Chapter 6: Redox Reactions
For most of the eighteenth century, chemists believed that anything which burns releases an invisible substance called phlogiston. It was a neat idea, and it was wrong. In Paris in the 1770s, Antoine Lavoisier decided to settle the question with a balance rather than an argument. He heated metals in sealed vessels and weighed everything before and after. The metal did not lose anything on burning — it gained mass, and the gain matched exactly the amount of air that disappeared. Something in the air was combining with the metal. Lavoisier named that gas oxygen, and declared that burning is nothing but combination with oxygen.
That single conclusion gave chemistry the word oxidation, and it still sits at the centre of this chapter. But it also created a puzzle that took another century to settle. Magnesium burns in fluorine gas with no oxygen anywhere near it. Sodium reacts with sulfur in the same violent way. Iron rusts slowly on a bicycle chain, food is burned inside your cells at body temperature, and a torch cell runs flat in a drawer — all of these behave like Lavoisier’s burning, yet oxygen is not always the partner.
So chemists widened the idea, step by step: from oxygen, to hydrogen, to the transfer of electrons, and finally to a simple bookkeeping number that works for every case.
Diagram
Fig. 6.0 – A single opening illustration with four panels sharing one caption “all of these are the same chemistry”: (i) a cut apple browning, (ii) a rusted iron nail, (iii) magnesium ribbon burning with bright flame, (iv) a torch cell connected to a bulb; a curved arrow labelled “transfer of electrons” linking all four panels.
You will be able to reason out each of these questions for yourself by the time you finish this chapter.
6.1 Introduction
The word redox is simply a short form made by joining two words that chemists always use together — oxidation and reduction. The two never happen alone. If one substance in a flask is being oxidised, some other substance in that same flask must be getting reduced at the very same moment, because the two changes are two halves of one event.
Once you learn to spot this pair, you begin to see it everywhere. Respiration inside your cells, the rusting of an iron gate, the burning of cooking gas, the browning of a cut fruit, the working of a torch cell — every one of these is a redox reaction. This is one of the few chapters in chemistry whose ideas run through biology, industry, and daily life at the same time.
6.1.1 Classical Ideas of Redox Reactions
The oldest and simplest meaning of oxidation is the one you can guess from the word itself. Classically, oxidation refers to the combination of an element or a substance with oxygen.
Take the burning of carbon. Carbon combines with oxygen from the air to give carbon dioxide.
\[\mathrm{C}\text{(s)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)} \qquad \text{...(6.1)}\]
Magnesium behaves in the same way. A burning magnesium ribbon combines with oxygen to give solid magnesium oxide.
\[2\,\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\,\mathrm{MgO}\text{(s)} \qquad \text{...(6.2)}\]
In reactions (6.1) and (6.2), the carbon and the magnesium are said to be oxidised, because each of them has picked up oxygen.
When oxygen is taken away
Now look at the opposite situation. When haematite is heated with coke in a furnace, oxygen is pulled away from the iron oxide.
\[2\,\mathrm{Fe_2O_3} + 3\,\mathrm{C}\text{(s)} \rightarrow 4\,\mathrm{Fe}\text{(s)} + 3\,\mathrm{CO_2}\text{(g)} \qquad \text{...(6.3)}\]
Here oxygen is removed from \(\mathrm{Fe_2O_3}\). A reaction in which oxygen is removed from a substance is called a reduction reaction. Notice something important about this single equation: the iron oxide is reduced, but the carbon that took the oxygen away is itself oxidised to \(\mathrm{CO_2}\). Oxidation and reduction have happened side by side in one equation, which is exactly why they are studied together.
Oxidation as removal of hydrogen
Oxygen is not the only marker of oxidation. Consider the burning of hydrogen sulfide in air.
\[2\,\mathrm{H_2S}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{S}\text{(s)} + 2\,\mathrm{H_2O}\text{(l)} \qquad \text{...(6.4)}\]
In this reaction hydrogen is removed from \(\mathrm{H_2S}\), and removal of hydrogen is also called oxidation. Here the sulfur in \(\mathrm{H_2S}\) loses hydrogen and undergoes oxidation, while the oxygen accepts that hydrogen and undergoes reduction. So the classical picture now has two tests for oxidation — addition of oxygen, and removal of hydrogen.
Before going further, it helps to have names for the two partners that make a redox reaction possible.
Oxidant / oxidising agent: A reagent or substance which itself undergoes reduction and causes oxidation of another species is called an oxidant or oxidising agent.
Notice the small trap in that definition, and mark it now, because examiners test it every year: the oxidising agent is not the substance that gets oxidised. It is the one that makes something else get oxidised, and it pays for that by being reduced itself.
Widening the idea of oxidation
Now consider two more examples, in which oxygen is nowhere in sight.
\[\mathrm{Mg}\text{(s)} + \mathrm{F_2}\text{(g)} \rightarrow \mathrm{MgF_2}\text{(s)} \qquad \text{...(6.5)}\]
\[\mathrm{Mg}\text{(s)} + \mathrm{S}\text{(s)} \rightarrow \mathrm{MgS}\text{(s)} \qquad \text{...(6.6)}\]
The above reactions are also examples of oxidation, even though no oxygen is involved, and thus the scope of oxidation can be expanded. Magnesium is behaving exactly as it did in reaction (6.2) — it is combining with a strongly electronegative partner. Combination with an electronegative element is oxidation.
An electronegative element, as you have already met in bonding, is one that pulls shared electrons strongly towards itself. Fluorine, oxygen, sulfur and chlorine all do this. From magnesium’s point of view, being attacked by fluorine and being attacked by oxygen are the same experience.
Oxidation can also be looked upon as the removal of an electropositive element. For example, when mercury(I) chloride breaks up, electropositive mercury is removed from it.
\[\mathrm{Hg_2Cl_2}\text{(s)} \rightarrow \mathrm{HgCl_2}\text{(s)} + \mathrm{Hg}\text{(s)} \qquad \text{...(6.7)}\]
Examples of reduction
If oxidation has grown four faces, reduction must grow the four opposite ones. Let us look at each in turn.
a. Removal of oxygen from mercuric oxide:
\[2\,\mathrm{HgO}\text{(s)} \rightarrow 2\,\mathrm{Hg}\text{(l)} + \mathrm{O_2}\text{(g)} \qquad \text{...(6.8)}\]
b. Removal of an electronegative element from \(\mathrm{FeCl_3}\):
\[2\,\mathrm{FeCl_3} + \mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{FeCl_2}\text{(aq)} + 2\,\mathrm{HCl} \qquad \text{...(6.9)}\]
c. Addition of hydrogen, as in the hydrogenation of ethene:
\[\mathrm{CH_2}\!=\!\mathrm{CH_2}\text{(g)} + \mathrm{H_2}\text{(g)} \rightarrow \mathrm{CH_3\!-\!CH_3}\text{(g)} \qquad \text{...(6.10)}\]
d. Addition of an electropositive element:
\[2\,\mathrm{HgCl_2}\text{(aq)} + \mathrm{SnCl_2}\text{(aq)} \rightarrow \mathrm{Hg_2Cl_2}\text{(s)} + \mathrm{SnCl_4}\text{(aq)} \qquad \text{...(6.11)}\]
Read the four together. In equation (6.8) there is removal of oxygen from mercuric oxide. Equation (6.9) shows removal of an electronegative element from \(\mathrm{FeCl_3}\). Equation (6.10) involves addition of hydrogen. Equation (6.11) involves addition of an electropositive element, mercury, to \(\mathrm{HgCl_2}\). All these reactions represent reduction.
Just as oxidation needed a partner, reduction needs one too.
Reductant / reducing agent: A reagent or reducing agent is defined as a substance or reagent which itself undergoes oxidation, bringing about the reduction of another species.
Now look again at equation (6.11), which is worth a second reading. This equation involves simultaneous oxidation and reduction. In this reaction \(\mathrm{HgCl_2}\) is reduced to \(\mathrm{Hg_2Cl_2}\) and \(\mathrm{SnCl_2}\) is oxidized to \(\mathrm{SnCl_4}\). Hence it is a redox reaction. Mercury(II) chloride is the oxidising agent here and tin(II) chloride is the reducing agent, and each one is changed by the other.
