Chapter 7: Redox Reactions
For a long time, chemists believed that anything which burned contained a mysterious, invisible substance called “phlogiston,” which escaped into the air as the material burned away. It was a neat idea — until someone decided to actually weigh things before and after burning.
In the late 1700s, the French chemist Antoine Lavoisier did exactly that. He heated metals in sealed containers and measured their mass with great care. To everyone’s surprise, a metal did not lose weight when it turned to a powdery calx (what we now call an oxide) — it gained weight. Something from the air was joining the metal, not leaving it. Lavoisier identified that “something” as a gas he named oxygen, and he showed that burning is really a combination of a substance with oxygen.
This careful weighing quietly demolished the phlogiston theory and gave chemistry its first clear meaning of the word oxidation: the addition of oxygen. From that single idea grew a whole family of reactions we now call redox reactions — the reactions behind rusting iron, burning fuels, working batteries, extracting metals from their ores, and even the energy your own body releases from food.
In this chapter you will watch this one idea grow from “adding oxygen” into a far more powerful concept built on the transfer of electrons.
Fig. 7.0 – The evolution of oxidation from oxygen addition to electron loss.
By the end of this chapter you will be able to reason your way to the answer to each of these questions.
7.1 Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
Why Redox Reactions Matter
Chemistry is the study of matter and of the changes that turn one kind of matter into another. A huge share of those changes belong to a single important family called redox reactions. The word “redox” is a short form made by joining reduction and oxidation, because — as you will soon see — these two processes always happen together.
Redox reactions are everywhere. They are central to the pharmaceutical, biological, industrial, metallurgical and agricultural fields. Burning fuels for cooking, transport and electricity is a redox process; so is the electrochemical extraction of highly reactive metals and non-metals from their ores, the manufacture of chemicals such as caustic soda, the working of dry and wet batteries, and the slow corrosion of metals. More recent concerns such as the Hydrogen Economy (using liquid hydrogen as a clean fuel) and the chemistry of the Ozone Hole also fall within redox phenomena. In short, understanding redox reactions lets you make sense of energy, materials and much of the living world.
Oxidation as the Addition of Oxygen
Originally, the term oxidation was used in a very simple sense: the addition of oxygen to an element or a compound. This meaning grew naturally out of experience, because the atmosphere is roughly 20% dioxygen (\(\mathrm{O_2}\)), so a great many elements combine with it. This is the main reason that so many elements are found in the earth’s crust in the form of their oxides — compounds of an element with oxygen.
The following reactions are examples of oxidation according to this limited “addition of oxygen” definition. When magnesium metal burns, and when sulphur burns, each element takes up oxygen:
\[2\,\mathrm{Mg}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{MgO}\text{(s)} \qquad (7.1)\]
\[\mathrm{S}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{SO_2}\text{(g)} \qquad (7.2)\]
In both reactions the element on the left gains oxygen, so both magnesium and sulphur are said to be oxidised. Notice something that surprised early chemists: adding oxygen means the product actually contains more mass than the starting element. Oxidation, in this sense, is a gaining process, not a losing one.
Methane, the main component of cooking gas, is oxidised in the same “addition of oxygen” way when it burns:
\[\mathrm{CH_4}\text{(g)} + 2\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)} \qquad (7.3)\]
Broadening Oxidation: Removing Hydrogen
A careful look at reaction (7.3) sparked a new idea. In methane the carbon was originally bonded to hydrogen; after the reaction that hydrogen has been replaced by oxygen. This prompted chemists to reinterpret oxidation more widely to also include the removal of hydrogen from a substance.
The following reaction shows removal of hydrogen being treated as oxidation. Here hydrogen is stripped away from hydrogen sulphide, leaving free sulphur:
\[2\,\mathrm{H_2S}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{S}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \qquad (7.4)\]
Broadening Further: Electronegative and Electropositive Elements
As chemists learned more, it seemed natural to extend “oxidation” to reactions like (7.1)–(7.4) that do not involve oxygen at all but involve other electronegative elements. An electronegative element is one that pulls bonding electrons strongly towards itself — the most electronegative elements are those like fluorine, oxygen and chlorine near the top-right of the periodic table.
Magnesium, for instance, is oxidised not only by oxygen but also by fluorine, chlorine and sulphur:
\[\mathrm{Mg}\text{(s)} + \mathrm{F_2}\text{(g)} \rightarrow \mathrm{MgF_2}\text{(s)} \qquad (7.5)\]
\[\mathrm{Mg}\text{(s)} + \mathrm{Cl_2}\text{(g)} \rightarrow \mathrm{MgCl_2}\text{(s)} \qquad (7.6)\]
\[\mathrm{Mg}\text{(s)} + \mathrm{S}\text{(s)} \rightarrow \mathrm{MgS}\text{(s)} \qquad (7.7)\]
Bringing reactions (7.5)–(7.7) into the family of oxidation reactions encouraged chemists to go one step more. If adding an electronegative element counts as oxidation, then removing an electropositive element should count too. An electropositive element is one that gives up electrons easily — the metals, especially those on the left of the periodic table such as sodium and potassium.
Consider this reaction, in which potassium ferrocyanide is changed to potassium ferricyanide:
\[2\,\mathrm{K_4[Fe(CN)_6]}\text{(aq)} + \mathrm{H_2O_2}\text{(aq)} \rightarrow 2\,\mathrm{K_3[Fe(CN)_6]}\text{(aq)} + 2\,\mathrm{KOH}\text{(aq)}\]
This is interpreted as an oxidation, because the electropositive element potassium is being removed from the ferrocyanide as it turns into ferricyanide.
Putting all of this together gives the full classical meaning of oxidation.
The Meaning of Reduction
In the beginning, reduction simply meant the removal of oxygen from a compound — the exact opposite of the earliest idea of oxidation. In just the same way that oxidation was broadened, the term reduction has since been widened.
The following reactions are examples of reduction under this broadened definition. First, the removal of oxygen from mercuric oxide on heating:
\[2\,\mathrm{HgO}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Hg}\text{(l)} + \mathrm{O_2}\text{(g)} \qquad (7.8)\]
Next, the removal of the electronegative element chlorine from ferric chloride:
\[2\,\mathrm{FeCl_3}\text{(aq)} + \mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{FeCl_2}\text{(aq)} + 2\,\mathrm{HCl}\text{(aq)} \qquad (7.9)\]
The addition of hydrogen, as when ethene is converted to ethane:
\[\mathrm{CH_2}\!=\!\mathrm{CH_2}\text{(g)} + \mathrm{H_2}\text{(g)} \rightarrow \mathrm{H_3C}\!-\!\mathrm{CH_3}\text{(g)} \qquad (7.10)\]
And finally, the addition of the electropositive element mercury to mercuric chloride:
\[2\,\mathrm{HgCl_2}\text{(aq)} + \mathrm{SnCl_2}\text{(aq)} \rightarrow \mathrm{Hg_2Cl_2}\text{(s)} + \mathrm{SnCl_4}\text{(aq)} \qquad (7.11)\]
Why the Two Always Go Together
Reaction (7.11) reveals something important. While mercuric chloride is being reduced (mercury is added to it), the stannous chloride is at the very same time being oxidised to stannic chloride — because the electronegative element chlorine is being added to it. Reduction and oxidation are happening in the same reaction, at the same time.
If you re-examine every equation above, you will find the same pattern: wherever something is oxidised, something else is reduced. Because the two processes are inseparable, chemists coined the single word “redox” to describe this whole class of reactions.
7.2 Redox Reactions in Terms of Electron Transfer Reactions
Sodium Reactions Revisited
The classical definitions of the previous section work well, but they lean on ideas like “electronegative” and “electropositive” that are not always easy to apply. A deeper and more powerful way of looking at redox reactions comes from watching what happens to electrons.
We already know that the following reactions are redox reactions:
\[2\,\mathrm{Na}\text{(s)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{NaCl}\text{(s)} \qquad (7.12)\]
\[4\,\mathrm{Na}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{Na_2O}\text{(s)} \qquad (7.13)\]
\[2\,\mathrm{Na}\text{(s)} + \mathrm{S}\text{(s)} \rightarrow \mathrm{Na_2S}\text{(s)} \qquad (7.14)\]
In each one sodium is oxidised (oxygen or a more electronegative element is added to it), while chlorine, oxygen and sulphur are simultaneously reduced (the electropositive sodium is added to them).
From our study of chemical bonding we also know that sodium chloride, sodium oxide and sodium sulphide are ionic compounds. It is more revealing to write them showing their ions: \(\mathrm{Na^+Cl^-}\text{(s)}\), \(\mathrm{(Na^+)_2O^{2-}}\text{(s)}\) and \(\mathrm{(Na^+)_2S^{2-}}\text{(s)}\).
The appearance of these charges is the key clue. A neutral sodium atom becomes a \(\mathrm{Na^+}\) ion — it must have lost an electron. A neutral chlorine atom becomes a \(\mathrm{Cl^-}\) ion — it must have gained an electron. The formation of the ionic compound is really the transfer of electrons from one species to another.
Splitting a Reaction into Half Reactions
For convenience, each of these reactions can be split into two separate steps: one step in which electrons are lost, and another in which they are gained. Take the formation of sodium chloride as an illustration:
\[2\,\mathrm{Na}\text{(s)} \rightarrow 2\,\mathrm{Na^+}\text{(g)} + 2e^-\]
\[\mathrm{Cl_2}\text{(g)} + 2e^- \rightarrow 2\,\mathrm{Cl^-}\text{(g)}\]
Each of these steps is called a half reaction, because it shows only half of the story — it explicitly displays the electrons that are lost or gained. Adding the two half reactions together (the electrons cancel, since the same number is lost by sodium and gained by chlorine) gives back the overall reaction:
\[2\,\mathrm{Na}\text{(s)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{Na^+Cl^-}\text{(s)} \ \ \text{or}\ \ 2\,\mathrm{NaCl}\text{(s)}\]
This split leads directly to the modern, electron-based definitions of oxidation and reduction. A half reaction that involves the loss of electrons is an oxidation reaction; a half reaction that involves the gain of electrons is a reduction reaction.
