Oxidation and reduction across the s-, p- and d-blocks

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This is A Level (IA2) content, and AS knowledge is assumed throughout: in particular the oxidation numbers, half-equations and redox vocabulary of Unit 2 Topic 8 (notes unit 8a).

Topic 16 is examined on the Unit 5 paper, WCH15, which may also set synoptic questions on Units 1, 2 and 4; the practical side of the topic, including core practicals 12, 13a and 13b, can also come up on the Unit 6 practical paper, WCH16.

Redox equilibria was the largest single topic on the 2024-2025 WCH15 papers, with about a fifth of the marks: every paper had a full electrochemical-cell question and a manganate(VII) titration.

How the marks work

In the 2024-2025 WCH15 schemes each scoring point earned one mark, written (1); guidance called the marks M1, M2, with no A, B or C codes. "Allow", "Ignore" and "Do not award" lines listed accepted and rejected wordings.

Calculations: a correct final answer usually scored full marks with or without working, and TE (transferred error) carried a wrong earlier value through later steps. Significant figures were ignored unless an answer was given to 1 s.f., or the stem asked for a number of s.f.

Equations: one mark for the correct species, one for balancing; leftover H⁺, H₂O or electrons lost the second mark. State symbols were marked only when the stem asked, and in cell diagrams.

E⦵cell: a value with the wrong sign scored at most part of the marks; a feasibility conclusion had to follow from a calculated E⦵cell.

One asterisked 6-mark item per paper: six indicative points gave up to 4 marks, plus up to 2 for a linked, sustained line of reasoning; wrong chemistry came off the reasoning marks only.

You need this from earlier units

Oxidation number rules and calculating oxidation numbers (notes unit 8a).

Writing ionic half-equations and combining them (notes unit 8a), and redox amounts from equations.

Moles, concentration in mol dm⁻³, volumetric analysis and percentage uncertainty (notes units 1 and 8b).

ΔStotal = ΔSsystem + ΔSsurroundings (notes unit 12a), and ΔStotal = R lnK (notes unit 13).

Le Chatelier's principle for a system at equilibrium (Unit 2, notes unit 9).

The idea: redox is the movement of electrons

A battery, a rusting nail, a breathalyser and the titration of iron tablets all have one thing in common: electrons move from one substance to another.

Topic 16 asks two questions about that movement.

Which way do the electrons want to go?

And how strongly?

The answers come as numbers, standard electrode potentials, that let you predict whether a redox reaction can happen without doing it.

Before you can use those numbers, you need to see any redox change, in any part of the Periodic Table, as a transfer of electrons.

You need to understand the terms 'oxidation' and 'reduction' in terms of electron transfer and changes in oxidation number, applied to s-, p- and d-block elements.

At AS you met the two definitions separately.

Here they are one idea seen two ways.

Two descriptions of the same change

Electron transfer. A species that loses electrons is oxidised; a species that gains electrons is reduced. "OIL RIG" (oxidation is loss, reduction is gain) is a useful memory aid.

Oxidation number. An increase in oxidation number is oxidation; a decrease is reduction.

Why do we need both?

Electron transfer is easy to see when ions form: a sodium atom becomes Na⁺ by losing one electron.

In a covalent species there are no whole ions, so you cannot point to one electron leaving.

Oxidation numbers solve that by bookkeeping: each atom is given the charge it would have if all the bonds were fully ionic.

When the oxidation number of an atom rises by 1, it has (on paper) lost one electron.

The two descriptions always agree, and the change in oxidation number tells you how many electrons appear in the half-equation.

\text{electrons transferred} = \text{change in oxidation number} \times \text{number of atoms that change}

s-block elements

The s-block metals have one or two electrons in their outer s orbital and lose them easily.

