0_{cell}\) is intensive: multiplying the equation changes both \(\Delta G^0\) and \(n\), leaving their ratio unchanged.
- A reaction is spontaneous when \(\Delta G\) is negative, that is when \(E_{cell}\) is positive.
- Combining with \(\Delta G^0 = -RT\ln K\):
\[E^0_{cell} = \frac{2.303\,RT}{nF}\log_{10}K = \frac{0.0592}{n}\log_{10}K \quad \text{at } 25\ ^\circ\text{C}\]
Reference electrodes
- A single electrode potential cannot be measured, so a reference electrode is needed: one whose potential is arbitrarily taken as zero or is exactly known.
- The standard hydrogen electrode, \(\mathrm{H^{\oplus}}\text{(1M)} \,|\, \mathrm{H_2}\text{(g, 1atm)} \,|\, \mathrm{Pt}\), is the primary reference, with \(E^0 = 0.000\ \text{V}\); standard potentials of all other electrodes are measured against it.
- It is difficult to obtain pure dry hydrogen, to hold the pressure at \(1\ \text{atm}\) and to keep \([\mathrm{H^{\oplus}}]\) at \(1\ \text{M}\), so calomel, silver–silver chloride and glass electrodes are used as secondary reference electrodes.
- For a hydrogen gas electrode away from standard conditions, \(E_{H_2} = -\dfrac{0.0592}{2}\log_{10}\dfrac{P_{H_2}}{[\mathrm{H^{\oplus}}]^2}\).
Cells in day-to-day life
- Primary cells cannot be recharged; secondary cells can, because their products are solids that stay on the electrodes. A secondary cell is galvanic while discharging and electrolytic while being recharged.
- Dry cell (Leclanché): zinc container anode, graphite cathode surrounded by \(\mathrm{MnO_2}\) and carbon, moist paste of \(\mathrm{NH_4Cl}\) and \(\mathrm{ZnCl_2}\) as electrolyte; \(\mathrm{MnO_2}\) oxidises the hydrogen formed at the cathode. The alkaline dry cell uses NaOH or KOH instead and lasts longer.
- Lead accumulator: spongy lead anode, \(\mathrm{PbO_2}\) cathode, 38% \(\mathrm{H_2SO_4}\); discharge coats both plates with \(\mathrm{PbSO_4}\) and consumes acid, lowering the density and the voltage:
\[\mathrm{Pb}\text{(s)} + \mathrm{PbO_2}\text{(s)} + 2\,\mathrm{H_2SO_4}\text{(aq)} \rightarrow 2\,\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)}\]
Recharging reverses this and regenerates the acid. Six \(2\ \text{V}\) cells in series give the \(12\ \text{V}\) battery used in automobiles and inverters. 45. NICAD cell: cadmium anode, \(\mathrm{NiO_2}\) cathode, basic electrolyte, about \(1.4\ \text{V}\); rechargeable because both products adhere to the electrodes. 46. Mercury battery: zinc amalgam anode, HgO reduced at the cathode, alkaline paste of KOH and ZnO; the electrolyte composition does not change, so it gives a constant \(1.35\ \text{V}\).
Fuel cells
- In a fuel cell the reactants are supplied continuously from outside instead of being stored in the cell.
- In the \(\mathrm{H_2}\)–\(\mathrm{O_2}\) fuel cell, porous carbon electrodes containing finely divided platinum dip in hot aqueous KOH; hydrogen is oxidised at the anode and oxygen reduced at the cathode, giving
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}, \qquad E^0_{cell} = 1.23\ \text{V}\]
- Fuel cells are not discarded when chemicals are consumed, are nonpolluting since the only product is water, and are about 70% efficient against 40% for thermal plants. Hydrogen is hazardous to handle and costly to prepare.
Electrochemical series
- Electrodes arranged in decreasing order of standard potential form the electrochemical series, with all half reactions written as reductions.
- The left-side species are oxidising agents, strongest at the top; the right-side species are reducing agents, strongest at the bottom.
- A redox reaction is spontaneous when the species of higher \(E^0\) is reduced and that of lower \(E^0\) is oxidised, giving a positive \(E^0_{cell}\).
- An oxidising agent can oxidise any reducing agent that appears below it in the series, and a reducing agent can reduce any oxidising agent above it.