The five tests, side by side
Gathering everything so far, the classical view gives five separate signals for each half of the story. The last of these five — the electron test — is the most powerful one, and the next subsection is devoted to it.
The pattern in this table is worth noticing rather than memorising: every row is a mirror image. If you are sure of any one row for one column, the other seven entries follow automatically.
To decide quickly what is happening to a given species, run the tests in the order shown in FC 6.1.
Flowchart
FC 6.1 – Classifying a species as oxidised or reduced: pick one species → has it gained O or lost H? → yes → oxidised; no → has it lost O or gained H? → yes → reduced; no → has it combined with an electronegative element or lost an electropositive one? → yes → oxidised; no → check electrons: electrons lost → oxidised; electrons gained → reduced; then branch: whichever species is oxidised is the reducing agent, whichever is reduced is the oxidising agent.
6.1.2 Redox Reaction in Terms of Electron Transfer
The classical tests work, but they are a collection of separate rules rather than one idea. A single idea does exist underneath them, and it is electron transfer. A redox reaction can be described as electron transfer, as shown below.
\[\mathrm{Mg} + \tfrac{1}{2}\,\mathrm{O_2} \rightarrow \mathrm{Mg^{2+}} + \mathrm{O^{2-}} \qquad \text{...(6.12)}\]
\[\mathrm{Mg} + \mathrm{F_2} \rightarrow \mathrm{Mg^{2+}} + 2\,\mathrm{F^-} \qquad \text{...(6.13)}\]
The development of charges on the species produced is the clue. Neutral atoms have gone in on the left, and charged ions have come out on the right — and charge cannot appear from nowhere. Whatever negative charge one species has gained, another must have lost. This suggests that reactions (6.12) and (6.13) can be written to show the electron bookkeeping directly, as drawn in Fig. 6.2.
Diagram
Fig. 6.2 – Two equations one below the other: Mg(s) + O₂(g) → Mg²⁺ + O²⁻ and Mg(s) + F₂(g) → Mg²⁺ + 2F⁻; above each, a bracket from Mg to Mg²⁺ labelled “Loss of 2e⁻”; below each, a bracket from O₂ (or F₂) to O²⁻ (or 2F⁻) labelled “Gain of 2e⁻”.
When Mg is oxidised to MgO, the neutral Mg atom loses electrons to form \(\mathrm{Mg^{2+}}\) in MgO. The elemental oxygen gains those electrons and forms \(\mathrm{O^{2-}}\) in MgO. Nothing is created and nothing is destroyed; two electrons have simply changed owner.
Half reactions
Each of the above steps represents a half reaction, which involves electron transfer — either a loss or a gain. A half reaction is not something you can carry out alone in a beaker, because free electrons do not float about in solution; it is a piece of bookkeeping that becomes real only when paired with its opposite. The sum of these two half reactions, or the overall reaction, is a redox reaction.
Now consider the following half reactions.
\[\mathrm{Fe}\text{(s)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + 2\,\mathrm{e^-} \qquad \text{...(6.14)}\]
\[\mathrm{Cu^{2+}}\text{(aq)} + 2\,\mathrm{e^-} \rightarrow \mathrm{Cu}\text{(s)} \qquad \text{...(6.15)}\]
Adding them, the two electrons cancel from both sides and the overall change is left behind:
\[\mathrm{Fe}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad \text{...(6.16)}\]
This is the change you can see for yourself when an iron nail is left standing in blue copper sulfate solution, as shown in Fig. 6.3.
Diagram
Fig. 6.3 – A beaker of blue copper sulfate solution with an iron nail dipped in it; label a reddish-brown copper deposit on the nail, the solution turning pale green, and a curved arrow from Fe to Cu²⁺ labelled “2 electrons”.
In the half reaction (6.14) Fe acts as a reducing agent, whereas \(\mathrm{Cu^{2+}}\) acts as an oxidising agent which accepts electrons from Fe. The half reaction involving loss of electrons is called an oxidation reaction and that involving gain of electrons is called reduction. Thus equation (6.14) is oxidation, equation (6.15) is reduction, and equation (6.16) is a redox reaction.
Many students find the pair easy to hold on to through the phrase OIL RIG — Oxidation Is Loss, Reduction Is Gain — where loss and gain always refer to electrons, never to oxygen.
The two agents can now be described in the same electron language, which is far sharper than the classical wording.
Displacement reactions as redox reactions
Displacement reactions can also be looked upon as redox reactions. In such reactions an ion (or atom) in a compound is replaced by an ion (or an atom) of another element. In general terms,
\[\mathrm{X} + \mathrm{YZ} \rightarrow \mathrm{XZ} + \mathrm{Y}\]
The reaction (6.16) is a displacement reaction. Iron has displaced copper from its salt, and it could do so because iron gives up electrons more readily than copper does. That is the hidden rule behind every displacement: a more reactive element pushes out a less reactive one, and “more reactive” here simply means “more willing to lose electrons.” You will meet this idea again in section 6.4 as a measured quantity rather than a general impression.
6.2 Oxidation Number
Describing a redox reaction as electron transfer works beautifully — but only for one type of reaction. The description of a redox reaction in terms of electron transfer suggests the involvement of only ionic species in it. Yet many reactions involving only covalent bonds also fulfil the classical definition of oxidation and reduction. For example:
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} \qquad \text{...(6.17)}\]
\[\mathrm{H_2}\text{(g)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{HCl}\text{(g)} \qquad \text{...(6.18)}\]
Equation (6.17) represents oxidation of \(\mathrm{H_2}\), as it is a combination with oxygen. In equation (6.18), an electronegative element, Cl, is added to \(\mathrm{H_2}\), and therefore equation (6.18) also represents oxidation of \(\mathrm{H_2}\).
The problem with covalent compounds
But now look closely at the products. As the products of both these oxidations are polar covalent molecules, there is an electron shift rather than a complete electron transfer from one species to the other. In water and in hydrogen chloride, the bonding electrons have not left hydrogen altogether — they have merely been dragged closer to the more electronegative partner. There is no \(\mathrm{H^+}\) ion and no \(\mathrm{O^{2-}}\) ion sitting in a molecule of water, as shown in Fig. 6.4.
Diagram
Fig. 6.4 – Two panels side by side. Left: an H–Cl molecule with the shared electron pair drawn nearer to Cl and a shaded electron cloud showing δ+ on H and δ− on Cl, captioned “actual: electron shift”. Right: the same molecule redrawn with the pair fully on Cl, labelled H⁺ Cl⁻, captioned “assumed for oxidation number: complete transfer”.
So a book-keeping device is needed — one that works for a genuine ion transfer and for a partial electron shift alike. A practical method based on oxidation number is developed to describe all the redox reactions, which are either ionic, involving complete electron transfer, or covalent, which refer to a shift of electrons. In this method the electron pair in a covalent bond is assumed to belong to the more electronegative element — that is, it is imagined to shift completely to the more electronegative element.
This is a deliberate pretence, and it is worth being honest about it. We know the electron pair in H–Cl is still shared. We simply agree to award it entirely to chlorine, because doing so lets us attach a definite number to every atom and then track how that number changes during a reaction.
The following rules have been formulated to determine the oxidation number of an element in a compound.
6.2.1 Rules to Assign Oxidation Number
1. The oxidation number of each atom of an element in the free state is zero. For example, each atom in \(\mathrm{H_2}\), \(\mathrm{Cl_2}\), \(\mathrm{O_3}\), \(\mathrm{S_8}\), \(\mathrm{P_4}\), \(\mathrm{O_2}\), \(\mathrm{Ca}\), etc., has an oxidation number of zero. The reason is easy to see: in \(\mathrm{Cl_2}\) both partners are identical, so neither can pull the shared pair away from the other, and neither gains or loses anything.