In their reactions they are always oxidised, to +1 (Group 1) or +2 (Group 2), so they act as reducing agents:

\text{2Na(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{2NaCl(s)} \qquad \text{Na: } 0 \rightarrow +1

\text{Mg(s)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{e}^- \qquad \text{Mg: } 0 \rightarrow +2

Their ions, Na⁺ and Mg²⁺, are very hard to reduce back to the metal.

You will see why in numbers soon: their standard electrode potentials, −2.71 V and −2.37 V in the Data Booklet, are among the most negative there are.

p-block elements

The p-block is more varied.

Most non-metals gain electrons and are reduced: chlorine to chloride, oxygen to oxide.

But p-block elements can also be oxidised, and some show several oxidation numbers:

Aluminium, a p-block metal, is oxidised: $\text{Al} \rightarrow \text{Al}^{3+} + 3\text{e}^-$ (0 → +3).

Iodide ions are oxidised to iodine: $2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$ (−1 → 0). You will use this in the iodine-thiosulfate titrations.

Sulfur runs from −2 (in H₂S) through 0 (S) and +4 (SO₂) to +6 (SO₄²⁻). In thiosulfate, S₂O₃²⁻, the average oxidation number of sulfur is +2; in the tetrathionate ion, S₄O₆²⁻, it is +2.5.

In hydrogen peroxide, H₂O₂, oxygen is −1. It can be reduced to −2 (in water) or oxidised to 0 (in oxygen gas), so H₂O₂ can act as an oxidising agent or a reducing agent.

d-block elements

The d-block metals are where redox chemistry gets rich, because most of them form ions with more than one oxidation number (the reasons are in notes unit 17a).

Iron is +2 or +3, copper +1 or +2, chromium +2, +3 or +6, manganese +2, +4 or +7, vanadium +2, +3, +4 and +5.

In an oxoanion such as manganate(VII), MnO₄⁻, the manganese is not a 7+ ion: the oxidation number +7 is bookkeeping.

When MnO₄⁻ becomes Mn²⁺ the oxidation number falls by 5, so the half-equation has five electrons:

\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)}

Block

Change

Oxidation number

Electrons per atom

Process

Na → Na⁺

0 → +1

1 lost

oxidation

Ca → Ca²⁺

0 → +2

2 lost

oxidation

Cl₂ → 2Cl⁻

0 → −1

1 gained

reduction

SO₂ → SO₄²⁻

+4 → +6

2 lost

oxidation

Fe²⁺ → Fe³⁺

+2 → +3

1 lost

oxidation

MnO₄⁻ → Mn²⁺

+7 → +2

5 gained

reduction

Cr₂O₇²⁻ → 2Cr³⁺

+6 → +3

3 gained (6 per ion)

reduction

Why the rule works

An oxidation number is defined so that the sum over all atoms equals the charge on the species.

If a species gains n electrons, its total charge falls by n, so the oxidation numbers inside it must fall by n in total.

That is why "change in oxidation number" and "electrons gained or lost" always match, even for an atom buried in an oxoanion.

It also gives you a check on any half-equation: the electrons written in it must equal the total change in oxidation number.

Where this leads

The next sections attach a number, the standard electrode potential, to each half-equation.

That number measures how strongly the oxidised form pulls in electrons.

It lets you predict feasibility, explains why some species disproportionate, underlies the redox titrations and runs fuel cells, all later in this unit.

In Unit 5 you will also use it for the oxidation states of vanadium and chromium (notes unit 17b).

Key terms

Oxidation: loss of electrons, or an increase in oxidation number.

Reduction: gain of electrons, or a decrease in oxidation number.

Oxidising agents and reducing agents: oxidising agents gain electrons and reducing agents lose electrons.

Oxidation number: the charge an atom would have if all the bonds in the compound or ion were fully ionic.

Learn each one word for word: "loss of electrons" and "increase in oxidation number" are the scoring words.

Worked example 1: a d-block and p-block change in one reaction

When copper(II) ions are added to excess iodide ions, a white precipitate of copper(I) iodide forms in a brown solution of iodine: $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2$ Explain, in terms of oxidation numbers and electrons, which species is oxidised and which is reduced.

