1. Combining half-equations
- Write and check the oxidation and reduction half-equations separately.
- Find the lowest common multiple of their electron numbers.
- Multiply every coefficient in each half-equation by the needed factor, including H⁺, H₂O and electrons.
- Add the left sides and add the right sides. Cancel electrons, and cancel any identical species appearing on both sides.
- Simplify to the smallest whole-number coefficients, then check every atom and total charge.
The number of electrons lost must equal the number gained. Electrons therefore do not appear in the final overall equation. Do not cancel different species just because they contain the same element: Fe²⁺ and Fe³⁺ are different.
2. Permanganate and iron(II)
The common electron number is five. Multiply the whole iron half-equation by five: 5Fe²⁺ → 5Fe³⁺ + 5e⁻. Add and cancel the five electrons:
| Quantity | Left | Right |
|---|---|---|
| Mn atoms | 1 | 1 |
| Fe atoms | 5 | 5 |
| O atoms | 4 | 4 |
| H atoms | 8 | 8 |
| Total charge | −1 + 8 + 5(+2) = +17 | +2 + 5(+3) = +17 |
Charge is conserved at +17 on each side; it does not have to be zero. MnO₄⁻ is the oxidising agent, and Fe²⁺ is the reducing agent. Spectator ions are omitted from the ionic equation.
3. Dichromate and sulfite
Multiply the sulfite half-equation by three. When the halves are added, cancel six electrons. There are also 6H⁺ on the product side to cancel against 14H⁺ on the reactant side, and 3H₂O on the reactant side to cancel against 7H₂O on the product side.
Check: two Cr, three S, sixteen O and eight H on each side. Charge: −2 − 6 + 8 = 0 on the left; +6 − 6 = 0 on the right. Six electrons were transferred before cancellation.
4. Oxalate and mole ratios
In C₂O₄²⁻, each carbon is +3: 2x + 4(−2) = −2. In CO₂ each carbon is +4. Two carbon atoms each lose one electron:
To combine with the five-electron permanganate reduction, use ten electrons: multiply the permanganate half by two and the oxalate half by five.
Check atoms: two Mn, ten C, twenty-eight O and sixteen H on each side. Charge: −2 + 16 − 10 = +4 on the left; +4 on the right.
5. Iodide and sulfuric acid
For the specified reduction product H₂S, sulfur changes +6 → −2, gaining eight electrons. Iodide oxidation is 2I⁻ → I₂ + 2e⁻, so multiply it by four.
The overall charge is zero on each side. This formal acidic ionic representation uses H₂SO₄ as the sulfur-containing starting species. If the supplied half-equation starts from SO₄²⁻ instead, the corresponding overall equation uses 10H⁺: SO₄²⁻ + 10H⁺ + 8I⁻ → H₂S + 4H₂O + 4I₂. Keep each version internally consistent.
6. Final checks
| Symptom | What to fix |
|---|---|
| Electrons remain in the overall equation | Match electron numbers before adding, then cancel |
| Atoms balance but charges differ | Recheck electron coefficients and ionic charges |
| Hydrogen or oxygen fails after multiplying | Multiply every term, not only the main reactant or electrons |
| An ion's charge was altered to make totals fit | Restore the correct ion; change coefficients instead |
| H⁺ or water occurs on both sides | Cancel equal amounts of the identical species |
| Correct chemistry but wrong species requested | Answer with the requested atom, ion, molecule or reagent |
Oxidation numbers help identify electron transfer. Electrical charges on the actual formulas are what you add for the final charge check. Include state symbols when requested or when they clarify the equation; never invent observations if the question only asks for balancing.
Quick checks
Original Finesse questions; indicative solutions, not official AQA mark allocations.
Q1. Combine Zn → Zn²⁺ + 2e⁻ and Ag⁺ + e⁻ → Ag.Show answer
Multiply the silver half by two and cancel two electrons:
Charge is +2 on both sides. Zn is the reducing agent; Ag⁺ is the oxidising agent.
Q2. Combine Cl₂ + 2e⁻ → 2Cl⁻ with Fe²⁺ → Fe³⁺ + e⁻.Show answer
Multiply the iron half by two:
Charge is +4 on the left and −2 + 6 = +4 on the right. No electrons remain.
Q3. How much Fe²⁺ reacts with 2.40 × 10⁻⁴ mol MnO₄⁻ in acid, forming Mn²⁺ and Fe³⁺?Show answer
The balanced ratio is one MnO₄⁻ to five Fe²⁺. Amount Fe²⁺ = 5 × 2.40 × 10⁻⁴ = 1.20 × 10⁻³ mol.
Q4. How many moles of dichromate oxidise 0.0180 mol sulfite using the equation above?Show answer
One mole Cr₂O₇²⁻ reacts with three moles SO₃²⁻. n(dichromate) = 0.0180/3 = 0.00600 mol.
Q5. A student doubles MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O but writes only 2MnO₄⁻ + 8H⁺ + 10e⁻ → 2Mn²⁺ + 4H₂O. Correct it.Show answer
Every coefficient must double, including H⁺ and water:
Both charges are +4, with two Mn, eight O and sixteen H on each side.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.7 Redox revision guide (N. Goalby) — Combining half-equations, oxalate and sulfuric-acid examples.
- AQA 7405 specification — 3.1.7 Oxidation, reduction and redox equations — 3.1.7: combining half-equations to construct redox equations.
- AQA 7404/1 mark scheme, 2021 (November archive; June header) — Q03.2: sulfite/dichromate half-equations and combination.
- AQA 7404/1 mark scheme, June 2023 — Q06.3: iodide oxidation and sulfuric-acid reduction.
- AQA 7404/1 examiner report, June 2023 — Q06.3: electron cancellation in combined equations.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.