AQA A-Level Chemistry: Oxidation, reduction and redox equations (7405)

This study guide covers section 3.1.7, Oxidation, reduction and redox equations, of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. All of 3.1.7 is AS content, so it is also part of the full A-level; nothing in this section is marked “A-level only”. The specification lists 3.1.7 under Paper 1 for both AS and A-level, and A-level Paper 3 can assess any content. No required practical is listed in this section, but the skills here are used again in inorganic chemistry and in titration calculations.

Use it with the revision notes and the practice questions. For the rest of the course, see the AQA A-Level Chemistry hub and the printable checklist.

What this section covers

Spec section	What you must be able to do	Status
3.1.7	Define oxidation as electron loss and reduction as electron gain	AS and A-level
3.1.7	Describe oxidising agents as electron acceptors and reducing agents as electron donors	AS and A-level
3.1.7	Apply the rules for assigning oxidation states	AS and A-level
3.1.7	Work out the oxidation state of an element in a compound or ion from the formula	AS and A-level
3.1.7	Write half-equations identifying the oxidation and reduction processes in redox reactions	AS and A-level
3.1.7	Combine half-equations to give an overall redox equation	AS and A-level

The specification sums up the topic like this: redox reactions involve a transfer of electrons from the reducing agent to the oxidising agent, and the change in oxidation state shows which element has been oxidised or reduced.

Oxidation, reduction and the two agents

In a redox reaction, electrons move from one species to another. The two processes always happen together.

• Oxidation is the process of electron loss.
• Reduction is the process of electron gain.
• An oxidising agent is an electron acceptor. It takes electrons, so it is itself reduced.
• A reducing agent is an electron donor. It gives electrons, so it is itself oxidised.

A memory aid: OIL RIG (Oxidation Is Loss, Reduction Is Gain). The agent definitions trip people up more than the processes do. The oxidising agent causes oxidation in something else, by removing its electrons.

Worked example 1: zinc in copper(II) sulfate solution

Zinc metal is added to blue copper(II) sulfate solution. A brown solid forms and the blue colour fades.

Ionic equation (sulfate ions are spectators):

Zn + Cu²⁺ → Zn²⁺ + Cu

1. Zinc atoms become Zn²⁺ ions. Each zinc atom loses two electrons. Zinc is oxidised.
2. Cu²⁺ ions become copper atoms. Each Cu²⁺ ion gains two electrons. Cu²⁺ is reduced.
3. Zinc gives electrons, so zinc is the reducing agent.
4. Cu²⁺ accepts electrons, so Cu²⁺ is the oxidising agent.

Name the species exactly. “Copper is reduced” is weaker than “Cu²⁺ ions are reduced”, because copper metal is the product, not the thing that gains electrons.

Rules for assigning oxidation states

An oxidation state is the charge an atom would have if all its bonds were treated as ionic. The specification requires you to know the rules for assigning them. Apply the rules in this order of priority:

Rule	Oxidation state
Uncombined element (Zn, O₂, Cl₂, S₈)	0
Sum of oxidation states in a neutral compound	0
Sum of oxidation states in an ion	the charge on the ion
Simple monatomic ion (Na⁺, Cl⁻, Al³⁺)	the charge on the ion
Group 1 metals in compounds	+1
Group 2 metals in compounds	+2
Aluminium in compounds	+3
Fluorine in compounds	−1 (always)
Hydrogen	+1, except in metal hydrides such as NaH, where it is −1
Oxygen	−2, except in peroxides such as H₂O₂ (−1) and in OF₂ (+2)
Chlorine	−1, except in compounds with oxygen or fluorine

Oxidation states are written with the sign first (+3, −2). A charge on an ion is written with the number first (3+, 2−). Keep these apart.

Roman numerals in names give the oxidation state: iron(III) chloride contains iron in the +3 state; manganate(VII) contains manganese in +7; dichromate(VI) contains chromium in +6; chlorate(I) contains chlorine in +1.

Worked example 2: oxidation states from formulae

Set up a simple equation: (oxidation state of unknown × number of atoms) + (known contributions) = overall charge.

(a) Sulfur in SO₄²⁻

x + 4(−2) = −2
x − 8 = −2
x = +6

(b) Chromium in Cr₂O₇²⁻

2x + 7(−2) = −2
2x = +12
x = +6

(c) Nitrogen in NH₄⁺

x + 4(+1) = +1
x = −3

(d) Oxygen in H₂O₂

Hydrogen ranks above oxygen in the rules, so H is +1:

2(+1) + 2x = 0
x = −1

(e) Hydrogen in NaH

Sodium, a Group 1 metal, is +1. So hydrogen must be −1. This is a metal hydride, the exception in the table.

(f) Sulfur in S₄O₆²⁻

4x + 6(−2) = −2
4x = +10
x = +2.5

A fractional value is an average. It means the sulfur atoms in the ion are not all in the same environment. You may be asked for it, and +2.5 is the accepted answer.

Using oxidation states to spot what is oxidised and reduced

An increase in oxidation state means oxidation. A decrease means reduction. This works even when no ions are obvious in the equation.