The chapter in one line of logic
5.14 Objective Questions
- Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
- 0.54
- 11.574
- 0.0864
- 1.852
- On diluting the solution of an electrolyte
- both \(\Lambda\) and \(k\) increase
- both \(\Lambda\) and \(k\) decrease
- \(\Lambda\) increases and \(k\) decreases
- \(\Lambda\) decreases and \(k\) increases
- \(1\ \text{S m}^2\text{mol}^{-1}\) is equal to
- \(10^{-4}\ \text{S m}^2\text{mol}^{-1}\)
- \(10^{4}\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
- \(10^{-2}\ \text{S cm}^2\text{mol}^{-1}\)
- \(10^{2}\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
- The standard potential of the cell in which the following reaction occurs, \(\mathrm{H_2}\text{(g, 1atm)} + \mathrm{Cu^{2\oplus}}\text{(1M)} \rightarrow 2\,\mathrm{H^{\oplus}}\text{(1M)} + \mathrm{Cu}\text{(s)}\), \((E^0_{Cu} = 0.34\ \text{V})\) is
- \(-0.34\ \text{V}\)
- \(0.34\ \text{V}\)
- \(0.17\ \text{V}\)
- \(-0.17\ \text{V}\)
- For the cell, \(\mathrm{Pb}\text{(s)}\,|\,\mathrm{Pb^{2\oplus}}\text{(1M)}\,\|\,\mathrm{Ag^{\oplus}}\text{(1M)}\,|\,\mathrm{Ag}\text{(s)}\), if the concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
- increase by \(10\ \text{V}\)
- increase by \(0.0296\ \text{V}\)
- decrease by \(10\ \text{V}\)
- decrease by \(0.0296\ \text{V}\)
- Consider the half reactions with standard potentials: i. \(\mathrm{Ag^{\oplus}}\text{(aq)} + \mathrm{e^{\ominus}} \rightarrow \mathrm{Ag}\text{(s)}\), \(E^0 = 0.8\ \text{V}\); ii. \(\mathrm{I_2}\text{(s)} + 2\,\mathrm{e^{\ominus}} \rightarrow 2\,\mathrm{I^{\ominus}}\text{(aq)}\), \(E^0 = 0.53\ \text{V}\); iii. \(\mathrm{Pb^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pb}\text{(s)}\), \(E^0 = -0.13\ \text{V}\); iv. \(\mathrm{Fe^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Fe}\text{(s)}\), \(E^0 = -0.44\ \text{V}\). The strongest oxidising and reducing agents respectively are
- Ag and \(\mathrm{Fe^{2\oplus}}\)
- \(\mathrm{Ag^{\oplus}}\) and Fe
- \(\mathrm{Pb^{2\oplus}}\) and \(\mathrm{I^{\ominus}}\)
- \(\mathrm{I_2}\) and \(\mathrm{Fe^{2\oplus}}\)
- For the reaction \(\mathrm{Ni}\text{(s)} + \mathrm{Cu^{2\oplus}}\text{(1M)} \rightarrow \mathrm{Ni^{2\oplus}}\text{(1M)} + \mathrm{Cu}\text{(s)}\), \(E^0_{cell} = 0.57\ \text{V}\), \(\Delta G^0\) of the reaction is
- \(110\ \text{kJ}\)
- \(-110\ \text{kJ}\)
- \(55\ \text{kJ}\)
- \(-55\ \text{kJ}\)
- Which of the following is not correct?
- Gibbs energy is an extensive property
- Electrode potential or cell potential is an intensive property
- Electrical work \(= -\Delta G\)
- If a half reaction is multiplied by a numerical factor, the corresponding \(E^0\) value is also multiplied by the same factor
- The oxidation reaction that takes place in a lead storage battery during discharge is
- \(\mathrm{Pb^{2\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow \mathrm{PbSO_4}\text{(s)}\)
- \(\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{PbO_2}\text{(s)} + 4\,\mathrm{H^{\oplus}}\text{(aq)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}}\)
- \(\mathrm{Pb}\text{(s)} + \mathrm{SO_4^{2\ominus}}\text{(aq)} \rightarrow \mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{e^{\ominus}}\)
- \(\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Pb}\text{(s)} + \mathrm{SO_4^{2\ominus}}\text{(aq)}\)
- Which of the following expressions represents the molar conductivity of \(\mathrm{Al_2(SO_4)_3}\)?