A handy way to remember this is the phrase OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).
Oxidising and Reducing Agents
In reactions (7.12) to (7.14), sodium is the species that is oxidised. Because it donates its electrons to each of the elements it reacts with, it helps those elements to be reduced. A species that behaves this way is called a reducing agent. Chlorine, oxygen and sulphur, on the other hand, are reduced — they accept the electrons from sodium — and so they act as oxidising agents.
Notice the neat twist that often confuses students at first: the reducing agent is itself oxidised, and the oxidising agent is itself reduced. This makes sense once you track the electrons — the donor loses electrons (so it is oxidised) even as it causes the other species to be reduced.
7.2.1 Competitive Electron Transfer Reactions
Zinc versus Copper
Electron transfer does not need an ionic solid to form; it can happen right in solution, and we can watch it. Place a strip of metallic zinc in an aqueous solution of copper nitrate for about an hour, as shown in Fig. 7.1. You will notice that the strip becomes coated with reddish metallic copper, while the blue colour of the solution slowly disappears.
Fig. 7.1 – Redox reaction between zinc and aqueous copper nitrate.
The disappearing blue colour (which comes from \(\mathrm{Cu^{2+}}\) ions) shows that copper ions are leaving the solution. The presence of the newly formed \(\mathrm{Zn^{2+}}\) ions can be confirmed separately: if hydrogen sulphide gas is passed through the now-colourless solution and it is made alkaline with ammonia, white zinc sulphide, \(\mathrm{ZnS}\), appears.
The reaction taking place is:
\[\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad (7.15)\]
Zinc has lost electrons to become \(\mathrm{Zn^{2+}}\), so zinc is oxidised. If zinc is releasing electrons, something must accept them and be reduced — and that is the copper ion, which gains those electrons to become copper metal. We can display this explicitly as two half reactions:
\[\mathrm{Zn}\text{(s)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + 2e^- \quad\text{(oxidation)}\]
\[\mathrm{Cu^{2+}}\text{(aq)} + 2e^- \rightarrow \mathrm{Cu}\text{(s)} \quad\text{(reduction)}\]
Which Way Does the Reaction Go?
We can also ask how far reaction (7.15) proceeds — where its equilibrium lies. To test this, place a strip of metallic copper in a zinc sulphate solution and watch: no visible reaction occurs. To check even more carefully for any \(\mathrm{Cu^{2+}}\) that might have formed, we pass \(\mathrm{H_2S}\) gas through the solution, hoping to produce the black colour of cupric sulphide, \(\mathrm{CuS}\) — a test so sensitive that even a tiny trace would show up, because \(\mathrm{CuS}\) is extremely insoluble. The test fails; no \(\mathrm{Cu^{2+}}\) can be detected.
We therefore conclude that the equilibrium of reaction (7.15) lies strongly to the right — it greatly favours the products (copper metal and zinc ions) over the reactants.
Copper versus Silver, Cobalt versus Nickel
Let us extend the same idea to copper metal placed in silver nitrate solution, arranged as shown in Fig. 7.2. The solution slowly develops a blue colour, which signals the formation of \(\mathrm{Cu^{2+}}\) ions:
\[\mathrm{Cu}\text{(s)} + 2\,\mathrm{Ag^+}\text{(aq)} \rightarrow \mathrm{Cu^{2+}}\text{(aq)} + 2\,\mathrm{Ag}\text{(s)} \qquad (7.16)\]
Fig. 7.2 – Redox reaction between copper and aqueous silver nitrate.
Here copper is oxidised to \(\mathrm{Cu^{2+}}\) and silver ions are reduced to silver metal. As with the zinc–copper case, the equilibrium greatly favours the products, \(\mathrm{Cu^{2+}}\) and \(\mathrm{Ag}\).
By way of contrast, consider metallic cobalt placed in a nickel sulphate solution:
\[\mathrm{Co}\text{(s)} + \mathrm{Ni^{2+}}\text{(aq)} \rightarrow \mathrm{Co^{2+}}\text{(aq)} + \mathrm{Ni}\text{(s)} \qquad (7.17)\]
When this system reaches equilibrium, chemical tests reveal that both \(\mathrm{Ni^{2+}}\text{(aq)}\) and \(\mathrm{Co^{2+}}\text{(aq)}\) are present at moderate concentrations. In this case neither the reactants nor the products are strongly favoured — the “tug of war” for electrons is much more evenly matched.
Ordering Metals by Electron-releasing Power
This competition for the release of electrons is remarkably like the competition for the release of protons that we see among acids. Just as we can rank acids by their strength, we can imagine ranking metals by their tendency to release electrons.
In fact we have already made the key comparisons. Zinc releases electrons to copper (reaction 7.15), and copper releases electrons to silver (reaction 7.16). So the electron-releasing tendency of these three metals runs in the order:
\[\mathrm{Zn} > \mathrm{Cu} > \mathrm{Ag}\]
If we extend this ranking to many metals, we build what is called the metal activity series or electrochemical series. The competition for electrons between different metals is also what lets us design a whole class of cells, called Galvanic cells, in which a redox reaction becomes a source of electrical energy. You will study these cells, and the full electrochemical series, in more detail in Class XII (and their electrode side is introduced in Section 7.4).
7.3 Oxidation Number
When Electron Transfer Is Not Obvious
The electron-transfer picture of Section 7.2 works beautifully for ionic reactions, where whole electrons move from one atom to another. But many redox reactions form covalent compounds, where the electrons are shared, not fully handed over. We need a bookkeeping tool that works even then. That tool is the oxidation number.
Consider the formation of water:
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} \qquad (7.18)\]
Though it is not obvious, we can picture the hydrogen atom going from a neutral (zero) state in \(\mathrm{H_2}\) to a positive state in \(\mathrm{H_2O}\), while the oxygen atom goes from a zero state in \(\mathrm{O_2}\) to a doubly-negative state in \(\mathrm{H_2O}\). Treating this as an electron transfer from H to O, we say \(\mathrm{H_2}\) is oxidised and \(\mathrm{O_2}\) is reduced.
In reality the charge transfer here is only partial — it is better described as an electron shift towards the more electronegative oxygen than a complete loss by hydrogen. The same is true for many reactions of covalent compounds, for example:
\[\mathrm{H_2}\text{(g)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{HCl}\text{(g)} \qquad (7.19)\]
\[\mathrm{CH_4}\text{(g)} + 4\,\mathrm{Cl_2}\text{(g)} \rightarrow \mathrm{CCl_4}\text{(l)} + 4\,\mathrm{HCl}\text{(g)} \qquad (7.20)\]
To keep track of electron shifts in such reactions, chemists developed the more practical method of oxidation number. In this method we deliberately pretend that the electron pair in every bond is transferred completely to the more electronegative atom — a convenient fiction used purely for bookkeeping. Doing so lets us assign a charge to each atom. Rewriting equations (7.18)–(7.20) this way (labelled 7.21–7.23):
In (7.21), each H is \(+1\) and O is \(-2\) in water (both are \(0\) as free elements). In (7.22), H is \(+1\) and Cl is \(-1\) in \(\mathrm{HCl}\). In (7.23), carbon is \(-4\) and hydrogen \(+1\) in \(\mathrm{CH_4}\), carbon \(+4\) and chlorine \(-1\) in \(\mathrm{CCl_4}\), and H \(+1\), Cl \(-1\) in \(\mathrm{HCl}\).
Remember that this assumed complete transfer is a bookkeeping device only; it will soon give us a very simple, unified description of all redox reactions.
Rules for Finding Oxidation Number
The oxidation number denotes the oxidation state of an element in a compound, worked out from a set of rules built on the idea that the shared electron pair in a covalent bond is assigned entirely to the more electronegative element.
It is not always easy to remember which element in a compound or ion is the more electronegative, so a set of rules is used instead. When two or more atoms of the same element appear in a species (as in \(\mathrm{Na_2S_2O_3}\) or \(\mathrm{Cr_2O_7^{2-}}\)), the oxidation number obtained is the average over all atoms of that element. The rules are:
In the free (uncombined) state, each atom of an element has an oxidation number of zero. Thus every atom in \(\mathrm{H_2}\), \(\mathrm{O_2}\), \(\mathrm{Cl_2}\), \(\mathrm{O_3}\), \(\mathrm{P_4}\), \(\mathrm{S_8}\), \(\mathrm{Na}\), \(\mathrm{Mg}\) and \(\mathrm{Al}\) has oxidation number zero.
For a monatomic ion, the oxidation number equals the charge on the ion. So \(\mathrm{Na^+}\) is \(+1\), \(\mathrm{Mg^{2+}}\) is \(+2\), \(\mathrm{Fe^{3+}}\) is \(+3\), \(\mathrm{Cl^-}\) is \(-1\), \(\mathrm{O^{2-}}\) is \(-2\), and so on. In their compounds all alkali metals have \(+1\) and all alkaline earth metals \(+2\); aluminium is taken as \(+3\) in all its compounds.