2. The oxidation number of an atom in a monoatomic ion is equal to its charge. Thus alkali metals have oxidation number +1 in all their compounds (\(\mathrm{NaCl}\), \(\mathrm{KCl}\), etc.). Alkaline earth metals have oxidation number +2 in all their compounds (\(\mathrm{CaCO_3}\), \(\mathrm{MgCl_2}\), etc.). Al is considered to have +3 as its oxidation number in all its compounds.
3. The oxidation number of O is usually −2 in all of its compounds, except in a peroxide or peroxide ion, where it has an oxidation number of −1, and in a superoxide, where each oxygen has an oxidation number of −1/2. The marked structures are shown in Fig. 6.5.
Diagram
Fig. 6.5 – Four small structures in a row with arrows pointing to each atom and the oxidation number written beneath: Ca–O (+2, −2); H–O–O–H (+1, −1, −1, +1); (O–O) as the peroxide ion (−1, −1); KO₂ (+1, −1/2).
The reason for the exceptions is structural, not arbitrary. In hydrogen peroxide the two oxygen atoms are joined to each other by an O–O bond, and a bond between two identical atoms is shared equally, so it contributes nothing to either atom’s oxidation number. Each oxygen is then left with only one bond to hydrogen, worth −1 instead of −2.
There is one more exception in the other direction: in \(\mathrm{OF_2}\) the oxidation number of oxygen is +2.
4. The oxidation number of the H atom is either +1 or −1. When the H atom is bonded to nonmetals, its oxidation number is +1. When it is bonded to metals, it possesses an oxidation number of −1. Four marked examples are given in Fig. 6.6.
Diagram
Fig. 6.6 – Four structures in a row with oxidation numbers written beneath each atom: (O–H)⁻ (−2, +1); H–O–H (+1, −2, +1); Li–H (+1, −1); H–Ca–H (−1, +2, −1).
Hydrogen is the one element that sits on the fence. Against a nonmetal such as oxygen or chlorine it is the weaker partner and is assigned +1; against a metal such as lithium or calcium it is the stronger partner and takes the shared pair, so it is assigned −1. Compounds of the second kind are the metal hydrides.
5. The oxidation number of F is −1 in all of its compounds. The other halogens Cl, Br and I usually exhibit an oxidation number of −1 in their halide compounds. However, in compounds in which the halogens Cl, Br and I are bonded to oxygen, the oxidation number of the halogens is +1. For example, see the marked structures in Fig. 6.7.
Diagram
Fig. 6.7 – Four structures in a row with oxidation numbers written beneath each atom: H–F (+1, −1); KBr (+1, −1); Cl–O–Cl (+1, −2, +1); H–O–Cl (+1, −2, +1).
Fluorine is the one element that never varies, because nothing outranks it in electronegativity. The heavier halogens are outranked by oxygen, which is why the value flips from −1 to +1 the moment a Cl–O bond appears.
6. The algebraic sum of the oxidation numbers of all the atoms in a neutral molecule is zero.
7. The algebraic sum of the oxidation numbers of all the atoms in a polyatomic ion is equal to the net charge of the ion.
8. When two or more atoms of an element are present in a molecule or ion, the oxidation number of the atom of that element will be the average oxidation number of all the atoms of that element in that molecule.
By applying rules 1 to 5 it is possible to determine the oxidation number(s) of atoms of various elements in molecules or ions. For doing this, rules 6, 7 and 8 are useful. In practice, rules 1 to 5 supply the values you already know, and rules 6 and 7 supply the single equation you solve for the one value you do not know. The working order is summarised in FC 6.2.
Flowchart
FC 6.2 – Assigning an unknown oxidation number: write the formula → is the species neutral or an ion? → neutral: set the sum equal to 0; ion: set the sum equal to the net charge → fill in the known values from rules 1–5 (O = −2, H = +1, alkali = +1, alkaline earth = +2, F = −1) → multiply each value by the number of such atoms → solve the resulting simple equation for the unknown → if more than one atom of that element is present, divide by their number to get the average value.
6.2.2 Stock Notation
The oxidation number represents the oxidation state of an atom, and it is also denoted by a Roman numeral in parentheses after the chemical symbol of the concerned element in the molecular formula. This representation is called Stock notation, after the German scientist Alfred Stock. For example:
- \(\mathrm{Au^{1+}Cl^{1-}} \rightarrow \mathrm{Au(I)Cl}\)
- \(\mathrm{Au^{3+}Cl_3^{1-}} \rightarrow \mathrm{Au(III)Cl_3}\)
- \(\mathrm{Sn^{4+}Cl_4^{1-}} \rightarrow \mathrm{Sn(IV)Cl_4}\)
- \(\mathrm{Sn^{2+}Cl_2^{1-}} \rightarrow \mathrm{Sn(II)Cl_2}\)
- \(\mathrm{Mn^{4+}O_2^{2-}} \rightarrow \mathrm{Mn(IV)O_2}\)
The Stock notation is used to specify the oxidation number of the metal.
Why bother with it at all? Because a name like “gold chloride” is ambiguous — from the formula \(\mathrm{AuCl}\) and \(\mathrm{AuCl_3}\) you can see the difference, but from the name alone you cannot. Writing gold(I) chloride and gold(III) chloride removes the doubt in one stroke, and this is the naming style you will meet throughout inorganic chemistry. Note also that the Roman numeral is written without a plus sign; Stock notation is used for the positive oxidation states of metals.
The idea of oxidation number is very convenient to define oxidation, reduction, and the substances like the oxidising agent (oxidant) and the reducing agent (reductant) of a redox reaction — which is precisely what the next subsection does.
6.2.3 Redox Reaction in Terms of Oxidation Number
With oxidation numbers in hand, the whole of section 6.1 can be restated in a single, uniform language — one that works for ionic and covalent reactions alike. Instead of hunting for oxygen, hydrogen or electrons, you now simply watch a number go up or down.
Oxidation: An increase in the oxidation number of an element in a given substance.
Reduction: A decrease in the oxidation number of an element in a given substance.
Oxidising agent: A substance which increases the oxidation number of an element in a given substance, and itself undergoes a decrease in the oxidation number of a constituent element.
Reducing agent: A substance that lowers the oxidation number of an element in a given substance, and itself undergoes an increase in the oxidation number of a constituent element in it.
This is consistent with everything said earlier, and you should check that for yourself rather than take it on trust. Losing electrons makes an atom’s assigned charge more positive, so loss of electrons and an increase in oxidation number are the same event seen from two angles.
Fractional oxidation numbers
Some elements in a particular compound may possess a fractional oxidation number. For example: \(\mathrm{C_3O_2}\), \(\mathrm{Br_3O_8}\), \(\mathrm{Na_2S_4O_6}\), \(\mathrm{C_8H_{18}}\), etc. In these compounds the oxidation numbers of C, Br, S and C are 4/3, 16/3, 2.5 and 9/4 respectively.
A fraction should not alarm you, because no atom is actually carrying two and a half charges. These oxidation numbers are actually the average oxidation number of all the atoms of that element in that compound — exactly what rule 8 warned about. Different atoms of the element in such species exhibit different oxidation states.
For example, the tetrathionate ion has two S atoms with oxidation number +5 and two with zero (0). Therefore, the average oxidation number of S in these species is \(10/4 = 2.5\). The structure is drawn in Fig. 6.8.
Diagram
Fig. 6.8 – Structure of the tetrathionate ion S₄O₆²⁻: ⁻O–S(=O)(=O)–S–S–S(=O)(=O)–O⁻, with oxidation numbers labelled +5, 0, 0, +5 above the four sulfur atoms from left to right; caption “structure of S₄O₆²⁻ (tetrathionate ion)”.
The two starred sulfur atoms in the middle are bonded only to other sulfur atoms. As you saw with the O–O bond in hydrogen peroxide, a bond between two identical atoms contributes nothing to either atom’s oxidation number, so these two are left at zero, while the two outer sulfur atoms, surrounded by oxygen, come out at +5. The average of \(+5\), \(0\), \(0\) and \(+5\) is \(10/4 = 2.5\).