- \(3\lambda^0_{\mathrm{Al^{3\oplus}}} + 2\lambda^0_{\mathrm{SO_4^{2\ominus}}}\)
- \(2\lambda^0_{\mathrm{Al^{3\oplus}}} + 3\lambda^0_{\mathrm{SO_4^{2\ominus}}}\)
- \(\tfrac{1}{3}\lambda^0_{\mathrm{Al^{3\oplus}}} + \tfrac{1}{2}\lambda^0_{\mathrm{SO_4^{2\ominus}}}\)
- \(\lambda^0_{\mathrm{Al^{3\oplus}}} + \lambda^0_{\mathrm{SO_4^{2\ominus}}}\)
5.15 Check Your Concepts
A student passes current through a copper wire and through a copper sulphate solution in turn. In one case the conductor is unchanged at the end and in the other it is not. Identify which is which, name the type of conduction in each, and explain the difference in terms of the particles that carry the charge.
Assertion: On dilution, the molar conductivity of acetic acid rises steeply while its conductivity falls. Reason: Dilution increases the degree of dissociation of a weak electrolyte. State whether the assertion and the reason are each true, and whether the reason correctly explains the assertion.
Why is alternating current, rather than direct current, used while measuring the resistance of an electrolytic solution on a Wheatstone bridge?
The cell constant of a conductivity cell is defined as \(l/a\), yet it is never obtained by measuring \(l\) and \(a\) with a scale. Explain why, and describe how it is actually determined.
The molar conductivity at zero concentration of \(\mathrm{KNO_3}\) can be found by extrapolating its \(\Lambda\) versus \(\sqrt{c}\) plot, but that of \(\mathrm{NH_4OH}\) cannot. Explain why, and state how the value for \(\mathrm{NH_4OH}\) is obtained instead.
Why is the electrode at which oxidation occurs marked positive in an electrolytic cell but negative in a galvanic cell, when in both cases it is called the anode?
Molten calcium chloride is electrolysed between inert electrodes. Predict the product at each electrode, write both half reactions and the overall cell reaction, and state which electrode carries the positive sign.
During the electrolysis of a moderately concentrated aqueous solution of sodium chloride, no sodium metal is ever obtained at the cathode. Explain this observation using the standard potentials of the two competing reductions.
What is meant by overpotential? Explain how it accounts for chlorine, rather than oxygen, being the gas collected at the anode during the electrolysis of aqueous NaCl.
State the three functions of a salt bridge, and explain what would be observed on the voltmeter if the salt bridge were removed from a working galvanic cell. Why must the electrolyte in the bridge be an inert one such as KCl or \(\mathrm{NH_4NO_3}\)?
For the cell \(\mathrm{Cd}\text{(s)}\,|\,\mathrm{Cd^{2\oplus}}\text{(1M)}\,\|\,\mathrm{Ag^{\oplus}}\text{(1M)}\,|\,\mathrm{Ag}\text{(s)}\), write the oxidation half reaction, the reduction half reaction and the balanced overall cell reaction, and state the direction in which electrons move through the external wire.
Assertion: Multiplying a cell reaction throughout by 3 leaves \(E^0_{cell}\) unchanged. Reason: \(\Delta G^0\) and \(n\) are both tripled, so their ratio is unaltered. State whether each is true and whether the reason explains the assertion.
Explain why the standard potential of the hydrogen electrode is taken as exactly zero, and state what the negative sign of \(E^0 = -0.763\ \text{V}\) for zinc actually tells you about zinc.
List three practical difficulties in setting up a standard hydrogen electrode, and explain how each of them threatens the very conditions that define the electrode’s potential as zero. Name two secondary reference electrodes used instead.
A garage mechanic judges whether a car battery needs charging by measuring the density of its electrolyte rather than by connecting a voltmeter. Explain the chemistry that makes this possible.
Explain, in terms of what happens to the products at the electrodes, why a lead accumulator and a NICAD cell can be recharged whereas a Leclanché dry cell cannot.
What is the role of \(\mathrm{MnO_2}\) in a dry cell? What would happen inside the cell if it were left out?
The mercury battery keeps a nearly constant voltage of \(1.35\ \text{V}\) throughout its working life, while the voltage of a Leclanché dry cell falls steadily. Account for this difference by examining what happens to the electrolyte in each cell.