Oxygen usually has an oxidation number of \(-2\). There are two kinds of exceptions. In peroxides (e.g. \(\mathrm{H_2O_2}\), \(\mathrm{Na_2O_2}\)), where oxygen atoms are joined to each other, each oxygen is \(-1\); in superoxides (e.g. \(\mathrm{KO_2}\), \(\mathrm{RbO_2}\)) each oxygen is \(-\tfrac{1}{2}\). The rarer exception is when oxygen is bonded to fluorine: in oxygen difluoride \(\mathrm{OF_2}\) oxygen is \(+2\) and in dioxygen difluoride \(\mathrm{O_2F_2}\) it is \(+1\). (Because fluorine is more electronegative than oxygen, here the number assigned to oxygen becomes positive.)
Hydrogen has an oxidation number of \(+1\), except when it is bonded to metals in binary (two-element) compounds. In metal hydrides such as \(\mathrm{LiH}\), \(\mathrm{NaH}\) and \(\mathrm{CaH_2}\), hydrogen is \(-1\).
Fluorine always has an oxidation number of \(-1\). The other halogens (Cl, Br, I) are also \(-1\) when they occur as halide ions; but when combined with oxygen, as in oxoacids and oxoanions, chlorine, bromine and iodine take positive oxidation numbers.
The algebraic sum of the oxidation numbers of all atoms in a neutral compound is zero. In a polyatomic ion, that sum equals the charge on the ion. For example, in the carbonate ion \(\mathrm{(CO_3)^{2-}}\), the three oxygens and one carbon must add up to \(-2\).
Applying these rules, we find that metallic elements have positive oxidation numbers, non-metals can be positive or negative, and transition elements commonly show several positive oxidation states. The highest oxidation number of a representative element equals its group number for the first two groups, and the group number minus 10 (long form of the periodic table) for the others. This means the highest oxidation number generally increases across a period; in the third period it rises steadily from \(+1\) to \(+7\), as the table below shows.
Table 7.1a: Highest oxidation state of third-period elements
| Group | 1 | 2 | 13 | 14 | 15 | 16 | 17 |
|---|---|---|---|---|---|---|---|
| Element | Na | Mg | Al | Si | P | S | Cl |
| Compound | \(\mathrm{NaCl}\) | \(\mathrm{MgSO_4}\) | \(\mathrm{AlF_3}\) | \(\mathrm{SiCl_4}\) | \(\mathrm{P_4O_{10}}\) | \(\mathrm{SF_6}\) | \(\mathrm{HClO_4}\) |
| Highest oxidation state | \(+1\) | \(+2\) | \(+3\) | \(+4\) | \(+5\) | \(+6\) | \(+7\) |
The term oxidation state is used interchangeably with oxidation number. Thus in \(\mathrm{CO_2}\) the oxidation state of carbon is \(+4\) (also its oxidation number) and the oxidation state of oxygen is \(-2\).
Stock Notation
The oxidation state of a metal in a compound is often shown using a notation introduced by the German chemist Alfred Stock, popularly called Stock notation. Here the oxidation number is written as a Roman numeral in parentheses right after the symbol of the metal.
Thus aurous chloride and auric chloride are written as \(\mathrm{Au(I)Cl}\) and \(\mathrm{Au(III)Cl_3}\); stannous and stannic chloride become \(\mathrm{Sn(II)Cl_2}\) and \(\mathrm{Sn(IV)Cl_4}\). Because the Roman numeral makes the oxidation state explicit, it tells us at a glance whether a species is in its oxidised or reduced form. For example, \(\mathrm{Hg_2(I)Cl_2}\) is the reduced form of \(\mathrm{Hg(II)Cl_2}\).
Redox Terms Redefined Through Oxidation Number
The idea of oxidation number lets us restate all the key redox terms in one simple, unified way, based purely on how oxidation numbers change.
7.3.1 Types of Redox Reactions
Redox reactions can be sorted into four broad classes: combination, decomposition, displacement and disproportionation. The overall test for each is summarised in FC 7.3, and the classes are compared in the table further below.
Flowchart
FC 7.3 – Classifying a redox reaction: Is it two species joining into one, with at least one an element (A + B → C)? → combination. Is it one compound breaking into two or more parts, at least one an element? → decomposition. Is an atom/ion in a compound replaced by an atom of another element (X + YZ → XZ + Y)? → displacement (metal or non-metal). Is a single element in one intermediate oxidation state both raised and lowered? → disproportionation.
Combination Reactions
A combination reaction is written as \(\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}\). For it to be a redox reaction, either A or B, or both, must be in the elemental form. All combustion reactions (which use elemental dioxygen) are redox, as are many other combinations involving free elements. Examples:
\[\mathrm{C}\text{(s)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{CO_2}\text{(g)} \qquad (7.24)\]
Here carbon goes from \(0\) to \(+4\) and oxygen from \(0\) to \(-2\).
\[3\,\mathrm{Mg}\text{(s)} + \mathrm{N_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{Mg_3N_2}\text{(s)} \qquad (7.25)\]
Here magnesium goes from \(0\) to \(+2\) and nitrogen from \(0\) to \(-3\).
\[\mathrm{CH_4}\text{(g)} + 2\,\mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{CO_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)}\]
In the combustion of methane, carbon goes from \(-4\) to \(+4\) and oxygen from \(0\) to \(-2\); notice the hydrogen stays at \(+1\) throughout.
Decomposition Reactions
Decomposition reactions are the opposite of combination: a single compound breaks down into two or more components, at least one of which must be an element. Examples:
\[2\,\mathrm{H_2O}\text{(l)} \xrightarrow{\ \Delta\ } 2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \qquad (7.26)\]
\[2\,\mathrm{NaH}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{Na}\text{(s)} + \mathrm{H_2}\text{(g)} \qquad (7.27)\]
\[2\,\mathrm{KClO_3}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)} \qquad (7.28)\]
It is worth noting that in reaction (7.28) the oxidation number of potassium does not change (it stays \(+1\)), just as the hydrogen in methane’s combination did not change; the redox action lies elsewhere in the molecule.
Not every decomposition is a redox reaction. When calcium carbonate is heated, no oxidation number changes at all, so it is not a redox reaction:
\[\mathrm{CaCO_3}\text{(s)} \xrightarrow{\ \Delta\ } \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\]
Here calcium stays \(+2\), carbon stays \(+4\) and oxygen stays \(-2\) on both sides.
Displacement Reactions
In a displacement reaction, an ion (or atom) in a compound is replaced by an ion (or atom) of another element: \(\mathrm{X} + \mathrm{YZ} \rightarrow \mathrm{XZ} + \mathrm{Y}\). These fall into two categories — metal displacement and non-metal displacement.
(a) Metal displacement. A metal in a compound can be displaced by another (uncombined) metal. We already met this idea in Section 7.2.1. Metal displacement is widely used in metallurgy to obtain pure metals from their compounds. Some examples:
\[\mathrm{CuSO_4}\text{(aq)} + \mathrm{Zn}\text{(s)} \rightarrow \mathrm{Cu}\text{(s)} + \mathrm{ZnSO_4}\text{(aq)} \qquad (7.29)\]
\[\mathrm{V_2O_5}\text{(s)} + 5\,\mathrm{Ca}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{V}\text{(s)} + 5\,\mathrm{CaO}\text{(s)} \qquad (7.30)\]
\[\mathrm{TiCl_4}\text{(l)} + 2\,\mathrm{Mg}\text{(s)} \xrightarrow{\ \Delta\ } \mathrm{Ti}\text{(s)} + 2\,\mathrm{MgCl_2}\text{(s)} \qquad (7.31)\]
\[\mathrm{Cr_2O_3}\text{(s)} + 2\,\mathrm{Al}\text{(s)} \xrightarrow{\ \Delta\ } \mathrm{Al_2O_3}\text{(s)} + 2\,\mathrm{Cr}\text{(s)} \qquad (7.32)\]
In each case the reducing metal is a better reducing agent than the metal it frees — that is, it loses electrons more readily than the one being reduced.
(b) Non-metal displacement. These include hydrogen displacement (common) and a rare oxygen displacement.
All alkali metals and some alkaline earth metals (Ca, Sr, Ba), being very good reductants, displace hydrogen from cold water:
\[2\,\mathrm{Na}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{NaOH}\text{(aq)} + \mathrm{H_2}\text{(g)} \qquad (7.33)\]
\[\mathrm{Ca}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{Ca(OH)_2}\text{(aq)} + \mathrm{H_2}\text{(g)} \qquad (7.34)\]
Less active metals such as magnesium and iron react only with steam:
\[\mathrm{Mg}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \xrightarrow{\ \Delta\ } \mathrm{Mg(OH)_2}\text{(s)} + \mathrm{H_2}\text{(g)} \qquad (7.35)\]
\[2\,\mathrm{Fe}\text{(s)} + 3\,\mathrm{H_2O}\text{(l)} \xrightarrow{\ \Delta\ } \mathrm{Fe_2O_3}\text{(s)} + 3\,\mathrm{H_2}\text{(g)} \qquad (7.36)\]
Many metals, including some that do not react with cold water or even steam (such as cadmium and tin), can still displace hydrogen from acids:
\[\mathrm{Zn}\text{(s)} + 2\,\mathrm{HCl}\text{(aq)} \rightarrow \mathrm{ZnCl_2}\text{(aq)} + \mathrm{H_2}\text{(g)} \qquad (7.37)\]
\[\mathrm{Mg}\text{(s)} + 2\,\mathrm{HCl}\text{(aq)} \rightarrow \mathrm{MgCl_2}\text{(aq)} + \mathrm{H_2}\text{(g)} \qquad (7.38)\]
\[\mathrm{Fe}\text{(s)} + 2\,\mathrm{HCl}\text{(aq)} \rightarrow \mathrm{FeCl_2}\text{(aq)} + \mathrm{H_2}\text{(g)} \qquad (7.39)\]
Reactions (7.37)–(7.39) are used to prepare hydrogen in the laboratory. The reactivity of the metal shows up in the rate of hydrogen evolution — slowest for the least active metal, iron, and fastest for the most reactive, magnesium. Very unreactive metals such as silver and gold, which often occur native, do not react even with hydrochloric acid.