6.3 Balancing of Redox Reactions
Every equation you have written so far in this chapter was already balanced for you. Redox equations, however, are notoriously awkward to balance by inspection, because water, hydrogen ions and hydroxide ions from the solvent join in without ever being written in the skeletal equation. Trial and error can take a very long time.
Fortunately, the oxidation number idea of section 6.2 supplies a shortcut. Two methods are used to balance chemical equations for redox processes: the oxidation number method and the half reaction method, also called the ion electron method. Both rest on the same single truth.
You have already seen this principle at work without being told its name. In Solved Example 6.6, three arsenic atoms lost 2 electrons each while one bromine atom gained 6. Balancing a redox equation simply means choosing coefficients that force those two totals to match.
6.3.1 The Oxidation Number Method Is Illustrated in the Following Steps
Step I: Write the unbalanced equation for the redox reaction. Balance the equation for all atoms in the reaction, except H and O. Identify the atoms which undergo a change in oxidation number and by how much. Draw the bracket to connect atoms of the elements that change the oxidation number.
Step II: Show an increase in oxidation number per atom of the oxidised species and hence the net increase in oxidation number. Similarly show a decrease in the oxidation number per atom of the reduced species and the net decrease in oxidation number. Determine the factors which will make the total increase and decrease equal. Insert the coefficients into the equation.
Step III: Balance oxygen atoms by adding \(\mathrm{H_2O}\) to the side containing less O atoms; one \(\mathrm{H_2O}\) is added for one O atom. Balance H atoms by adding \(\mathrm{H^+}\) ions to the side having less H atoms.
Step IV: If the reaction occurs in basic medium, then add \(\mathrm{OH^-}\) ions, equal to the number of \(\mathrm{H^+}\) ions added in step III, on both the sides of the equation. The \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) on the same side of the reaction are combined to give \(\mathrm{H_2O}\) molecules.
Step V: Check the equation with respect to both, the number of atoms of each element and the charges. It is balanced.
Note: For acidic medium step IV is omitted.
The whole method is summarised in FC 6.3.
Flowchart
FC 6.3 – Oxidation number method: write the skeletal equation → balance all atoms except H and O → assign oxidation numbers and bracket the atoms that change → find increase per atom and decrease per atom → multiply by whole-number factors so total increase = total decrease → insert these coefficients → add H₂O to the side short of O (one H₂O per O atom) → add H⁺ to the side short of H → branch: acidic medium → skip to the check; basic medium → add OH⁻ equal to the H⁺ added, to both sides, and combine H⁺ + OH⁻ on the same side into H₂O → check atoms and charges → balanced equation.
6.3.2 Ion Electron Method (Half Reaction Method)
In this method two half equations are balanced separately and then added together to give the balanced equation. The advantage is that you never have to keep track of the whole reaction at once — each half is a small, manageable problem, and the electrons written explicitly do the matching for you. The following steps are involved.
Step I: Write the unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in the reactants and products. Divide the equation into two half equations. One half equation involves an increase in oxidation number and another involves a decrease in oxidation number (write the two half equations separately).
Step II: Balance the atoms except O and H in each half equation. Balance the oxygen atoms by adding \(\mathrm{H_2O}\) to the side with less O atoms.
Step III: Balance the H atom by adding \(\mathrm{H^+}\) ions to the side with less H atoms.
Step IV: Balance the charges by adding an appropriate number of electrons to the right side of the oxidation half equation and to the left of the reduction half equation.
Step V: Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations. Add the two half equations and cancel the number of electrons on both sides of the equation.
Step VI: If the reaction occurs in basic medium, then add \(\mathrm{OH^-}\) ions, equal to the number of \(\mathrm{H^+}\) ions, on both sides of the equation. The \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions on the same side of the equation combine to give \(\mathrm{H_2O}\) molecules.
Step VII: Check that the equation is balanced in both, the atoms and the charges.
Notice the logic behind step IV, which students often apply mechanically. Oxidation is loss of electrons, so the electrons must appear as a product — on the right of the oxidation half equation. Reduction is gain of electrons, so they must appear as a reactant — on the left of the reduction half equation. If you remember what the words mean, you will never place the electrons on the wrong side. The complete route is set out in FC 6.4.
Flowchart
FC 6.4 – Ion electron (half reaction) method: write the skeletal equation and assign oxidation numbers → split into an oxidation half equation (oxidation number rises) and a reduction half equation (oxidation number falls) → in each half: balance all atoms except O and H → add H₂O to the side short of O → add H⁺ to the side short of H → add electrons to balance charge, on the right for oxidation and on the left for reduction → multiply one or both halves so the electrons are equal in number → add the two halves and cancel the electrons → branch: acidic medium → go to the final check; basic medium → add OH⁻ equal to the H⁺ present, to both sides, combine H⁺ + OH⁻ into H₂O on the same side and cancel any excess H₂O → check atoms and charges.
Both methods must give the same final equation. If they do not, one of them has been applied carelessly, and the atom-and-charge check in the last step will show where.
Working in basic medium
Neither of the two worked problems above needed step VI, because both were carried out in acidic medium. Since basic-medium questions are common in examinations, it is worth seeing the extra step in action once.
6.4 Redox Reaction and Electrode Potential
We noted in section 6.1 that a displacement reaction can be looked upon as a redox reaction. Consider the following displacement reaction, in which the electron bookkeeping is shown by the brackets in Fig. 6.14.
\[\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)} \qquad \text{...(6.17)}\]
Diagram
Fig. 6.14 – The equation Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) with a bracket above running from Zn to Zn²⁺ labelled “Loss of 2e⁻” and a bracket below running from Cu²⁺ to Cu labelled “Gain of 2e⁻”.
This reaction can be brought about in two ways, and the difference between those two ways is the whole point of this section.
Letting the electrons travel by wire
The simpler method to observe this reaction is to take copper sulfate solution in a container and dip a zinc rod in it. The redox reaction (6.17) takes place in that container. Zinc gets oxidized to \(\mathrm{Zn^{2+}}\) ion and \(\mathrm{Cu^{2+}}\) ions get reduced to metallic Cu. A direct transfer of electrons from a zinc atom to cupric ions takes place in this case.
Nothing useful comes out of this arrangement except heat. The electrons jump across at the moment a \(\mathrm{Cu^{2+}}\) ion touches the zinc surface, travelling a distance smaller than an atom, and there is no way to make them do work on the way.
The electron transfer from a Zn atom to \(\mathrm{Cu^{2+}}\) ions can be demonstrated by carrying out two half reactions in two separate containers as shown in Fig. 6.1.
Diagram
Fig. 6.1 – Daniel Cell: two beakers, the left with a zinc electrode (anode) in zinc salt solution and the right with a copper electrode (cathode) in copper salt solution; the two electrodes joined by a wire through a switch and a voltmeter marked V, with arrows labelled “electron flow” running from zinc to copper in the external wire; an inverted U-tube salt bridge joining the two solutions with an arrow labelled “ion flow”; electrode reactions written below — Zn(s) → Zn²⁺(aq) + 2e⁻, oxidation (loss of electrons), and Cu²⁺(aq) + 2e⁻ → Cu(s), reduction (gain of electrons).
The set up in Fig. 6.1 is that of a Daniel cell. The zinc and copper plates are connected by an electric wire through a switch and voltmeter. The solutions in the two containers are connected by a salt bridge, which is a U-shaped glass tube containing a gel of \(\mathrm{KCl}\) or \(\mathrm{NH_4NO_3}\) in agar-agar. When the switch is on, the electrical circuit is complete, as indicated by the deflection in the voltmeter.
The circuit has two parts. One is called the external circuit, in the form of an electrical wire which allows the flow of electrons, and the other is in the form of two solutions joined by the salt bridge. In the solution part of the circuit, the electric current is carried by movement of ions. This is an important difference to hold on to: metals conduct by moving electrons, solutions conduct by moving ions, and a working cell needs both.
When the circuit is complete, the zinc atoms on the zinc plate spontaneously lose electrons, which are picked up in the external circuit. The electrons flow from the zinc plate to the copper plate through the wire. \(\mathrm{Cu^{2+}}\) ions in the second container receive these electrons through the copper plate and are reduced to copper atoms, which get deposited on the copper plate.