Worked example 3: chlorine and sodium bromide solution

Cl₂ + 2NaBr → 2NaCl + Br₂

Element	Before	After	Change	Process
Cl	0 (in Cl₂)	−1 (in NaCl)	down by 1	reduced
Br	−1 (in NaBr)	0 (in Br₂)	up by 1	oxidised
Na	+1	+1	none	spectator

Chlorine is the oxidising agent. Bromide ions are the reducing agent. Sodium ions take no part, so the ionic equation is:

Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

Writing half-equations

A half-equation shows one process only, with the electrons written in. Electrons appear on the left in a reduction and on the right in an oxidation.

Simple examples from worked example 3:

Reduction:  Cl₂ + 2e⁻ → 2Cl⁻
Oxidation:  2Br⁻ → Br₂ + 2e⁻

Many redox reactions happen in acidic solution and involve oxygen-containing ions. Use this sequence:

1. Balance the atoms of the element that changes oxidation state.
2. Balance oxygen by adding H₂O.
3. Balance hydrogen by adding H⁺.
4. Balance charge by adding electrons (e⁻) to the more positive side.
5. Check: atoms balance, charges balance, and the number of electrons equals the change in oxidation state × number of atoms changing.

Worked example 4: manganate(VII) to manganese(II) in acid

MnO₄⁻ → Mn²⁺

1. Mn is balanced (one each side).
2. Four O on the left, so add 4H₂O on the right: MnO₄⁻ → Mn²⁺ + 4H₂O
3. Eight H on the right, so add 8H⁺ on the left: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
4. Charge on left = −1 + 8 = +7. Charge on right = +2. Add 5e⁻ to the left:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

5. Check: Mn goes from +7 to +2, a change of 5, and five electrons are gained. Electrons on the left, so this is a reduction.

Worked example 5: sulfur dioxide to sulfate ions

SO₂ → SO₄²⁻

1. S is balanced.
2. Two O on the left, four on the right, so add 2H₂O on the left: SO₂ + 2H₂O → SO₄²⁻
3. Four H on the left, so add 4H⁺ on the right: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺
4. Charge on left = 0. Charge on right = −2 + 4 = +2. Add 2e⁻ to the right:

SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻

5. Check: S goes from +4 to +6, a change of 2, and two electrons are lost. This is an oxidation.

Combining half-equations

To get the overall equation:

1. Write the oxidation and the reduction half-equations.
2. Multiply one or both so that the number of electrons lost equals the number gained.
3. Add the two half-equations.
4. Cancel the electrons, and cancel any H⁺ or H₂O that appear on both sides.
5. Check atoms and charge.

Worked example 6: manganate(VII) and iron(II) ions

Reduction:  MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation:  Fe²⁺ → Fe³⁺ + e⁻

The reduction needs 5 electrons, so multiply the oxidation by 5:

5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add and cancel 5e⁻:

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Charge check: left = −1 + 8 + 10 = +17; right = +2 + 15 = +17. Balanced.

This 1 : 5 ratio of MnO₄⁻ to Fe²⁺ is what you would use in a titration calculation, which links to amount of substance (see the mole revision notes for moles from volumes and concentrations).

Worked example 7: manganate(VII) and sulfur dioxide (with cancelling)

Reduction:  MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation:  SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻

Lowest common multiple of 5 and 2 is 10. Multiply the reduction by 2 and the oxidation by 5:

2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
5SO₂ + 10H₂O → 5SO₄²⁻ + 20H⁺ + 10e⁻

Add, then cancel. There are 16H⁺ on the left and 20H⁺ on the right, leaving 4H⁺ on the right. There are 10H₂O on the left and 8H₂O on the right, leaving 2H₂O on the left:

2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺

Charge check: left = −2; right = +4 − 10 + 4 = −2. Balanced.

Common errors

• Saying the oxidising agent is oxidised. It is reduced, because it accepts electrons.
• Writing an oxidation state as 2+ instead of +2, or a charge as +2 instead of 2+.
• Forgetting the exceptions: H is −1 in NaH, O is −1 in H₂O₂, Cl is positive in ClO⁻ and ClO₃⁻.
• Balancing charge with H⁺ instead of electrons, or putting electrons on the wrong side.
• Leaving electrons in the overall equation. If any e⁻ remain, the multipliers are wrong.
• Not cancelling H⁺ and H₂O that appear on both sides (worked example 7).
• Stating “copper is reduced” when the species reduced is Cu²⁺.
• Treating spectator ions such as Na⁺ or SO₄²⁻ as if they change oxidation state.

Where to go next

• Condensed summary and a quick self-test: revision notes
• Original exam-style questions with mark-by-mark answers: practice questions
• Ionic charges and why ions form: AQA A-Level Chemistry bonding
• Electron arrangements behind ion charges: atomic structure study guide
• Mole calculations for titrations: relative mass and the mole
• Paper formats and timing: AQA A-Level Chemistry exam preparation
• Check your gaps across the course: free diagnostics

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, section 3.1.7 Oxidation, reduction and redox equations. Published by AQA.