As we saw in Section 7.2.1, the reducing tendency of these metals runs \(\mathrm{Zn} > \mathrm{Cu} > \mathrm{Ag}\). An activity series exists for the halogens too. Their power as oxidising agents decreases as we go down group 17 from fluorine to iodine. Fluorine is so reactive that it can even attack water and displace its oxygen:
\[2\,\mathrm{H_2O}\text{(l)} + 2\,\mathrm{F_2}\text{(g)} \rightarrow 4\,\mathrm{HF}\text{(aq)} + \mathrm{O_2}\text{(g)} \qquad (7.40)\]
Because of this, fluorine’s displacement reactions are not usually carried out in water. Chlorine, however, can displace bromide and iodide ions in aqueous solution:
\[\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{KBr}\text{(aq)} \rightarrow 2\,\mathrm{KCl}\text{(aq)} + \mathrm{Br_2}\text{(l)} \qquad (7.41)\]
\[\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{KI}\text{(aq)} \rightarrow 2\,\mathrm{KCl}\text{(aq)} + \mathrm{I_2}\text{(s)} \qquad (7.42)\]
Since \(\mathrm{Br_2}\) and \(\mathrm{I_2}\) are coloured and dissolve in \(\mathrm{CCl_4}\), they can be spotted by the colour of the solution. In ionic form:
\[\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{Br^-}\text{(aq)} \rightarrow 2\,\mathrm{Cl^-}\text{(aq)} + \mathrm{Br_2}\text{(l)} \qquad (7.41a)\]
\[\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{I^-}\text{(aq)} \rightarrow 2\,\mathrm{Cl^-}\text{(aq)} + \mathrm{I_2}\text{(s)} \qquad (7.42b)\]
Reactions (7.41) and (7.42) are the basis of the laboratory ‘Layer Test’ for identifying \(\mathrm{Br^-}\) and \(\mathrm{I^-}\). Likewise, bromine can displace iodide:
\[\mathrm{Br_2}\text{(l)} + 2\,\mathrm{I^-}\text{(aq)} \rightarrow 2\,\mathrm{Br^-}\text{(aq)} + \mathrm{I_2}\text{(s)} \qquad (7.43)\]
Halogen displacement has a direct industrial use: recovering a halogen from its halide requires an oxidation,
\[2\,\mathrm{X^-} \rightarrow \mathrm{X_2} + 2e^- \qquad (7.44)\]
where X is a halogen. Chemical oxidants are available to oxidise \(\mathrm{Cl^-}\), \(\mathrm{Br^-}\) and \(\mathrm{I^-}\), but since fluorine is the strongest oxidising agent of all, there is no chemical reagent able to oxidise \(\mathrm{F^-}\) to \(\mathrm{F_2}\); fluorine can be obtained from fluoride only by electrolytic oxidation, which you will study later.
Disproportionation Reactions
Disproportionation is a special type of redox reaction in which the same element, in one oxidation state, is simultaneously oxidised and reduced. The reacting substance always contains an element that can exist in at least three oxidation states; that element starts in the intermediate state, and both a higher and a lower oxidation state of it are produced.
The decomposition of hydrogen peroxide is a familiar example, where oxygen disproportionates:
\[2\,\mathrm{H_2O_2}\text{(aq)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)} \qquad (7.45)\]
Here the peroxide oxygen, which starts in the \(-1\) state, is partly raised to \(0\) (in \(\mathrm{O_2}\)) and partly lowered to \(-2\) (in \(\mathrm{H_2O}\)).
Phosphorus, sulphur and chlorine also disproportionate in alkaline medium:
\[\mathrm{P_4}\text{(s)} + 3\,\mathrm{OH^-}\text{(aq)} + 3\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{PH_3}\text{(g)} + 3\,\mathrm{H_2PO_2^-}\text{(aq)} \qquad (7.46)\]
Phosphorus goes from \(0\) down to \(-3\) (in \(\mathrm{PH_3}\)) and up to \(+1\) (in \(\mathrm{H_2PO_2^-}\)).
\[\mathrm{S_8}\text{(s)} + 12\,\mathrm{OH^-}\text{(aq)} \rightarrow 4\,\mathrm{S^{2-}}\text{(aq)} + 2\,\mathrm{S_2O_3^{2-}}\text{(aq)} + 6\,\mathrm{H_2O}\text{(l)} \qquad (7.47)\]
Sulphur goes from \(0\) down to \(-2\) and up to \(+2\).
\[\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{ClO^-}\text{(aq)} + \mathrm{Cl^-}\text{(aq)} + \mathrm{H_2O}\text{(l)} \qquad (7.48)\]
Chlorine goes from \(0\) up to \(+1\) (in \(\mathrm{ClO^-}\)) and down to \(-1\) (in \(\mathrm{Cl^-}\)).
Bromine and iodine follow the same trend as chlorine in reaction (7.48), but fluorine behaves differently with alkali:
\[2\,\mathrm{F_2}\text{(g)} + 2\,\mathrm{OH^-}\text{(aq)} \rightarrow 2\,\mathrm{F^-}\text{(aq)} + \mathrm{OF_2}\text{(g)} + \mathrm{H_2O}\text{(l)} \qquad (7.49)\]
(Fluorine will also attack the water itself to release some oxygen.) This departure is expected: being the most electronegative element, fluorine cannot show any positive oxidation state, so among the halogens fluorine does not disproportionate.
The Paradox of Fractional Oxidation Number
Sometimes we meet compounds in which the oxidation number of an element comes out as a fraction. For example, carbon is \(\tfrac{4}{3}\) in \(\mathrm{C_3O_2}\), bromine is \(\tfrac{16}{3}\) in \(\mathrm{Br_3O_8}\), and sulphur is \(2.5\) in \(\mathrm{Na_2S_4O_6}\).
A fractional oxidation number is puzzling, because electrons are never shared or transferred in fractions. The resolution is that a fractional value is only the average oxidation state of that element; the true structure shows the atoms sitting in different whole-number states. The bonding pictures are drawn in Fig. 7.3a, Fig. 7.3b and Fig. 7.3c.
Fig. 7.3a – Structure and oxidation states of carbon suboxide.
Fig. 7.3b – Structure and oxidation states of tribromooctaoxide.
Fig. 7.3c – Structure and oxidation states of the tetrathionate ion.
In \(\mathrm{C_3O_2}\), two carbons are \(+2\) each and the middle one is \(0\), so the average is \(\tfrac{4}{3}\). In \(\mathrm{Br_3O_8}\), the two terminal bromines are \(+6\) and the middle one is \(+4\), averaging \(\tfrac{16}{3}\). In \(\mathrm{S_4O_6^{2-}}\), the two outer sulphurs are \(+5\) and the two inner ones \(0\), averaging \(2.5\).
So whenever a fractional oxidation state turns up, understand it as an average only; the reality (revealed by structure) is a set of whole-number states. Mixed oxides such as \(\mathrm{Fe_3O_4}\), \(\mathrm{Mn_3O_4}\) and \(\mathrm{Pb_3O_4}\) are further examples where the metal shows more than one whole-number state. The oxidation state can even be a genuine fraction for the ions \(\mathrm{O_2^+}\) and \(\mathrm{O_2^-}\), where oxygen is \(+\tfrac{1}{2}\) and \(-\tfrac{1}{2}\) respectively.
7.3.2 Balancing of Redox Reactions
Two methods are used to balance redox equations. One is based on the change in oxidation number of the reducing and oxidising agents; the other splits the reaction into two half reactions — one oxidation, one reduction — and is called the ion–electron (half reaction) method. Both are in use, and the choice is a matter of preference.
(a) Oxidation Number Method
To balance a redox equation by the oxidation number method, we use the following steps, summarised in FC 7.1.
Step 1: Write the correct formula for each reactant and product.
Step 2: Identify the atoms that change oxidation number by assigning oxidation numbers to all elements.
Step 3: Calculate the increase and the decrease in oxidation number per atom, and for the whole molecule or ion. If these are not equal, multiply by suitable numbers to make them equal. (If you find that two substances are reduced and nothing oxidised, or the reverse, something is wrong — either a formula is incorrect or the oxidation numbers were not assigned properly.)
Step 4: If the reaction takes place in water, balance the ionic charges by adding \(\mathrm{H^+}\) ions (in acidic solution) or \(\mathrm{OH^-}\) ions (in basic solution) to the appropriate side.
Step 5: Make the number of hydrogen atoms equal by adding water (\(\mathrm{H_2O}\)) molecules, then check that the oxygen atoms also balance. If they do, the equation is balanced.
Flowchart
FC 7.1 – Oxidation number method: Step 1 write correct formulae → Step 2 assign oxidation numbers, mark atoms that change → Step 3 find total increase and total decrease in oxidation number, multiply to equalise → Step 4 add H⁺ (acidic) or OH⁻ (basic) to balance charge → Step 5 add H₂O to balance H, then check O → balanced equation.
(b) Half Reaction (Ion–Electron) Method
In this method the two half equations are balanced separately and then added to give the balanced overall equation. The full procedure is summarised in FC 7.2. As an example, we balance the oxidation of \(\mathrm{Fe^{2+}}\) to \(\mathrm{Fe^{3+}}\) by dichromate ions in acidic medium, where \(\mathrm{Cr_2O_7^{2-}}\) is reduced to \(\mathrm{Cr^{3+}}\).
