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AQA A-Level Chemistry 7405 · 3.1.7 Oxidation, reduction and redox equations

Part 2: Writing and balancing half-equations

All three parts available · worked equations and charge checks included. Reviewed 1 October 2026.

1. Simple half-equations

A half-equation shows either oxidation or reduction. Electrons are on the right for oxidation and on the left for reduction. Each electron has charge −1. Balance both atoms and total electrical charge; the charge total need not be zero.

Simple half-equations
ChangeHalf-equationCharge check
Magnesium oxidisedMg → Mg²⁺ + 2e⁻0 = +2 − 2
Iron(III) reducedFe³⁺ + e⁻ → Fe²⁺+3 − 1 = +2
Bromide oxidised2Br⁻ → Br₂ + 2e⁻−2 = 0 − 2
Chlorine reducedCl₂ + 2e⁻ → 2Cl⁻0 − 2 = −2

Remember diatomic elements such as Br₂, I₂ and Cl₂. Balance by changing coefficients, never by changing the correct chemical formula or the charge of a named ion.

2. Balancing in acid

  1. Write the reactant and product species specified.
  2. Balance atoms other than oxygen and hydrogen.
  3. Balance oxygen by adding H₂O.
  4. Balance hydrogen by adding H⁺.
  5. Balance total charge by adding electrons to the more positive side.
  6. Check every atom, charge and the electron count from oxidation-number changes.

3. Worked complex half-equations

4. Why the specified product matters

Different sulfur products require different electron counts
Reduction in acidSulfur oxidation-number changeElectrons gained
SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O+6 → +42
SO₄²⁻ + 10H⁺ + 8e⁻ → H₂S + 4H₂O+6 → −28

If the question writes H₂SO₄ as the starting species instead of SO₄²⁻, the hydrogen balance changes:

H₂SO₄ + 8H⁺ + 8e⁻ → H₂S + 4H₂O

These are alternative ways of representing the specified acidic reduction. Do not take the H⁺ coefficient from one version and combine it with the starting formula from the other. Do not substitute SO₂ when the question explicitly asks for sulfur or H₂S.

5. Alkaline conditions: an extension of the method

The species formed can depend on the reaction medium. In alkaline conditions, use the stated product and finish with no H⁺ in the equation. One method is to balance in acid first, then add enough OH⁻ to both sides to neutralise every H⁺, form water and cancel water appearing on both sides.

Quick checks

Original Finesse questions; indicative worked solutions, not official AQA mark allocations.

Q1. Write the oxidation half-equation for aluminium forming Al³⁺.Show answer
Al → Al³⁺ + 3e⁻

One Al atom each side; charge 0 = +3 − 3. Electrons are products because oxidation is electron loss.

Q2. Balance IO₃⁻ → I₂ in acidic solution.Show answer

Use 2IO₃⁻ for two iodine atoms; add 6H₂O on the right and 12H⁺ on the left. Left charge is initially +10, so add ten electrons.

2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O

Final charge is zero on both sides; iodine +5 → 0 confirms ten electrons for two atoms.

Q3. Balance H₂O₂ → O₂ in acidic solution and classify the change.Show answer
H₂O₂ → O₂ + 2H⁺ + 2e⁻

Oxygen changes −1 → 0: oxidation. Two oxygen atoms lose two electrons in total. Charge is zero on both sides.

Q4. Why does Cr₂O₇²⁻ reduction to Cr³⁺ use six electrons rather than three?Show answer

Each Cr goes from +6 to +3, gaining three electrons. There are two chromium atoms in each dichromate ion, so six electrons are gained overall.

Q5. Correct MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5e⁻.Show answer

Atoms balance, but electrons are on the wrong side. Left charge is +7 while the shown right charge is −3.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Now both charges are +2; electron gain agrees with reduction.

Sources

Sources and examiner guidance (reviewed 1 October 2026)

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