Match each cell in Column I with the correct electrolyte in Column II:
| a. Leclanché dry cell |
i. hot aqueous KOH |
| b. lead accumulator |
ii. moist paste of \(\mathrm{NH_4Cl}\) and \(\mathrm{ZnCl_2}\) |
| c. NICAD cell |
iii. alkaline paste of KOH and ZnO |
| d. mercury battery |
iv. 38% \(\mathrm{H_2SO_4}\) |
| e. \(\mathrm{H_2}\)–\(\mathrm{O_2}\) fuel cell |
v. a basic solution |
In what way does a fuel cell resemble an ordinary galvanic cell, and in what two ways does it differ? Explain why a fuel cell converts a larger fraction of the available chemical energy into electricity than a thermal power plant does.
Iron objects are protected from rusting by coating them with zinc, and blocks of magnesium are bolted to underground steel pipelines. Using the electrochemical series, explain the principle common to both practices, and state why a copper coating would not protect iron in the same way.
Using Table 5.1, state with reasons which member of each pair is the stronger oxidising agent: (i) \(\mathrm{F_2}\) or \(\mathrm{Cl_2}\); (ii) \(\mathrm{Ag^{\oplus}}\) or \(\mathrm{Cu^{2\oplus}}\); (iii) \(\mathrm{Fe^{3\oplus}}\) or \(\mathrm{I_2}\).
Three metals — copper, zinc and silver — are each placed in dilute hydrochloric acid. Predict which of them liberate hydrogen gas, justify your prediction from their positions in the electrochemical series, and write the equation for any reaction that occurs.
A solution of a strong electrolyte at high concentration shows a high conductivity but a comparatively low molar conductivity. Explain how both statements can be true of the same solution.
Write the relationship between conductivity and molar conductivity, and hence the unit of molar conductivity.
Write the electrode reactions during electrolysis of molten KCl.
Write the Nernst equation. What part of it represents the correction factor for nonstandard state conditions?
Under what conditions is the cell potential called the standard cell potential?
Formulate a cell from the following electrode reactions: \(\mathrm{Au^{3\oplus}}\text{(aq)} + 3\,\mathrm{e^{\ominus}} \rightarrow \mathrm{Au}\text{(s)}\) and \(\mathrm{Mg}\text{(s)} \rightarrow \mathrm{Mg^{2\oplus}}\text{(aq)} + 2\,\mathrm{e^{\ominus}}\).
What is the significance of the single vertical line and the double vertical line in the formulation of a galvanic cell?
Explain the effect of dilution of solution on conductivity.
What is a salt bridge?
Write electrode reactions for the electrolysis of aqueous NaCl.
Using the relationship between \(\Delta G^0\) of a cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property?
Derive the relationship between standard cell potential and the equilibrium constant of a cell reaction.
It is impossible to measure the potential of a single electrode. Comment.
Why does the cell potential of a lead accumulator decrease when it generates electricity? How can the cell potential be increased?
Write the electrode reactions and the net cell reaction in a NICAD battery.
What is Kohlrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte? Explain with an example.
Explain electrolysis of molten NaCl.
What is a cell constant? What are its units? How is it determined experimentally?
How will you calculate the moles of electrons passed and the mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry?
Write the electrode reactions when a lead storage cell generates electricity. What are the anode and cathode, and the electrode reactions, during its recharging?
What are the anode and cathode of an \(\mathrm{H_2}\)–\(\mathrm{O_2}\) fuel cell? Name the electrolyte used in it. Write the electrode reactions and the net cell reaction taking place in the fuel cell.
What are the anode and cathode for a Leclanché dry cell? Write the electrode reactions and the overall cell reaction when it generates electricity.
Identify the oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis: Al \((-1.66\ \text{V})\), \(\mathrm{Al^{3\oplus}}\) \((-1.66\ \text{V})\), \(\mathrm{Cl_2}\) \((1.36\ \text{V})\), \(\mathrm{Cd^{2\oplus}}\) \((-0.4\ \text{V})\), Fe \((-0.44\ \text{V})\), \(\mathrm{I_2}\) \((0.54\ \text{V})\), \(\mathrm{Br^{\ominus}}\) \((1.09\ \text{V})\).
Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis: K \((-2.93\ \text{V})\), \(\mathrm{Br_2}\) \((1.09\ \text{V})\), Mg \((-2.36\ \text{V})\), \(\mathrm{Ce^{3\oplus}}\) \((1.61\ \text{V})\), \(\mathrm{Ti^{2\oplus}}\) \((-0.37\ \text{V})\), \(\mathrm{Ag^{\oplus}}\) \((0.8\ \text{V})\), Ni \((-0.23\ \text{V})\).