Here, the zinc plate acts as the anode (negative electrode) and the copper plate acts as the cathode (positive electrode). Thus, when two half reactions, namely oxidation and reduction, are allowed to take place in separate containers, and a provision is made for completing the electrical circuit, electron transfer takes place through the external circuit. This results in a flow of electric current in the circuit, as indicated by deflection in the voltmeter. At the same time ions move within the solution, resulting in completion of the electrical circuit. This is a simple electrochemical cell called a Daniel cell, in which electricity is generated by a redox reaction. The complete path is traced in FC 6.5.
Flowchart
FC 6.5 – Current path in a Daniel cell (one closed loop): Zn(s) at the anode loses 2e⁻ → Zn²⁺ enters the solution → electrons travel through the external wire and voltmeter to the copper plate → Cu²⁺ in the right beaker accepts 2e⁻ at the cathode → Cu deposits on the plate → meanwhile in the salt bridge, anions migrate towards the zinc half-cell and cations towards the copper half-cell, keeping both solutions electrically neutral → loop closes back at the zinc anode.
What the electrodes are doing
In an electrochemical cell electrons flow in the external circuit from anode to cathode, while the conventional current flows from cathode to anode. This apparent contradiction is only a matter of an old convention: conventional current was defined as the flow of positive charge long before the electron was discovered, so it is drawn opposite to the actual electron drift.
An electrical potential is said to be established at the two electrodes of an electrochemical cell. It is called electrode potential. The magnitude and direction of the electrode potential depend upon the nature of the metal and ions, the concentration of ions, and the temperature.
The reaction associated with an electrode is called an electrode reaction. The two chemical species linked by transfer of electrons form a redox couple.
6.4.1 Standard Electrode Potential
A single electrode potential cannot be measured on its own, because a voltmeter always needs two terminals — you can only ever measure a difference between two electrodes. Chemists solved this by fixing conditions and choosing one electrode as a zero mark, exactly as sea level is chosen as the zero for measuring the height of a mountain.
When the concentration of each species taking part in the electrode reaction is unity and the temperature is 298 K, the observed electrode potential is called the standard electrode potential (\(E^0\)).
By convention the standard electrode potential of the hydrogen electrode is 0.00 Volt. The standard hydrogen electrode is used as the reference electrode for the determination of \(E^0\) values of various redox couples, and its construction is shown in Fig. 6.15.
Diagram
Fig. 6.15 – Standard hydrogen electrode: a glass tube containing a platinised platinum foil dipped in an acid solution of unit H⁺ concentration, with H₂ gas at 1 bar bubbled over the foil through an inlet, the temperature marked 298 K, and the half reaction 2H⁺(aq) + 2e⁻ ⇌ H₂(g) written alongside; a terminal at the top for connection to the other electrode.
Significance of \(E^0\) value
The electrode reaction considered for the electrode potential is the reduction reaction. Therefore, the value and sign of the standard electrode potential is a measure of the tendency of the active species in the electrode reaction to remain in the oxidized or reduced form. A negative \(E^0\) means the redox couple is a stronger reducing agent than the \(\mathrm{H^+/H_2}\) couple. Table 6.1 shows standard electrode potentials of the reduction process of some redox couples.
Table 6.1: Standard electrode potentials of some redox couples
| Oxidised form + \(n\mathrm{e^-}\) \(\rightarrow\) reduced form | \(E^0\)/V |
|---|---|
| \(\mathrm{F_2}\text{(g)} + 2\mathrm{e^-} \rightarrow 2\mathrm{F^-}\) | 2.87 |
| \(\mathrm{H_2O_2} + 2\mathrm{H^+} + 2\mathrm{e^-} \rightarrow 2\mathrm{H_2O}\) | 1.78 |
| \(\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}\) | 1.51 |
| \(\mathrm{Cl_2}\text{(g)} + 2\mathrm{e^-} \rightarrow 2\mathrm{Cl^-}\) | 1.36 |
| \(\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}\) | 1.33 |
| \(\mathrm{O_2}\text{(g)} + 4\mathrm{H^+} + 4\mathrm{e^-} \rightarrow 2\mathrm{H_2O}\) | 1.23 |
| \(\mathrm{Br_2} + 2\mathrm{e^-} \rightarrow 2\mathrm{Br^-}\) | 1.09 |
| \(2\mathrm{Hg^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Hg_2^{2+}}\) | 0.92 |
| \(\mathrm{Fe^{3+}} + \mathrm{e^-} \rightarrow \mathrm{Fe^{2+}}\) | 0.77 |
| \(\mathrm{I_2}\text{(s)} + 2\mathrm{e^-} \rightarrow 2\mathrm{I^-}\) | 0.54 |
| \(2\mathrm{H^+} + 2\mathrm{e^-} \rightarrow \mathrm{H_2}\text{(g)}\) | 0.00 |
| \(\mathrm{Zn^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Zn}\text{(s)}\) | −0.76 |
| \(\mathrm{Al^{3+}} + 3\mathrm{e^-} \rightarrow \mathrm{Al}\text{(s)}\) | −1.66 |
| \(\mathrm{Mg^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Mg}\text{(s)}\) | −2.36 |
| \(\mathrm{Na^{+}} + \mathrm{e^-} \rightarrow \mathrm{Na}\text{(s)}\) | −2.71 |
| \(\mathrm{Ca^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Ca}\text{(s)}\) | −2.87 |
| \(\mathrm{K^{+}} + \mathrm{e^-} \rightarrow \mathrm{K}\text{(s)}\) | −2.93 |
| \(\mathrm{Li^{+}} + \mathrm{e^-} \rightarrow \mathrm{Li}\text{(s)}\) | −3.05 |
A large negative value of \(E^0\) means that the redox couple is a strong reducing agent. On the other hand, a more positive value of \(E^0\) means the redox couple is a strong oxidizing agent.
From Table 6.1 it is seen that alkali metals have high negative values of \(E^0\), which means that they are strong reducing agents. The alkali metals have a great tendency to give away an electron and form cations. On the other hand, fluorine has a highly positive value of \(E^0\) and thus a great tendency to gain an electron, and is therefore a very strong oxidizing agent.
Diagram
Fig. 6.16 – A vertical scale from +3.0 V at the top to −3.5 V at the bottom, with the couples of Table 6.1 marked at their E⁰ positions (F₂/F⁻ at 2.87, MnO₄⁻/Mn²⁺ at 1.51, Cl₂/Cl⁻ at 1.36, Fe³⁺/Fe²⁺ at 0.77, H⁺/H₂ at 0.00 marked as the reference line, Zn²⁺/Zn at −0.76, Mg²⁺/Mg at −2.36, Li⁺/Li at −3.05); an arrow down the left side labelled “oxidising power increases upward” and an arrow up the right side labelled “reducing power increases downward”.
We shall learn more about electrode potential and electrochemical cells in the 12th standard.
Predicting whether a reaction will occur
The most useful thing \(E^0\) values do is let you decide, before mixing anything, whether a reaction will happen on its own. Write one species as a reduction half reaction and the other as an oxidation half reaction, reverse the sign of \(E^0\) for the one written as oxidation, and add the two values. A positive total means the combination is spontaneous.
6.5 Summary
Redox is a short form for oxidation and reduction taken together. The two never occur separately: whenever one species in a reaction mixture is oxidised, another is being reduced at the same time. Respiration, rusting and the combustion of fuels are all redox processes.
Classically, oxidation means the addition of oxygen, the addition of an electronegative element, the removal of hydrogen, the removal of an electropositive element, or the loss of electrons by any species.
Reduction is the exact mirror image: the removal of oxygen, the removal of an electronegative element, the addition of hydrogen, the addition of an electropositive element, or the gain of electrons by any species.
An oxidant, or oxidising agent, is a substance that itself undergoes reduction and causes the oxidation of another species. It is an electron acceptor.
A reductant, or reducing agent, is a substance that itself undergoes oxidation and brings about the reduction of another species. It is an electron donor.