Step 1: Write the unbalanced ionic equation: \[\mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cr_2O_7^{2-}}\text{(aq)} \rightarrow \mathrm{Fe^{3+}}\text{(aq)} + \mathrm{Cr^{3+}}\text{(aq)} \qquad (7.50)\]
Step 2: Separate into half reactions (iron rises \(+2 \to +3\); chromium falls \(+6 \to +3\)): \[\text{Oxidation half: } \mathrm{Fe^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{3+}}\text{(aq)} \qquad (7.51)\] \[\text{Reduction half: } \mathrm{Cr_2O_7^{2-}}\text{(aq)} \rightarrow \mathrm{Cr^{3+}}\text{(aq)} \qquad (7.52)\]
Step 3: Balance atoms other than O and H. The oxidation half is already balanced in Fe. For the reduction half, multiply \(\mathrm{Cr^{3+}}\) by \(2\) to balance Cr: \[\mathrm{Cr_2O_7^{2-}}\text{(aq)} \rightarrow 2\,\mathrm{Cr^{3+}}\text{(aq)} \qquad (7.53)\]
Step 4: For an acidic medium, add \(\mathrm{H_2O}\) to balance O and \(\mathrm{H^+}\) to balance H: \[\mathrm{Cr_2O_7^{2-}}\text{(aq)} + 14\,\mathrm{H^+}\text{(aq)} \rightarrow 2\,\mathrm{Cr^{3+}}\text{(aq)} + 7\,\mathrm{H_2O}\text{(l)} \qquad (7.54)\]
Step 5: Add electrons to one side of each half reaction to balance charge. The oxidation half becomes: \[\mathrm{Fe^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{3+}}\text{(aq)} + e^- \qquad (7.55)\] In the reduction half there are net twelve positive charges on the left and only six on the right, so add six electrons on the left: \[\mathrm{Cr_2O_7^{2-}}\text{(aq)} + 14\,\mathrm{H^+}\text{(aq)} + 6e^- \rightarrow 2\,\mathrm{Cr^{3+}}\text{(aq)} + 7\,\mathrm{H_2O}\text{(l)} \qquad (7.56)\] To equalise the electrons, multiply the oxidation half by \(6\): \[6\,\mathrm{Fe^{2+}}\text{(aq)} \rightarrow 6\,\mathrm{Fe^{3+}}\text{(aq)} + 6e^- \qquad (7.57)\]
Step 6: Add the two half reactions and cancel the electrons, giving the net ionic equation: \[6\,\mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cr_2O_7^{2-}}\text{(aq)} + 14\,\mathrm{H^+}\text{(aq)} \rightarrow 6\,\mathrm{Fe^{3+}}\text{(aq)} + 2\,\mathrm{Cr^{3+}}\text{(aq)} + 7\,\mathrm{H_2O}\text{(l)} \qquad (7.58)\]
Step 7: Verify that the same type and number of atoms and the same total charge appear on both sides. This check confirms the equation is fully balanced.
For a reaction in basic medium, first balance it as if it were acidic. Then, for each \(\mathrm{H^+}\) ion, add an equal number of \(\mathrm{OH^-}\) ions to both sides. Wherever \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) end up on the same side, combine them into \(\mathrm{H_2O}\).
Flowchart
FC 7.2 – Half-reaction (ion–electron) method: write skeletal ionic equation → split into oxidation half and reduction half → balance atoms other than O and H → (acidic) add H₂O for O and H⁺ for H → add electrons to balance charge in each half → multiply halves to make electrons equal → add halves and cancel electrons → (if basic) add OH⁻ to each side per H⁺ and combine H⁺ + OH⁻ into H₂O → verify atoms and charge.
7.3.3 Redox Reactions as the Basis for Titrations
Just as acid–base titrations use a pH-sensitive indicator to find the strength of one solution against another, redox titrations use a redox-sensitive indicator to find the strength of a reductant or an oxidant. Indicators are used in three different ways.
- Self-indicator. Sometimes the reagent is itself intensely coloured, so it needs no separate indicator. The best example is permanganate ion, \(\mathrm{MnO_4^-}\), which is deep purple. Here \(\mathrm{MnO_4^-}\) acts as its own indicator: the end point is reached when the last of the reductant (such as \(\mathrm{Fe^{2+}}\) or \(\mathrm{C_2O_4^{2-}}\)) has been oxidised, and the very next drop leaves a first lasting tinge of pink. This pink shows up at a \(\mathrm{MnO_4^-}\) concentration as low as \(10^{-6}\ \text{mol dm}^{-3}\) (\(10^{-6}\ \text{mol L}^{-1}\)), so the “overshoot” beyond the equivalence point (where reductant and oxidant are present in their exact mole ratio) is minimal.
- Indicator oxidised sharply at the end. If there is no dramatic auto-colour change, an indicator is used that is oxidised right after the last of the reactant is consumed, producing a sudden colour change. The best example uses \(\mathrm{Cr_2O_7^{2-}}\), which is not a self-indicator but oxidises the added substance diphenylamine just after the equivalence point to give an intense blue, signalling the end point.
- Iodine method. A third, very common approach is used for reagents that can oxidise \(\mathrm{I^-}\) ions — for example copper(II):
\[2\,\mathrm{Cu^{2+}}\text{(aq)} + 4\,\mathrm{I^-}\text{(aq)} \rightarrow \mathrm{Cu_2I_2}\text{(s)} + \mathrm{I_2}\text{(aq)} \qquad (7.59)\]
This method relies on two facts: iodine gives an intense blue colour with starch, and iodine reacts very specifically (and redox-wise) with thiosulphate ions, \(\mathrm{S_2O_3^{2-}}\):
\[\mathrm{I_2}\text{(aq)} + 2\,\mathrm{S_2O_3^{2-}}\text{(aq)} \rightarrow 2\,\mathrm{I^-}\text{(aq)} + \mathrm{S_4O_6^{2-}}\text{(aq)} \qquad (7.60)\]
Iodine, though insoluble in water, stays in solution in the presence of \(\mathrm{KI}\) as \(\mathrm{KI_3}\). When starch is added after iodine has been liberated (by \(\mathrm{Cu^{2+}}\) acting on \(\mathrm{I^-}\)), an intense blue appears; this colour vanishes the instant the iodine is used up by the thiosulphate. The end point is thus easy to track, and the rest is a stoichiometric calculation.
7.3.4 Limitations of the Concept of Oxidation Number
As you have seen, the concept of redox has kept evolving with time, and it continues to do so. In more recent thinking, the oxidation process is pictured as a decrease in electron density, and the reduction process as an increase in electron density, around the atom or atoms taking part in the reaction. This finer, density-based view goes beyond the whole-number bookkeeping of oxidation numbers, which is only an approximate model.
7.4 Redox Reactions and Electrode Processes
Direct versus Indirect Electron Transfer
We can revisit the reaction of Section 7.2.1 in a slightly different form. If a zinc rod is dipped in copper sulphate solution, the same redox reaction takes place: zinc is oxidised to zinc ions and copper ions are reduced to metallic copper, because electrons pass directly from the zinc to the copper ions. During this reaction heat is also given off — the energy is released as warmth and simply lost to the surroundings.
Now let us cleverly modify the experiment so that the same redox reaction runs with the electrons travelling indirectly, through an external wire, where their flow can be put to use. To do this, the zinc metal must be kept apart from the copper sulphate solution.
We take copper sulphate solution in one beaker with a copper rod in it, and zinc sulphate solution in a separate beaker with a zinc rod in it. At the interface of each metal and its own salt solution, a reaction occurs, and both the reduced form and the oxidised form of that species are present together. These two forms are exactly the species that appear in a reduction or an oxidation half reaction. The pairing of the two forms has a special name — a redox couple.
A redox couple is written by separating the oxidised form from the reduced form with a vertical line or a slash, which stands for the interface (for example, solid touching solution), with the oxidised form written first. In this experiment the two redox couples are \(\mathrm{Zn^{2+}}/\mathrm{Zn}\) and \(\mathrm{Cu^{2+}}/\mathrm{Cu}\).
The Daniell Cell
Now we place the two beakers side by side and connect them, as shown in Fig. 7.4. The electrical connection between the solutions is made by a salt bridge — a U-tube filled with a solution of potassium chloride or ammonium nitrate, set into a jelly by boiling with agar agar and cooling. The salt bridge lets charge pass between the two solutions without letting them mix. The zinc and copper rods are joined by a metallic wire fitted with an ammeter and a switch. This whole set-up is called a Daniell cell.
Fig. 7.4 – The Daniell cell and its electrode processes.
When the switch is in the off position, no reaction takes place in either beaker and no current flows through the wire. The interesting behaviour begins the moment the switch is turned on.
What We Observe When the Switch Is On
Two things happen once the circuit is closed.
First, the transfer of electrons no longer goes directly from Zn to \(\mathrm{Cu^{2+}}\). Instead, the electrons released as zinc is oxidised travel through the metallic wire connecting the two rods, which is why the ammeter registers a flow of current. The electrode where oxidation happens (the zinc rod, which sends out electrons) is called the anode and is the negative terminal; the electrode where reduction happens (the copper rod, where \(\mathrm{Cu^{2+}}\) picks up electrons) is called the cathode and is the positive terminal.
Second, the circuit is completed inside the cell: electricity passes from the solution in one beaker to the solution in the other by the migration of ions through the salt bridge. A current can flow only because there is a potential difference between the copper and zinc rods, which act as electrodes here.