(3)

1. Cu goes from +2 in Cu²⁺ to +1 in CuI: a decrease, so copper(II) is reduced; each Cu²⁺ gains one electron (1)

2. Two of the four iodide ions go from −1 to 0 in I₂: an increase, so iodide is oxidised; each loses one electron (1)

3. Electrons balance: 2 gained by two Cu²⁺ and 2 lost by two I⁻; the other two I⁻ stay at −1 in CuI (1)

Answer: Cu²⁺ is reduced (+2 to +1) and I⁻ is oxidised (−1 to 0).

Not every iodide ion changes: count only the atoms whose oxidation number changes.

Worked example 2: oxidising agent and reducing agent from oxidation numbers

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ Identify the oxidising agent and the reducing agent, and show that the electrons balance.

(3)

1. Mn: +7 → +2, a decrease of 5, so MnO₄⁻ gains 5 electrons: MnO₄⁻ is the oxidising agent (1)

2. Fe: +2 → +3 for each of 5 Fe²⁺, 5 electrons lost in total: Fe²⁺ is the reducing agent (1)

3. H stays +1 and O stays −2, so the 5 electrons gained equal the 5 lost (1)

Answer: oxidising agent MnO₄⁻; reducing agent Fe²⁺.

The H⁺ ions are not oxidised or reduced: they are needed to remove the oxygen of MnO₄⁻ as water.

Worked example 3: s-, p- and d-block in one reaction

In the thermite reaction, aluminium powder reduces iron(III) oxide: $2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$ .

State the change in oxidation number of each metal and the total number of electrons transferred for the equation as written.

(2)

1. Al: 0 → +3 (oxidised); Fe: +3 → 0 (reduced) (1)

2. two Al atoms each lose 3 electrons and two Fe atoms each gain 3: 6 electrons transferred (1)

Answer: Al 0 → +3, Fe +3 → 0; 6 electrons.

Oxygen stays at −2 throughout, so it plays no part in the electron count.

Worked example 4: an exam-length chain with hydrogen peroxide

Hydrogen peroxide oxidises iron(II) ions in acidic solution.
(a) State the oxidation number of oxygen in H₂O₂. (1)
(b) Write the two half-equations. (2)
(c) Write the overall ionic equation. (2)
(d) Calculate the volume of 0.0500 mol dm⁻³ H₂O₂ that exactly oxidises the iron(II) ions in 25.0 cm³ of 0.120 mol dm⁻³ iron(II) sulfate. (2)

1. (a) −1 (a peroxide) (1)

2. (b) $\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow 2\text{H}_2\text{O}$ (1); $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$ (1)

3. (c) species: H₂O₂, H⁺, Fe²⁺ → Fe³⁺, H₂O (1); balanced: $\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$ (1)

4. (d) n(Fe²⁺) = 25.0/1000 × 0.120 = 3.00 × 10⁻³ mol, so n(H₂O₂) = 1.50 × 10⁻³ mol (1)

5. V = 1.50 × 10⁻³ ÷ 0.0500 = 0.0300 dm³ = 30.0 cm³ (1)

Answer: (d) 30.0 cm³.

The 1 : 2 ratio comes from the electrons: H₂O₂ takes two, each Fe²⁺ gives one.

Exam tip

When a question asks you to explain a redox change "in terms of oxidation numbers", give the element, both oxidation numbers with signs, and the word increase or decrease, then name the process.

"Iron is oxidised" alone does not show the change; "Fe: +2 → +3, an increase, so iron(II) is oxidised" does.

Common mistakes

Treating an oxidation number as a real charge. It is tempting to read MnO₄⁻ as containing an Mn⁷⁺ ion. It does not: the bonds are covalent, and +7 is bookkeeping. Write "+7", never "7+", for an oxidation number.

Counting every atom of an element as changing. In 2Cu²⁺ + 4I⁻ only two of the four iodide ions are oxidised. Count the atoms that actually change.