A redox reaction can be described as electron transfer. Each loss or gain of electrons is written as a half reaction; the sum of an oxidation half reaction and a reduction half reaction is the overall redox reaction. For example, \(\mathrm{Fe}\text{(s)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + 2\,\mathrm{e^-}\) added to \(\mathrm{Cu^{2+}}\text{(aq)} + 2\,\mathrm{e^-} \rightarrow \mathrm{Cu}\text{(s)}\) gives \(\mathrm{Fe}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)}\).
Displacement reactions, of the general form \(\mathrm{X} + \mathrm{YZ} \rightarrow \mathrm{XZ} + \mathrm{Y}\), are redox reactions, since an ion or atom in a compound is replaced by an ion or atom of another element.
Electron transfer describes ionic reactions well, but reactions between covalent molecules involve only a shift of electrons. The oxidation number method covers both cases, because the electron pair of a covalent bond is assumed to belong completely to the more electronegative element.
The oxidation number of an element in a compound is the number of electrical charges it carries, assuming complete electron transfer in the case of a covalent bond.
Rules for assigning oxidation number: (1) each atom of an element in the free state is zero; (2) an atom in a monoatomic ion carries its own charge, so alkali metals are +1, alkaline earth metals +2 and Al is +3 in their compounds; (3) O is usually −2, but −1 in peroxides and the peroxide ion, −1/2 in superoxides, and +2 in \(\mathrm{OF_2}\); (4) H is +1 when bonded to nonmetals and −1 when bonded to metals; (5) F is −1 in all its compounds, while Cl, Br and I are usually −1 in halides but +1 when bonded to oxygen.
Three further rules complete the working method: (6) the algebraic sum of the oxidation numbers of all atoms in a neutral molecule is zero; (7) in a polyatomic ion the sum equals the net charge of the ion; (8) when two or more atoms of an element are present, the value obtained is the average oxidation number of all those atoms.
An oxidation number may be positive or negative, and a whole number or a fraction. A fractional value, such as 2.5 for S in \(\mathrm{Na_2S_4O_6}\), is always an average; in the tetrathionate ion two sulfur atoms are at +5 and two at 0, giving \(10/4 = 2.5\).
Stock notation writes the oxidation number of a metal as a Roman numeral in parentheses after its symbol, as in \(\mathrm{Mn(IV)O_2}\) or \(\mathrm{Sn(II)Cl_2}\). It is named after the German scientist Alfred Stock.
In terms of oxidation number, oxidation is an increase in the oxidation number of an element and reduction is a decrease. An oxidising agent increases the oxidation number of an element in another substance while a constituent element in it undergoes a decrease; a reducing agent lowers the oxidation number of an element in another substance while a constituent element in it undergoes an increase.
Redox equations are balanced by two methods, the oxidation number method and the ion electron or half reaction method. Both rest on the same requirement: the total increase in oxidation number must equal the total decrease, which is the same as saying electrons lost equal electrons gained.
In the oxidation number method, all atoms other than H and O are balanced first, the increase and decrease per atom are made equal by inserting coefficients, oxygen is then balanced with \(\mathrm{H_2O}\) and hydrogen with \(\mathrm{H^+}\); in basic medium, \(\mathrm{OH^-}\) equal to the \(\mathrm{H^+}\) added is put on both sides and combined with \(\mathrm{H^+}\) to give water. This last step is omitted for acidic medium. The equation is finally checked for both atoms and charges.
In the ion electron method, the equation is split into two half equations, each is balanced for atoms other than O and H, then for O with \(\mathrm{H_2O}\) and for H with \(\mathrm{H^+}\); electrons are added to the right of the oxidation half and to the left of the reduction half to balance charges. The halves are multiplied so that the electrons are equal, then added and the electrons cancelled. In basic medium, \(\mathrm{OH^-}\) equal to the \(\mathrm{H^+}\) present is added to both sides. The final equation is checked for atoms and charges.
The overall route of the chapter, from recognising a redox reaction to predicting whether it will occur, is condensed in FC 6.6.
Flowchart
FC 6.6 – The chapter at a glance: a given reaction → test it by the classical rules (O, H, electronegative or electropositive element added or removed) → or by electron transfer (electrons lost or gained) → or by oxidation number (value rises = oxidation, falls = reduction) → identify the oxidant (itself reduced) and the reductant (itself oxidised) → balance the equation, branching into either the oxidation number method or the ion electron method, each branch further splitting into acidic medium (H⁺ and H₂O only) and basic medium (add OH⁻ as the last step) → separate the two half reactions into two containers with a salt bridge → an electrochemical cell such as the Daniel cell → measure E⁰ against the standard hydrogen electrode → add the two E⁰ values to predict whether the reaction is spontaneous.
A displacement reaction such as \(\mathrm{Zn}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)}\) can be carried out either by direct contact, in which electrons pass straight from zinc to \(\mathrm{Cu^{2+}}\), or by separating the two half reactions into two containers joined by a salt bridge and an external wire.
The second arrangement is the Daniel cell, a simple electrochemical cell in which electricity is generated by a redox reaction. The zinc plate is the anode, the negative electrode where oxidation occurs, and the copper plate is the cathode, the positive electrode where reduction occurs. The salt bridge is a U-shaped tube holding a gel of \(\mathrm{KCl}\) or \(\mathrm{NH_4NO_3}\) in agar-agar.
The circuit has two parts. In the external wire the current is carried by electrons, which flow from anode to cathode; in the solutions it is carried by the movement of ions. Conventional current flows the other way, from cathode to anode.
The electrical potential established at the electrodes of an electrochemical cell is called the electrode potential. It depends on the nature of the metal and its ions, the concentration of the ions, and the temperature. The reaction taking place at an electrode is the electrode reaction, and two species linked by transfer of electrons form a redox couple.
When the concentration of every species in the electrode reaction is unity and the temperature is 298 K, the measured value is the standard electrode potential, \(E^0\), expressed in volts. By convention the standard hydrogen electrode is assigned 0.00 V and is used as the reference for all other couples.
The tabulated \(E^0\) values refer to reduction reactions. A negative \(E^0\) means the couple is a stronger reducing agent than the \(\mathrm{H^+/H_2}\) couple. A large negative value marks a strong reducing agent, and a more positive value marks a strong oxidising agent.
On this scale the alkali metals have high negative \(E^0\) values and are strong reducing agents, having a great tendency to give away an electron and form cations, while fluorine has a highly positive \(E^0\) and is a very strong oxidising agent.
Spontaneity can be predicted without performing the reaction. Write one species as a reduction half reaction and the other as an oxidation half reaction, reverse the sign of \(E^0\) for the half reaction written as oxidation, and add the two values. If the sum is positive, the reaction between the two species is spontaneous.
6.6 Check Your Concepts
Explain why oxidation can never take place alone in a reaction vessel. Illustrate your answer using the reaction \(2\,\mathrm{Fe_2O_3} + 3\,\mathrm{C}\text{(s)} \rightarrow 4\,\mathrm{Fe}\text{(s)} + 3\,\mathrm{CO_2}\text{(g)}\).
The classical meaning of oxidation was widened more than once in this chapter. State each widened definition and give one reaction from the chapter that made the widening necessary.
Assertion (A): In the reaction \(\mathrm{Fe}\text{(s)} + \mathrm{Cu^{2+}}\text{(aq)} \rightarrow \mathrm{Fe^{2+}}\text{(aq)} + \mathrm{Cu}\text{(s)}\), iron acts as the reducing agent. Reason (R): A reducing agent is itself reduced during the reaction. State whether A and R are true, and whether R is the correct explanation of A.
Why is the oxidation number of oxygen +2 in \(\mathrm{OF_2}\) but −2 in \(\mathrm{CaO}\)? Answer in terms of electronegativity.
Explain why every atom of an element in the free state is assigned an oxidation number of zero, whether the element exists as \(\mathrm{H_2}\), \(\mathrm{O_3}\), \(\mathrm{P_4}\) or \(\mathrm{S_8}\).