Flowchart
FC 7.4 – Working of the Daniell cell (one full cycle): Zn oxidised at the anode (Zn → Zn²⁺ + 2e⁻) → electrons enter the external wire → flow through the wire and ammeter to the cathode → at the cathode Cu²⁺ + 2e⁻ → Cu (copper deposits) → meanwhile ions migrate through the salt bridge to keep both solutions neutral → circuit complete, current continues until reactants are used up.
Electrode Potential and Standard Electrode Potential
The potential associated with each electrode is called its electrode potential. To compare electrodes fairly, chemists fix a standard set of conditions.
Since a single electrode’s potential cannot be measured in isolation, one electrode is chosen as a reference. By convention, the standard electrode potential of the hydrogen electrode is taken as exactly \(0.00\) volts. Every other electrode potential is then measured against it.
The value of \(E^{\ominus}\) for an electrode process is a measure of the relative tendency of the active species to stay in its oxidised or reduced form. A negative \(E^{\ominus}\) means the redox couple is a stronger reducing agent than the \(\mathrm{H^+}/\mathrm{H_2}\) couple; a positive \(E^{\ominus}\) means it is a weaker reducing agent than the \(\mathrm{H^+}/\mathrm{H_2}\) couple.
The standard electrode potentials are very important, and a great deal of other useful information can be drawn from them. The values for some selected electrode processes (written as reduction reactions) are collected in Table 7.1. You will study electrode reactions and cells in much more detail in Class XII.
Table 7.1: The Standard Electrode Potentials at 298 K (Ions are aqueous species and \(\mathrm{H_2O}\) is liquid; gases and solids are shown by (g) and (s).)
| Reaction (Oxidised form + \(ne^-\) → Reduced form) | \(E^{\ominus}\,/\,\text{V}\) |
|---|---|
| \(\mathrm{F_2}\text{(g)} + 2e^- \rightarrow 2\,\mathrm{F^-}\) | \(2.87\) |
| \(\mathrm{Co^{3+}} + e^- \rightarrow \mathrm{Co^{2+}}\) | \(1.81\) |
| \(\mathrm{H_2O_2} + 2\,\mathrm{H^+} + 2e^- \rightarrow 2\,\mathrm{H_2O}\) | \(1.78\) |
| \(\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}\) | \(1.51\) |
| \(\mathrm{Au^{3+}} + 3e^- \rightarrow \mathrm{Au}\text{(s)}\) | \(1.40\) |
| \(\mathrm{Cl_2}\text{(g)} + 2e^- \rightarrow 2\,\mathrm{Cl^-}\) | \(1.36\) |
| \(\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^+} + 6e^- \rightarrow 2\,\mathrm{Cr^{3+}} + 7\,\mathrm{H_2O}\) | \(1.33\) |
| \(\mathrm{O_2}\text{(g)} + 4\,\mathrm{H^+} + 4e^- \rightarrow 2\,\mathrm{H_2O}\) | \(1.23\) |
| \(\mathrm{MnO_2}\text{(s)} + 4\,\mathrm{H^+} + 2e^- \rightarrow \mathrm{Mn^{2+}} + 2\,\mathrm{H_2O}\) | \(1.23\) |
| \(\mathrm{Br_2} + 2e^- \rightarrow 2\,\mathrm{Br^-}\) | \(1.09\) |
| \(\mathrm{NO_3^-} + 4\,\mathrm{H^+} + 3e^- \rightarrow \mathrm{NO}\text{(g)} + 2\,\mathrm{H_2O}\) | \(0.97\) |
| \(2\,\mathrm{Hg^{2+}} + 2e^- \rightarrow \mathrm{Hg_2^{2+}}\) | \(0.92\) |
| \(\mathrm{Ag^+} + e^- \rightarrow \mathrm{Ag}\text{(s)}\) | \(0.80\) |
| \(\mathrm{Fe^{3+}} + e^- \rightarrow \mathrm{Fe^{2+}}\) | \(0.77\) |
| \(\mathrm{O_2}\text{(g)} + 2\,\mathrm{H^+} + 2e^- \rightarrow \mathrm{H_2O_2}\) | \(0.68\) |
| \(\mathrm{I_2}\text{(s)} + 2e^- \rightarrow 2\,\mathrm{I^-}\) | \(0.54\) |
| \(\mathrm{Cu^+} + e^- \rightarrow \mathrm{Cu}\text{(s)}\) | \(0.52\) |
| \(\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}\text{(s)}\) | \(0.34\) |
| \(\mathrm{AgCl}\text{(s)} + e^- \rightarrow \mathrm{Ag}\text{(s)} + \mathrm{Cl^-}\) | \(0.22\) |
| \(\mathrm{AgBr}\text{(s)} + e^- \rightarrow \mathrm{Ag}\text{(s)} + \mathrm{Br^-}\) | \(0.10\) |
| \(2\,\mathrm{H^+} + 2e^- \rightarrow \mathrm{H_2}\text{(g)}\) | \(0.00\) |
| \(\mathrm{Pb^{2+}} + 2e^- \rightarrow \mathrm{Pb}\text{(s)}\) | \(-0.13\) |
| \(\mathrm{Sn^{2+}} + 2e^- \rightarrow \mathrm{Sn}\text{(s)}\) | \(-0.14\) |
| \(\mathrm{Ni^{2+}} + 2e^- \rightarrow \mathrm{Ni}\text{(s)}\) | \(-0.25\) |
| \(\mathrm{Fe^{2+}} + 2e^- \rightarrow \mathrm{Fe}\text{(s)}\) | \(-0.44\) |
| \(\mathrm{Cr^{3+}} + 3e^- \rightarrow \mathrm{Cr}\text{(s)}\) | \(-0.74\) |
| \(\mathrm{Zn^{2+}} + 2e^- \rightarrow \mathrm{Zn}\text{(s)}\) | \(-0.76\) |
| \(2\,\mathrm{H_2O} + 2e^- \rightarrow \mathrm{H_2}\text{(g)} + 2\,\mathrm{OH^-}\) | \(-0.83\) |
| \(\mathrm{Al^{3+}} + 3e^- \rightarrow \mathrm{Al}\text{(s)}\) | \(-1.66\) |
| \(\mathrm{Mg^{2+}} + 2e^- \rightarrow \mathrm{Mg}\text{(s)}\) | \(-2.36\) |
| \(\mathrm{Na^+} + e^- \rightarrow \mathrm{Na}\text{(s)}\) | \(-2.71\) |
| \(\mathrm{Ca^{2+}} + 2e^- \rightarrow \mathrm{Ca}\text{(s)}\) | \(-2.87\) |
| \(\mathrm{K^+} + e^- \rightarrow \mathrm{K}\text{(s)}\) | \(-2.93\) |
| \(\mathrm{Li^+} + e^- \rightarrow \mathrm{Li}\text{(s)}\) | \(-3.05\) |
Notes on Table 7.1: 1. A negative \(E^{\ominus}\) means the redox couple is a stronger reducing agent than the \(\mathrm{H^+}/\mathrm{H_2}\) couple. 2. A positive \(E^{\ominus}\) means the redox couple is a weaker reducing agent than the \(\mathrm{H^+}/\mathrm{H_2}\) couple.
7.5 Summary
Redox reactions are an important class of reactions in which oxidation and reduction always occur together — whenever one species is oxidised, another is reduced.
The idea of oxidation and reduction has been built up in three tiers, and this evolution is condensed in FC 7.5. Each tier gives a matching pair of definitions for oxidation, reduction, oxidising agent (oxidant) and reducing agent (reductant).
Flowchart
FC 7.5 – Three-tier view of oxidation and reduction: (1) Classical idea → oxidation = addition of oxygen/electronegative element or removal of hydrogen/electropositive element; reduction is the reverse. → (2) Electronic idea → oxidation = loss of electrons; reduction = gain of electrons; oxidant accepts electrons, reductant donates them. → (3) Oxidation-number idea → oxidation = increase in oxidation number; reduction = decrease in oxidation number.
In the classical view, oxidation is the addition of oxygen (or an electronegative element) or the removal of hydrogen (or an electropositive element); reduction is the opposite.
In the electronic view, oxidation is the loss of electrons and reduction is the gain of electrons. The oxidising agent accepts electrons (and is itself reduced); the reducing agent donates electrons (and is itself oxidised).
In the oxidation-number view, oxidation is an increase in oxidation number and reduction is a decrease. Oxidation numbers are assigned using a consistent set of rules, and a fractional value simply means an average over atoms in different whole-number states.
Two methods are used to write and balance redox equations: the oxidation number method and the ion–electron (half reaction) method. Both give the same balanced result.
Redox reactions are classified into four categories: combination, decomposition, displacement (metal and non-metal), and disproportionation. In disproportionation, a single element in one intermediate oxidation state is both oxidised and reduced.
Redox reactions are the basis of redox titrations, using self-indicators (such as \(\mathrm{MnO_4^-}\)), added indicators (such as diphenylamine with \(\mathrm{Cr_2O_7^{2-}}\)), or the iodine–thiosulphate–starch method.
The chapter introduces the redox couple (the oxidised and reduced forms of a substance in a half reaction) and electrode processes. In a Daniell cell the two half reactions are separated so that electrons flow through an external wire, turning a redox reaction into a source of electrical energy.
The standard electrode potential, \(E^{\ominus}\) (measured at 298 K, unit concentrations, 1 atm for gases, with the \(\mathrm{H^+}/\mathrm{H_2}\) couple taken as \(0.00\ \text{V}\)), measures the tendency of a couple to stay oxidised or reduced. A negative \(E^{\ominus}\) marks a stronger reducing agent than \(\mathrm{H^+}/\mathrm{H_2}\); a positive \(E^{\ominus}\) a weaker one.
Redox reactions find wide applications in combustion, metallurgy, batteries, titrations, corrosion, and the broader study of electrode processes and cells.