Predict whether the following reactions would occur spontaneously under standard state conditions (use the information of Table 5.1): a. \(\mathrm{Ca}\text{(s)} + \mathrm{Cd^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Ca^{2\oplus}}\text{(aq)} + \mathrm{Cd}\text{(s)}\); b. \(2\,\mathrm{Br^{\ominus}}\text{(s)} + \mathrm{Sn^{2\oplus}}\text{(aq)} \rightarrow \mathrm{Br_2}\text{(l)} + \mathrm{Sn}\text{(s)}\); c. \(2\,\mathrm{Ag}\text{(s)} + \mathrm{Ni^{2\oplus}}\text{(aq)} \rightarrow 2\,\mathrm{Ag^{\oplus}}\text{(aq)} + \mathrm{Ni}\text{(s)}\).
Write the electrode reactions and the net cell reaction in the electrolysis of molten barium chloride.
\(k_1\) and \(k_2\) are conductivities of two solutions and \(c_1\) and \(c_2\) are their concentrations. Establish the relationship between \(k_1\), \(k_2\), \(c_1\), \(c_2\) and the molar conductivities \(\Lambda_1\) and \(\Lambda_2\) of the two solutions.
5.16 Practice With Numericals
The resistance of a \(0.01\ \text{M}\) solution of an electrolyte, measured at \(298\ \text{K}\) in a cell of cell constant \(0.98\ \text{cm}^{-1}\), is \(245\ \Omega\). Calculate the conductivity and the molar conductivity of the solution.
A conductivity cell filled with \(0.1\ \text{M}\) KCl, whose conductivity is \(0.0129\ \Omega^{-1}\text{cm}^{-1}\) at \(298\ \text{K}\), shows a resistance of \(85\ \Omega\). The same cell filled with a \(0.02\ \text{M}\) solution of a salt shows a resistance of \(320\ \Omega\). Calculate (i) the cell constant, (ii) the conductivity of the salt solution and (iii) its molar conductivity.
The molar conductivity of a \(0.025\ \text{M}\) solution of a strong electrolyte is \(138.0\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) at \(298\ \text{K}\). Calculate its conductivity, and express your answer in \(\text{S m}^{-1}\) as well as \(\Omega^{-1}\text{cm}^{-1}\).
The molar ionic conductivities at zero concentration are \(189\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) for \(\mathrm{Al^{3\oplus}}\) and \(160\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) for \(\mathrm{SO_4^{2\ominus}}\). Calculate \(\Lambda_0\) for \(\mathrm{Al_2(SO_4)_3}\).
The values of \(\Lambda_0\) are \(426.2\), \(126.5\) and \(91.0\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\) for HCl, NaCl and \(\mathrm{CH_3COONa}\) respectively. Calculate \(\Lambda_0\) for acetic acid.
The molar conductivity of a \(0.05\ \text{M}\) solution of a weak monobasic acid is \(7.36\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\), and its molar conductivity at zero concentration is \(368\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\). Calculate the degree of dissociation of the acid at this concentration and its dissociation constant.
A current of \(5\ \text{A}\) is passed for 3 hours through molten \(\mathrm{Al_2O_3}\). Calculate (i) the quantity of electricity passed, (ii) the moles of electrons passed and (iii) the mass of aluminium deposited at the cathode. (Molar mass of Al \(= 27\ \text{g mol}^{-1}\))
For how long must a current of \(2.5\ \text{A}\) be passed through a silver salt solution to deposit \(5.0\ \text{g}\) of silver at the cathode? (Molar mass of Ag \(= 108\ \text{g mol}^{-1}\))
What current strength is required to deposit \(1.0\ \text{g}\) of zinc from a \(\mathrm{ZnSO_4}\) solution in 30 minutes? (Molar mass of Zn \(= 65.4\ \text{g mol}^{-1}\))
How many moles of electrons, and how many faradays of electricity, are required to reduce \(0.5\) mole of \(\mathrm{Cr^{3\oplus}}\) ions completely to chromium metal?
Two cells, one containing \(\mathrm{AgNO_3}\) solution and the other containing \(\mathrm{CuSO_4}\) solution, are connected in series. When \(2.0\ \text{g}\) of silver has been deposited in the first cell, calculate the mass of copper deposited in the second. (Molar masses: Ag \(= 108\ \text{g mol}^{-1}\), Cu \(= 63.5\ \text{g mol}^{-1}\))
Using the standard potentials \(E^0_{Cd} = -0.403\ \text{V}\) and \(E^0_{Ag} = +0.799\ \text{V}\), identify the anode and the cathode, calculate \(E^0_{cell}\), and write the cell notation for the galvanic cell built from these two electrodes.