A student states that the value +2.5 for sulfur in \(\mathrm{Na_2S_4O_6}\) shows that half an electron has been transferred to each sulfur atom. Explain what is wrong with this statement, and what the fractional value actually indicates.
Distinguish between complete electron transfer and electron shift, giving one reaction of each type from this chapter. Why does the oxidation number method work for both?
In the ion electron method, why must electrons be written on the right side of the oxidation half equation and on the left side of the reduction half equation? What would go wrong if this were reversed?
Why is step IV of the oxidation number method (and step VI of the ion electron method) omitted when the reaction takes place in acidic medium?
While balancing a redox equation, all atoms except hydrogen and oxygen are balanced first. Give the chemical reason for treating H and O differently from every other element.
Acidified potassium dichromate is normally prepared using dilute sulphuric acid and never hydrochloric acid. Give the reason.
When acidified potassium permanganate solution is added to a solution containing \(\mathrm{Fe^{2+}}\) ions, the deep purple colour fades. State the change in oxidation number of manganese and of iron, and explain why no separate indicator is needed in such a titration.
Arrange the following in increasing order of oxidising power, using the values in Table 6.1, and give the reason for your order: \(\mathrm{I_2}\), \(\mathrm{Br_2}\), \(\mathrm{Cl_2}\), \(\mathrm{F_2}\).
Assertion (A): Alkali metals are strong reducing agents. Reason (R): Alkali metals have large negative values of standard electrode potential. State whether A and R are true, and whether R is the correct explanation of A.
Match each species in Column I with the oxidation number of its central atom in Column II.
| Column I | Column II |
|---|---|
| a. \(\mathrm{MnO_4^{-}}\) | i. +3 |
| b. \(\mathrm{Cr_2O_7^{2-}}\) | ii. −3 |
| c. \(\mathrm{NH_4^{+}}\) | iii. +7 |
| d. \(\mathrm{NO_2^{-}}\) | iv. +6 |
What happens if the salt bridge of a working Daniel cell is lifted out of the two beakers? Explain your answer in terms of the charge that builds up in each half-cell.
In a Daniel cell, state which plate loses mass and which gains mass, and write the electrode reaction occurring at each plate. Name the redox couple present in each beaker.
Electrons flow from anode to cathode in the external circuit, yet conventional current is said to flow from cathode to anode. Explain why the two statements do not contradict each other.
Why is it impossible to measure the electrode potential of a single electrode, and how does the standard hydrogen electrode solve this problem?
Iron displaces copper from copper sulfate solution, but copper does not displace iron from ferrous sulfate solution. Explain this observation using the idea of reducing power.
A block of zinc is bolted to the steel hull of a ship. Using the position of zinc in Table 6.1, explain why the zinc corrodes instead of the iron.
The five steps of the oxidation number method are given below in a jumbled order. Arrange them in the correct sequence.
- Balance oxygen by adding \(\mathrm{H_2O}\) and hydrogen by adding \(\mathrm{H^+}\).
- Check the equation for both atoms and charges.
- Balance all atoms except H and O in the skeletal equation.
- If the medium is basic, add \(\mathrm{OH^-}\) equal to the \(\mathrm{H^+}\) added, on both sides.
- Make the total increase and total decrease in oxidation number equal by inserting coefficients.
Complete the following reaction and state which species is reduced and which is the reducing agent: \(2\,\mathrm{FeCl_3} + \mathrm{H_2}\text{(g)} \rightarrow\) ……
Explain why a substance in which an element is present in its highest possible oxidation number, such as \(\mathrm{ClO_4^{-}}\) or \(\mathrm{MnO_4^{-}}\), is expected to act as an oxidising agent.
Write the formula for the following compounds
- Mercury(II) chloride
- Thallium(I) sulphate
- Tin(IV) oxide
- Chromium(III) oxide
In which chemical reaction does carbon exhibit a variation of oxidation state from −4 to +4? Write the balanced chemical reaction.
In which reaction does nitrogen exhibit a variation of oxidation state from −3 to +5?
Calculate the oxidation number of the underlined atoms.
- \(\mathrm{H_2\underline{S}O_4}\)
- \(\mathrm{H\underline{N}O_3}\)
- \(\mathrm{H_3\underline{P}O_3}\)
- \(\mathrm{K_2\underline{C}_2O_4}\)
- \(\mathrm{H_2\underline{S}_4O_6}\)
- \(\mathrm{\underline{Cr}_2O_7^{2-}}\)
- \(\mathrm{NaH_2\underline{P}O_4}\)
Justify that the following reactions are redox reactions; identify the species oxidized/reduced, which acts as an oxidant and which acts as a reductant.
- \(2\,\mathrm{Cu_2O}\text{(s)} + \mathrm{Cu_2S}\text{(s)} \rightarrow 6\,\mathrm{Cu}\text{(s)} + \mathrm{SO_2}\text{(g)}\)
- \(\mathrm{HF}\text{(aq)} + \mathrm{OH^-}\text{(aq)} \rightarrow \mathrm{H_2O}\text{(l)} + \mathrm{F^-}\text{(aq)}\)
- \(\mathrm{I_2}\text{(aq)} + 2\,\mathrm{S_2O_3^{2-}}\text{(aq)} \rightarrow \mathrm{S_4O_6^{2-}}\text{(aq)} + 2\,\mathrm{I^-}\text{(aq)}\)
What is oxidation? Which one of the following pairs of species is in its oxidized state?
- \(\mathrm{Mg}/\mathrm{Mg^{2+}}\)
- \(\mathrm{Cu}/\mathrm{Cu^{2+}}\)
- \(\mathrm{O_2}/\mathrm{O^{2-}}\)
- \(\mathrm{Cl_2}/\mathrm{Cl^-}\)
Justify the following reaction as a redox reaction.
\[2\,\mathrm{Na}\text{(s)} + \mathrm{S}\text{(s)} \rightarrow \mathrm{Na_2S}\text{(s)}\]
Find out the oxidizing and reducing agents.
Provide the Stock notation for the following compounds: \(\mathrm{HAuCl_4}\), \(\mathrm{Tl_2O}\), \(\mathrm{FeO}\), \(\mathrm{Fe_2O_3}\), \(\mathrm{MnO}\) and \(\mathrm{CuO}\).
Assign oxidation number to each atom in the following species.
- \(\mathrm{Cr(OH)_4^-}\)
- \(\mathrm{Na_2S_2O_3}\)
- \(\mathrm{H_3BO_3}\)
Which of the following redox couples is the stronger oxidizing agent?
- \(\mathrm{Cl_2}\) (\(E^0 = 1.36\) V) and \(\mathrm{Br_2}\) (\(E^0 = 1.09\) V)
- \(\mathrm{MnO_4^-}\) (\(E^0 = 1.51\) V) and \(\mathrm{Cr_2O_7^{2-}}\) (\(E^0 = 1.33\) V)
Which of the following redox couples is the stronger reducing agent?
- Li (\(E^0 = -3.05\) V) and Mg (\(E^0 = -2.36\) V)
- Zn (\(E^0 = -0.76\) V) and Fe (\(E^0 = -0.44\) V)
Balance the following reactions by the oxidation number method.
- \(\mathrm{Cr_2O_7^{2-}}\text{(aq)} + \mathrm{SO_3^{2-}}\text{(aq)} \rightarrow \mathrm{Cr^{3+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)}\) (acidic)
- \(\mathrm{MnO_4^{-}}\text{(aq)} + \mathrm{Br^{-}}\text{(aq)} \rightarrow \mathrm{MnO_2}\text{(s)} + \mathrm{BrO_3^{-}}\text{(aq)}\) (basic)
- \(\mathrm{H_2SO_4}\text{(aq)} + \mathrm{C}\text{(s)} \rightarrow \mathrm{CO_2}\text{(g)} + \mathrm{SO_2}\text{(g)} + \mathrm{H_2O}\text{(l)}\) (acidic)
- \(\mathrm{Bi(OH)_3}\text{(g)} + \mathrm{Sn(OH)_3^{-}}\text{(aq)} \rightarrow \mathrm{Bi}\text{(s)} + \mathrm{Sn(OH)_6^{2-}}\text{(aq)}\) (basic)
Balance the following redox equations by the half reaction method.