7.6 NCERT Questions
Assign the oxidation number to the underlined element in each of the following species:
- \(\mathrm{NaH_2PO_4}\) (P)
- \(\mathrm{NaHSO_4}\) (S)
- \(\mathrm{H_4P_2O_7}\) (P)
- \(\mathrm{K_2MnO_4}\) (Mn)
- \(\mathrm{CaO_2}\) (O)
- \(\mathrm{NaBH_4}\) (B)
- \(\mathrm{H_2S_2O_7}\) (S)
- \(\mathrm{KAl(SO_4)_2\cdot 12H_2O}\) (S)
What are the oxidation numbers of the underlined elements in each of the following, and how do you rationalise your results?
- \(\mathrm{KI_3}\) (I)
- \(\mathrm{H_2S_4O_6}\) (S)
- \(\mathrm{Fe_3O_4}\) (Fe)
- \(\mathrm{CH_3CH_2OH}\) (C)
- \(\mathrm{CH_3COOH}\) (C)
Justify that the following reactions are redox reactions:
- \(\mathrm{CuO}\text{(s)} + \mathrm{H_2}\text{(g)} \rightarrow \mathrm{Cu}\text{(s)} + \mathrm{H_2O}\text{(g)}\)
- \(\mathrm{Fe_2O_3}\text{(s)} + 3\,\mathrm{CO}\text{(g)} \rightarrow 2\,\mathrm{Fe}\text{(s)} + 3\,\mathrm{CO_2}\text{(g)}\)
- \(4\,\mathrm{BCl_3}\text{(g)} + 3\,\mathrm{LiAlH_4}\text{(s)} \rightarrow 2\,\mathrm{B_2H_6}\text{(g)} + 3\,\mathrm{LiCl}\text{(s)} + 3\,\mathrm{AlCl_3}\text{(s)}\)
- \(2\,\mathrm{K}\text{(s)} + \mathrm{F_2}\text{(g)} \rightarrow 2\,\mathrm{K^+F^-}\text{(s)}\)
- \(4\,\mathrm{NH_3}\text{(g)} + 5\,\mathrm{O_2}\text{(g)} \rightarrow 4\,\mathrm{NO}\text{(g)} + 6\,\mathrm{H_2O}\text{(g)}\)
Fluorine reacts with ice and results in the change \(\mathrm{H_2O}\text{(s)} + \mathrm{F_2}\text{(g)} \rightarrow \mathrm{HF}\text{(g)} + \mathrm{HOF}\text{(g)}\). Justify that this reaction is a redox reaction.
Calculate the oxidation number of sulphur, chromium and nitrogen in \(\mathrm{H_2SO_5}\), \(\mathrm{Cr_2O_7^{2-}}\) and \(\mathrm{NO_3^-}\). Suggest the structure of these compounds. Account for the fallacy.
Write formulas for the following compounds:
- Mercury(II) chloride
- Nickel(II) sulphate
- Tin(IV) oxide
- Thallium(I) sulphate
- Iron(III) sulphate
- Chromium(III) oxide
Suggest a list of substances where carbon can exhibit oxidation states from \(-4\) to \(+4\) and nitrogen from \(-3\) to \(+5\).
While sulphur dioxide and hydrogen peroxide can act both as oxidising and as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Consider the reactions:
- \(6\,\mathrm{CO_2}\text{(g)} + 6\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{C_6H_{12}O_6}\text{(aq)} + 6\,\mathrm{O_2}\text{(g)}\)
- \(\mathrm{O_3}\text{(g)} + \mathrm{H_2O_2}\text{(l)} \rightarrow \mathrm{H_2O}\text{(l)} + 2\,\mathrm{O_2}\text{(g)}\) Why is it more appropriate to write these reactions as:
- \(6\,\mathrm{CO_2}\text{(g)} + 12\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{C_6H_{12}O_6}\text{(aq)} + 6\,\mathrm{H_2O}\text{(l)} + 6\,\mathrm{O_2}\text{(g)}\)
- \(\mathrm{O_3}\text{(g)} + \mathrm{H_2O_2}\text{(l)} \rightarrow \mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)} + \mathrm{O_2}\text{(g)}\) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
The compound \(\mathrm{AgF_2}\) is unstable. However, if formed, it acts as a very strong oxidising agent. Why?
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess, and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
How do you account for the following observations?
- Though both alkaline potassium permanganate and acidic potassium permanganate are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
- When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get a colourless pungent-smelling gas, HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Identify the substance oxidised, the substance reduced, the oxidising agent and the reducing agent for each of the following reactions:
- \(2\,\mathrm{AgBr}\text{(s)} + \mathrm{C_6H_6O_2}\text{(aq)} \rightarrow 2\,\mathrm{Ag}\text{(s)} + 2\,\mathrm{HBr}\text{(aq)} + \mathrm{C_6H_4O_2}\text{(aq)}\)
- \(\mathrm{HCHO}\text{(l)} + 2[\mathrm{Ag(NH_3)_2}]^+\text{(aq)} + 3\,\mathrm{OH^-}\text{(aq)} \rightarrow 2\,\mathrm{Ag}\text{(s)} + \mathrm{HCOO^-}\text{(aq)} + 4\,\mathrm{NH_3}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)}\)
- \(\mathrm{HCHO}\text{(l)} + 2\,\mathrm{Cu^{2+}}\text{(aq)} + 5\,\mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{Cu_2O}\text{(s)} + \mathrm{HCOO^-}\text{(aq)} + 3\,\mathrm{H_2O}\text{(l)}\)
- \(\mathrm{N_2H_4}\text{(l)} + 2\,\mathrm{H_2O_2}\text{(l)} \rightarrow \mathrm{N_2}\text{(g)} + 4\,\mathrm{H_2O}\text{(l)}\)
- \(\mathrm{Pb}\text{(s)} + \mathrm{PbO_2}\text{(s)} + 2\,\mathrm{H_2SO_4}\text{(aq)} \rightarrow 2\,\mathrm{PbSO_4}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)}\)
Consider the reactions: \[2\,\mathrm{S_2O_3^{2-}}\text{(aq)} + \mathrm{I_2}\text{(s)} \rightarrow \mathrm{S_4O_6^{2-}}\text{(aq)} + 2\,\mathrm{I^-}\text{(aq)}\] \[\mathrm{S_2O_3^{2-}}\text{(aq)} + 2\,\mathrm{Br_2}\text{(l)} + 5\,\mathrm{H_2O}\text{(l)} \rightarrow 2\,\mathrm{SO_4^{2-}}\text{(aq)} + 4\,\mathrm{Br^-}\text{(aq)} + 10\,\mathrm{H^+}\text{(aq)}\] Why does the same reductant, thiosulphate, react differently with iodine and with bromine?
Justify, giving reactions, that among the halogens fluorine is the best oxidant, and among the hydrohalic compounds hydroiodic acid is the best reductant.
Why does the following reaction occur? \[\mathrm{XeO_6^{4-}}\text{(aq)} + 2\,\mathrm{F^-}\text{(aq)} + 6\,\mathrm{H^+}\text{(aq)} \rightarrow \mathrm{XeO_3}\text{(g)} + \mathrm{F_2}\text{(g)} + 3\,\mathrm{H_2O}\text{(l)}\] What conclusion about the compound \(\mathrm{Na_4XeO_6}\) (of which \(\mathrm{XeO_6^{4-}}\) is a part) can be drawn from the reaction?
Consider the reactions:
- \(\mathrm{H_3PO_2}\text{(aq)} + 4\,\mathrm{AgNO_3}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_3PO_4}\text{(aq)} + 4\,\mathrm{Ag}\text{(s)} + 4\,\mathrm{HNO_3}\text{(aq)}\)
- \(\mathrm{H_3PO_2}\text{(aq)} + 2\,\mathrm{CuSO_4}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_3PO_4}\text{(aq)} + 2\,\mathrm{Cu}\text{(s)} + \mathrm{H_2SO_4}\text{(aq)}\)
- \(\mathrm{C_6H_5CHO}\text{(l)} + 2[\mathrm{Ag(NH_3)_2}]^+\text{(aq)} + 3\,\mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{C_6H_5COO^-}\text{(aq)} + 2\,\mathrm{Ag}\text{(s)} + 4\,\mathrm{NH_3}\text{(aq)} + 2\,\mathrm{H_2O}\text{(l)}\)
- \(\mathrm{C_6H_5CHO}\text{(l)} + 2\,\mathrm{Cu^{2+}}\text{(aq)} + 5\,\mathrm{OH^-}\text{(aq)} \rightarrow\) No change observed. What inference do you draw about the behaviour of \(\mathrm{Ag^+}\) and \(\mathrm{Cu^{2+}}\) from these reactions?