Calculate the emf at \(25\ ^\circ\text{C}\) of the cell \(\mathrm{Zn}\text{(s)}\,|\,\mathrm{Zn^{2\oplus}}\text{(0.01M)}\,\|\,\mathrm{Ag^{\oplus}}\text{(0.01M)}\,|\,\mathrm{Ag}\text{(s)}\), given \(E^0_{Zn} = -0.763\ \text{V}\) and \(E^0_{Ag} = +0.799\ \text{V}\). Write the overall cell reaction and state the value of \(n\) you have used.
Calculate the potential of the electrode \(\mathrm{Cu^{2\oplus}}\text{(0.001M)}\,|\,\mathrm{Cu}\text{(s)}\) at \(25\ ^\circ\text{C}\), given \(E^0_{Cu} = 0.337\ \text{V}\).
A hydrogen gas electrode operates at \(25\ ^\circ\text{C}\) with hydrogen at \(1\ \text{atm}\) and a solution in which \([\mathrm{H^{\oplus}}] = 0.001\ \text{M}\). Calculate the potential of the electrode.
For a certain cell reaction in which 3 moles of electrons are transferred, \(E^0_{cell} = 0.72\ \text{V}\) at \(25\ ^\circ\text{C}\). Calculate \(\Delta G^0\) for the reaction and state whether it is spontaneous.
The standard cell potential of a cell reaction involving the transfer of 2 moles of electrons is \(0.46\ \text{V}\) at \(25\ ^\circ\text{C}\). Calculate the equilibrium constant of the cell reaction.
The equilibrium constant of a cell reaction in which one mole of electrons is transferred is \(2.0\times10^{4}\) at \(25\ ^\circ\text{C}\). Calculate \(E^0_{cell}\) and \(\Delta G^0\) for the reaction.
What is the standard cell potential for the reaction \(2\,\mathrm{Al}\text{(s)} + 3\,\mathrm{Ni^{2\oplus}}\text{(1M)} \rightarrow 2\,\mathrm{Al^{3\oplus}}\text{(1M)} + 3\,\mathrm{Ni}\text{(s)}\) if \(E^0_{Ni} = -0.25\ \text{V}\) and \(E^0_{Al} = -1.66\ \text{V}\)?
How many electrons would have a total charge of 1 coulomb?
How many moles of electrons are passed when \(0.8\) ampere current is passed for 1 hour through molten \(\mathrm{CaCl_2}\)?
Construct a galvanic cell from the electrodes \(\mathrm{Co^{3\oplus}}\,|\,\mathrm{Co}\) and \(\mathrm{Mn^{2\oplus}}\,|\,\mathrm{Mn}\). \(E^0_{Co} = 1.82\ \text{V}\), \(E^0_{Mn} = -1.18\ \text{V}\). Calculate \(E^0_{cell}\).
What current strength in amperes will be required to produce \(2.4\ \text{g}\) of Cu from \(\mathrm{CuSO_4}\) solution in 1 hour? Molar mass of Cu \(= 63.5\ \text{g mol}^{-1}\).
The equilibrium constant of the reaction \(2\,\mathrm{Cu^{\oplus}}\text{(aq)} \rightarrow \mathrm{Cu^{2\oplus}}\text{(aq)} + \mathrm{Cu}\text{(s)}\) is \(1.2\times10^{6}\). What is the standard potential of the cell in which the reaction takes place?
Calculate the emf of the cell \(\mathrm{Zn}\text{(s)}\,|\,\mathrm{Zn^{2\oplus}}\text{(0.2M)}\,\|\,\mathrm{H^{\oplus}}\text{(1.6M)}\,|\,\mathrm{H_2}\text{(g, 1.8 atm)}\,|\,\mathrm{Pt}\) at \(25\ ^\circ\text{C}\).
Calculate the emf of the following cell at \(25\ ^\circ\text{C}\): \(\mathrm{Zn}\text{(s)}\,|\,\mathrm{Zn^{2\oplus}}\text{(0.08M)}\,\|\,\mathrm{Cr^{3\oplus}}\text{(0.1M)}\,|\,\mathrm{Cr}\), given \(E^0_{Zn} = -0.76\ \text{V}\) and \(E^0_{Cr} = -0.74\ \text{V}\).