- \(\mathrm{H_2C_2O_4}\text{(aq)} + \mathrm{MnO_4^{-}}\text{(aq)} \rightarrow \mathrm{CO_2}\text{(g)} + \mathrm{Mn^{2+}}\text{(aq)}\) (acidic)
- \(\mathrm{Bi(OH)_3}\text{(s)} + \mathrm{SnO_2^{2-}}\text{(aq)} \rightarrow \mathrm{SnO_3^{2-}}\text{(aq)} + \mathrm{Bi}\text{(s)}\) (basic)
6.7 Practice With Numericals
Calculate the oxidation number of the underlined element in each of the following:
- \(\mathrm{Na\underline{Cl}O_3}\)
- \(\mathrm{K_2\underline{Mn}O_4}\)
- \(\mathrm{H_4\underline{P}_2O_7}\)
- \(\mathrm{Ca(\underline{Cl}O_4)_2}\)
- \(\mathrm{\underline{N}_2H_4}\)
Determine the average oxidation number of carbon in acetic acid, \(\mathrm{CH_3COOH}\), and state whether the value obtained is the actual oxidation number of every carbon atom in the molecule.
The oxidation number of sulfur in a certain oxoacid \(\mathrm{H_2S_xO_6}\) is found to be +2.5. Given that the compound contains 6 oxygen atoms and 2 hydrogen atoms, calculate the value of x.
Balance the following equation by the oxidation number method, showing the increase and decrease per atom at each stage:
\[\mathrm{Cu}\text{(s)} + \mathrm{H_2SO_4}\text{(aq)} \rightarrow \mathrm{CuSO_4}\text{(aq)} + \mathrm{SO_2}\text{(g)} + \mathrm{H_2O}\text{(l)}\]
- Balance the following equation in acidic medium by the ion electron (half reaction) method, writing both half equations separately:
\[\mathrm{Cr_2O_7^{2-}}\text{(aq)} + \mathrm{I^{-}}\text{(aq)} \rightarrow \mathrm{Cr^{3+}}\text{(aq)} + \mathrm{I_2}\text{(s)}\]
- Balance the following equation in basic medium by the ion electron (half reaction) method:
\[\mathrm{MnO_4^{-}}\text{(aq)} + \mathrm{SO_3^{2-}}\text{(aq)} \rightarrow \mathrm{MnO_2}\text{(s)} + \mathrm{SO_4^{2-}}\text{(aq)}\]
For each of the following reductions in acidic medium, state the number of electrons gained per formula unit of the oxidising agent:
- \(\mathrm{MnO_4^{-}} \rightarrow \mathrm{Mn^{2+}}\)
- \(\mathrm{Cr_2O_7^{2-}} \rightarrow 2\,\mathrm{Cr^{3+}}\)
- \(\mathrm{ClO_4^{-}} \rightarrow \mathrm{ClO_2^{-}}\)
In the balanced net ionic equation for the reaction of \(\mathrm{MnO_4^{-}}\) with \(\mathrm{Fe^{2+}}\) in acidic medium, verify the balance of charge by adding up the total charge on each side of the equation.
Using the values in Table 6.1, write the two half reactions with their \(E^0\) values, find the sum, and state whether the following reaction is spontaneous:
\[\mathrm{Zn}\text{(s)} + 2\,\mathrm{H^{+}}\text{(aq)} \rightarrow \mathrm{Zn^{2+}}\text{(aq)} + \mathrm{H_2}\text{(g)}\]
Using Table 6.1, decide whether iodine can oxidise \(\mathrm{Fe^{2+}}\) to \(\mathrm{Fe^{3+}}\). Show both half reactions with their \(E^0\) values and the sum obtained.
Given \(E^0\) for \(\mathrm{Cu^{2+}/Cu}\) as \(+0.34\) V, use Table 6.1 to determine whether copper metal will dissolve in dilute acid with the evolution of hydrogen gas. Support your answer with the sum of the two \(E^0\) values.
From the values in Table 6.1, arrange Li, K, Ca, Na, Mg, Al and Zn in decreasing order of reducing power, and state the difference in \(E^0\) between the strongest and the weakest reducing agent in your list.
6.8 Objective Questions
- Oxidation numbers of Cl atoms marked as Cl\(^a\) and Cl\(^b\) in \(\mathrm{CaOCl_2}\) (bleaching powder) are
Diagram
Fig. 6.17 – Structure of bleaching powder for this question: a Ca atom joined by one bond to a chlorine labelled Clᵃ and by another bond to an O, which is in turn joined to a chlorine labelled Clᵇ.
a. zero in each
b. −1 in Cl$^a$ and +1 in Cl$^b$
c. +1 in Cl$^a$ and −1 in Cl$^b$
d. 1 in each- Which of the following is not an example of a redox reaction?
- \(\mathrm{CuO} + \mathrm{H_2} \rightarrow \mathrm{Cu} + \mathrm{H_2O}\)
- \(\mathrm{Fe_2O_3} + 3\,\mathrm{CO_2} \rightarrow 2\,\mathrm{Fe} + 3\,\mathrm{CO_2}\)
- \(2\,\mathrm{K} + \mathrm{F_2} \rightarrow 2\,\mathrm{KF}\)
- \(\mathrm{BaCl_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{BaSO_4} + 2\,\mathrm{HCl}\)
- A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is −2, the compound is possibly represented by
- \(\mathrm{A_2(BC_3)_2}\)
- \(\mathrm{A_3(BC_4)_2}\)
- \(\mathrm{A_3(B_4C)_2}\)
- \(\mathrm{ABC_2}\)
- The coefficients p, q, r, s in the reaction
\[p\,\mathrm{Cr_2O_7^{2-}} + q\,\mathrm{Fe^{2+}} \rightarrow r\,\mathrm{Cr^{3+}} + s\,\mathrm{Fe^{3+}} + \mathrm{H_2O}\]
respectively are: a. 1, 2, 6, 6 b. 6, 1, 2, 4 c. 1, 6, 2, 6 d. 1, 2, 4, 6
- For the following redox reaction, find the correct statement.
\[\mathrm{Sn^{2+}} + 2\,\mathrm{Fe^{3+}} \rightarrow \mathrm{Sn^{4+}} + 2\,\mathrm{Fe^{2+}}\] a. \(\mathrm{Sn^{2+}}\) is undergoing oxidation b. \(\mathrm{Fe^{3+}}\) is undergoing oxidation c. It is not a redox reaction d. Both \(\mathrm{Sn^{2+}}\) and \(\mathrm{Fe^{3+}}\) are oxidised
- Oxidation number of carbon in \(\mathrm{H_2CO_3}\) is
- +1
- +2
- +3
- +4
- Which is the correct Stock notation for manganese dioxide?
- \(\mathrm{Mn(I)O_2}\)
- \(\mathrm{Mn(II)O_2}\)
- \(\mathrm{Mn(III)O_2}\)
- \(\mathrm{Mn(IV)O_2}\)
- Oxidation number of oxygen in a superoxide is
- −2
- −1
- \(-\dfrac{1}{2}\)
- 0
- Which of the following halogens always shows oxidation state −1?
- F
- Cl
- Br
- I
- The process \(\mathrm{SO_2} \rightarrow \mathrm{S_2Cl_2}\) is
- Reduction
- Oxidation
- Neither oxidation nor reduction
- Oxidation and reduction
- Complete the following table
- Assign oxidation number to the underlined species and write the Stock notation of the compound.
| Compound | Oxidation number | Stock notation |
|---|---|---|
| \(\mathrm{\underline{Au}Cl_3}\) | ||
| \(\mathrm{\underline{Sn}Cl_2}\) | ||
| \(\mathrm{\underline{V}_2O_7^{4-}}\) | ||
| \(\mathrm{\underline{Pt}Cl_6^{2-}}\) | ||
| \(\mathrm{H_3\underline{As}O_3}\) |