Balance the following redox reactions by the ion–electron method:
- \(\mathrm{MnO_4^-}\text{(aq)} + \mathrm{I^-}\text{(aq)} \rightarrow \mathrm{MnO_2}\text{(s)} + \mathrm{I_2}\text{(s)}\) (in basic medium)
- \(\mathrm{MnO_4^-}\text{(aq)} + \mathrm{SO_2}\text{(g)} \rightarrow \mathrm{Mn^{2+}}\text{(aq)} + \mathrm{HSO_4^-}\text{(aq)}\) (in acidic solution)
- \(\mathrm{H_2O_2}\text{(aq)} + \mathrm{Fe^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{3+}}\text{(aq)} + \mathrm{H_2O}\text{(l)}\) (in acidic solution)
- \(\mathrm{Cr_2O_7^{2-}} + \mathrm{SO_2}\text{(g)} \rightarrow \mathrm{Cr^{3+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)}\) (in acidic solution)
Balance the following equations in basic medium by the ion–electron method and the oxidation number method, and identify the oxidising agent and the reducing agent:
- \(\mathrm{P_4}\text{(s)} + \mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{PH_3}\text{(g)} + \mathrm{HPO_2^-}\text{(aq)}\)
- \(\mathrm{N_2H_4}\text{(l)} + \mathrm{ClO_3^-}\text{(aq)} \rightarrow \mathrm{NO}\text{(g)} + \mathrm{Cl^-}\text{(g)}\)
- \(\mathrm{Cl_2O_7}\text{(g)} + \mathrm{H_2O_2}\text{(aq)} \rightarrow \mathrm{ClO_2^-}\text{(aq)} + \mathrm{O_2}\text{(g)} + \mathrm{H^+}\)
What sorts of information can you draw from the following reaction? \[(\mathrm{CN})_2\text{(g)} + 2\,\mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{CN^-}\text{(aq)} + \mathrm{CNO^-}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
The \(\mathrm{Mn^{3+}}\) ion is unstable in solution and undergoes disproportionation to give \(\mathrm{Mn^{2+}}\) \(\mathrm{MnO_2}\) and \(\mathrm{H^+}\) ion. Write a balanced ionic equation for the reaction.
Consider the elements Cs, Ne, I and F.
- Identify the element that exhibits only a negative oxidation state.
- Identify the element that exhibits only a positive oxidation state.
- Identify the element that exhibits both positive and negative oxidation states.
- Identify the element that exhibits neither a negative nor a positive oxidation state.
Chlorine is used to purify drinking water. Excess chlorine is harmful and is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Refer to the periodic table given in your book and answer the following:
- Select the possible non-metals that can show a disproportionation reaction.
- Select three metals that can show a disproportionation reaction.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
Using the standard electrode potentials given in Table 7.1, predict whether the reaction between the following is feasible:
- \(\mathrm{Fe^{3+}}\text{(aq)}\) and \(\mathrm{I^-}\text{(aq)}\)
- \(\mathrm{Ag^+}\text{(aq)}\) and \(\mathrm{Cu}\text{(s)}\)
- \(\mathrm{Fe^{3+}}\text{(aq)}\) and \(\mathrm{Cu}\text{(s)}\)
- \(\mathrm{Ag}\text{(s)}\) and \(\mathrm{Fe^{3+}}\text{(aq)}\)
- \(\mathrm{Br_2}\text{(aq)}\) and \(\mathrm{Fe^{2+}}\text{(aq)}\)
Predict the products of electrolysis in each of the following:
- An aqueous solution of \(\mathrm{AgNO_3}\) with silver electrodes.
- An aqueous solution of \(\mathrm{AgNO_3}\) with platinum electrodes.
- A dilute solution of \(\mathrm{H_2SO_4}\) with platinum electrodes.
- An aqueous solution of \(\mathrm{CuCl_2}\) with platinum electrodes.
Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.
Given the standard electrode potentials \(\mathrm{K^+}/\mathrm{K} = -2.93\ \text{V}\), \(\mathrm{Ag^+}/\mathrm{Ag} = 0.80\ \text{V}\), \(\mathrm{Hg^{2+}}/\mathrm{Hg} = 0.79\ \text{V}\), \(\mathrm{Mg^{2+}}/\mathrm{Mg} = -2.37\ \text{V}\) and \(\mathrm{Cr^{3+}}/\mathrm{Cr} = -0.74\ \text{V}\), arrange these metals in their increasing order of reducing power.
Depict the galvanic cell in which the reaction \(\mathrm{Zn}\text{(s)} + 2\,\mathrm{Ag^+}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + 2\,\mathrm{Ag}\text{(s)}\) takes place. Further show:
- which electrode is negatively charged,
- the carriers of the current in the cell, and
- the individual reaction at each electrode.
7.7 Check Your Concepts
Using oxidation numbers, explain why the reaction of an element with oxygen is always a redox reaction, whereas the thermal decomposition of calcium carbonate (\(\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}\)) is not.
Assertion–Reason. Assertion: Among the halogens, fluorine does not undergo a disproportionation reaction with alkali. Reason: Fluorine is the most electronegative element and cannot exhibit a positive oxidation state. State whether each statement is correct, and whether the reason correctly explains the assertion.
When dilute hydrochloric acid is added separately to magnesium and to iron, hydrogen is evolved much faster with magnesium. Explain this difference in terms of the reducing tendency of the two metals.
In a Daniell cell, no current flows when the salt bridge is removed, even though the two electrodes are still joined by the external wire. Explain why the salt bridge is essential.
A student claims that “oxidation always means the addition of oxygen.” Give one reaction from this chapter that shows this claim is incomplete, and restate the definition of oxidation using the electron-transfer idea.
Using the standard electrode potential values in Table 7.1, arrange \(\mathrm{F_2}\), \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\) and \(\mathrm{I_2}\) in decreasing order of oxidising strength, and justify the order.
Would it be safe to store a copper sulphate solution in a container made of zinc? Predict what would happen and justify your answer using a redox reaction.
The oxidation number of sulphur in \(\mathrm{Na_2S_4O_6}\) works out to \(+2.5\). Explain why this fractional value is called an “average,” and state what the actual structure reveals about the individual sulphur atoms.
Match each reaction in Column I with its correct type in Column II.
| Column I | Column II |
|---|---|
| a. \(\mathrm{N_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{NO}\text{(g)}\) | i. combination |
| b. \(2\,\mathrm{KClO_3}\text{(s)} \rightarrow 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}\) | ii. decomposition |
| c. \(\mathrm{Zn}\text{(s)} + \mathrm{CuSO_4}\text{(aq)} \rightarrow \mathrm{ZnSO_4}\text{(aq)} + \mathrm{Cu}\text{(s)}\) | iii. displacement |
| d. \(\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{ClO^-}\text{(aq)} + \mathrm{Cl^-}\text{(aq)} + \mathrm{H_2O}\text{(l)}\) | iv. disproportionation |
Explain why an oxidising agent is itself reduced during a redox reaction. Illustrate your answer with the reaction \(\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)}\).
Complete and balance the reaction \(\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{OH^-}\text{(aq)} \rightarrow\) (cold, dilute alkali). Name the type of redox reaction, and name the household product formed.
Assertion–Reason. Assertion: In the reaction \(2\,\mathrm{Na} + \mathrm{H_2} \rightarrow 2\,\mathrm{NaH}\), hydrogen is reduced. Reason: In a metal hydride the oxidation number of hydrogen is \(-1\). State whether each statement is correct, and whether the reason correctly explains the assertion.
Explain why permanganate ion can act as its own indicator in a redox titration, whereas dichromate ion cannot, and name the indicator commonly used with dichromate.
Using the classical definition, identify the substance oxidised and the substance reduced in \(\mathrm{Fe_2O_3}\text{(s)} + 2\,\mathrm{Al}\text{(s)} \rightarrow \mathrm{Al_2O_3}\text{(s)} + 2\,\mathrm{Fe}\text{(s)}\), and explain why this reaction is exploited in the thermite process.
A redox couple with a negative standard electrode potential is said to be a strong reducing agent. Explain, in words, what a negative \(E^{\ominus}\) means physically, and give one example of such a couple from Table 7.1.
7.8 Practice With Numericals
Calculate the oxidation number of the indicated element in each species:
- Mn in \(\mathrm{KMnO_4}\),
- Cr in \(\mathrm{K_2CrO_4}\),
- S in \(\mathrm{H_2SO_3}\), and
- both nitrogen atoms in \(\mathrm{NH_4NO_3}\).
In the first step of Ostwald’s process, ammonia is oxidised as \(4\,\mathrm{NH_3}\text{(g)} + 5\,\mathrm{O_2}\text{(g)} \rightarrow 4\,\mathrm{NO}\text{(g)} + 6\,\mathrm{H_2O}\text{(g)}\). Calculate the mass of nitric oxide obtained from 34 g of ammonia when oxygen is present in excess. (Molar masses: \(\mathrm{NH_3} = 17\ \text{g mol}^{-1}\), \(\mathrm{NO} = 30\ \text{g mol}^{-1}\).)
For the balanced redox reaction \(6\,\mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cr_2O_7^{2-}}\text{(aq)} + 14\,\mathrm{H^+}\text{(aq)} \rightarrow 6\,\mathrm{Fe^{3+}}\text{(aq)} + 2\,\mathrm{Cr^{3+}}\text{(aq)} + 7\,\mathrm{H_2O}\text{(l)}\), determine how many moles of \(\mathrm{Fe^{2+}}\) are oxidised by one mole of dichromate ion, and how many moles of electrons are transferred.
In acidic medium, permanganate is reduced as \(\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}\). Calculate the number of moles of electrons that 0.1 mol of \(\mathrm{KMnO_4}\) can accept in acidic solution.
Calculate the percentage by mass of chromium in potassium dichromate, \(\mathrm{K_2Cr_2O_7}\). (Atomic masses: \(\mathrm{K} = 39\), \(\mathrm{Cr} = 52\), \(\mathrm{O} = 16\ \text{g mol}^{-1}\).)
When copper is added to silver nitrate solution, the reaction \(\mathrm{Cu}\text{(s)} + 2\,\mathrm{AgNO_3}\text{(aq)} \rightarrow \mathrm{Cu(NO_3)_2}\text{(aq)} + 2\,\mathrm{Ag}\text{(s)}\) occurs. Calculate the mass of silver deposited when 6.35 g of copper reacts completely. (Atomic masses: \(\mathrm{Cu} = 63.5\), \(\mathrm{Ag} = 108\ \text{g mol}^{-1}\).)