Forgetting the number of atoms. Cr₂O₇²⁻ → 2Cr³⁺ is a change of 3 per chromium atom but 6 electrons per dichromate ion; writing 3e⁻ in the half-equation leaves it unbalanced for charge.

Calling H⁺ an oxidising agent in a manganate(VII) reaction. The H⁺ ions keep their oxidation number of +1; they react with the oxygen of MnO₄⁻ to form water.

Check your understanding

1. In which change is the transition element reduced? (1)

A Cr₂O₇²⁻ → CrO₄²⁻
B Fe²⁺ → Fe³⁺
C MnO₄⁻ → MnO₂
D Cu → Cu²⁺ Total [1]

2. How many electrons are gained when one dichromate(VI) ion, Cr₂O₇²⁻, is reduced to chromium(III) ions? (1)

A 3
B 6
C 7
D 12 Total [1]

3. Complete the table. (4)

Change

Element whose oxidation number changes

Oxidation numbers (before → after)

Oxidised or reduced?

Ca → Ca²⁺

SO₃²⁻ → SO₄²⁻

VO²⁺ → VO₂⁺

H₂O₂ → H₂O

Total [4]

4. Acidified potassium dichromate(VI) oxidises iodide ions to iodine.
(a) State the oxidation number of chromium before and after the reaction. (2)
(b) Write the half-equation for the oxidation of iodide ions. (1)
(c) Write the half-equation for the reduction of Cr₂O₇²⁻ in acid. (2)
(d) Write the overall ionic equation. (2)
(e) Calculate the amount, in mol, of iodine formed from 20.0 cm³ of 0.0150 mol dm⁻³ potassium dichromate(VI) with excess iodide. (1) Total [8]

5. A student writes: "In MnO₄⁻ the manganese atom has lost seven electrons, so the manganese carries a charge of 7+." Explain why the student is wrong. (2) Total [2]

6. Explain why the s-block metals act only as reducing agents in their reactions, while iron(II) ions can act as either a reducing agent or an oxidising agent. (3) Total [3]

Answers

1. C (1). Mn falls from +7 to +4, a decrease, so it is reduced. A is not correct because Cr is +6 in both ions: this is not a redox change. B is not correct because Fe rises from +2 to +3, which is oxidation. D is not correct because copper loses electrons (0 to +2), which is oxidation.

2. B (1). Each Cr falls from +6 to +3 (3 electrons) and there are two Cr atoms: 6 electrons. A is not correct because it counts only one chromium atom. C is not correct because 7 is the number of water molecules in the half-equation, not the electrons. D is not correct because it doubles the electron count again (6 per Cr).

3. Ca: 0 → +2, oxidised (1); S: +4 → +6, oxidised (1); V: +4 → +5, oxidised (1); O: −1 → −2, reduced (1). Each row needs the element, both numbers and the process.

4. (a) +6 before (1); +3 after (1). (b) $2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$ (1). (c) species with 6e⁻ (1); balanced: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ (1). (d) species (1); balanced with no electrons left: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$ (1). (e) n(Cr₂O₇²⁻) = 3.00 × 10⁻⁴ mol, so n(I₂) = 3 × 3.00 × 10⁻⁴ = 9.00 × 10⁻⁴ mol (1).

5. The bonds between Mn and O are covalent, so no electrons have actually been lost to form a 7+ ion (1); +7 is the oxidation number, the charge Mn would have if all the bonds were fully ionic (1).

6. s-block metals have one or two outer electrons that they lose easily, and their only stable oxidation numbers are 0 and +1 or +2 (1), so they can only lose electrons/be oxidised, which makes them reducing agents (1). Iron(II) can lose an electron to become Fe³⁺ (acting as a reducing agent) or gain two electrons to become Fe (acting as an oxidising agent), because iron has more than one stable oxidation number (1).

Oxidation and reduction across the s-, p- and d-blocks

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