The big idea: Every redox reaction is really two things happening at once: one species loses electrons and another gains them.
A half-equation shows just one of these halves on its own, with the electrons (e⁻) written in.
- Oxidation half-equation — the species loses electrons, so e⁻ appear on the right. - Reduction half-equation — the species gains electrons, so e⁻ appear on the left.
OIL RIG — and where the electrons go: OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
- Oxidation: species → product + e⁻ (electrons on the product side). - Reduction: species + e⁻ → product (electrons on the reactant side).
The two halves must transfer the same number of electrons — the electrons lost by one are exactly the electrons gained by the other.
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A correct half-equation must balance twice: the atoms balance, and then the charge balances. You fix the charge last, by adding the right number of electrons.
The method
- Write the species changing on each side (e.g. Fe²⁺ and Fe³⁺).
- Balance the atoms of that element first.
- Add up the total charge on each side.
- Add electrons (e⁻) to the more positive side until the charges are equal — that is the balanced half-equation.
Worked example — oxidation of iron(II)
Write the half-equation for iron(II) ions, Fe²⁺, being oxidised to iron(III) ions, Fe³⁺.
Solution
- Atoms: one Fe on each side — already balanced.
- Charge: left = +2, right = +3. The right side is more positive, so add one electron to the right to make both sides +2:
- Electrons are on the product side, so this is an oxidation (loss of electrons).
Final answer
Fe²⁺ → Fe³⁺ + e⁻ (an oxidation).
| Process | Half-equation | Type |
|---|---|---|
| Zinc → zinc ion | Zn → Zn²⁺ + 2e⁻ | oxidation (loses e⁻) |
| Copper ion → copper | Cu²⁺ + 2e⁻ → Cu | reduction (gains e⁻) |
| Iron(II) → iron(III) | Fe²⁺ → Fe³⁺ + e⁻ | oxidation |
| Chlorine → chloride | Cl2 + 2e⁻ → 2Cl⁻ | reduction |
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To get the overall redox equation, combine the oxidation and reduction halves so the electrons cancel out. First multiply each half-equation so both have the same number of electrons, then add them and cancel the e⁻.
Why multiply?: Electrons are never left over in the final equation — every electron lost must be gained. If one half releases 2 e⁻ and the other accepts 3 e⁻, scale them to a common total (here 6 e⁻) so the electrons cancel exactly.
Worked example — zinc displacing copper
Combine the half-equations for zinc and copper(II) into the overall equation for zinc reacting with copper(II) ions.
Solution
- Oxidation (Zn loses electrons):
- Reduction (Cu²⁺ gains electrons):
- Both halves transfer 2 electrons, so no multiplying is needed. Add them and cancel the 2e⁻ on each side:
Final answer
Zn + Cu²⁺ → Zn²⁺ + Cu — the electrons cancel, leaving a balanced overall equation.
Worked example — when you must multiply
Combine Al → Al³⁺ + 3e⁻ with Ag⁺ + e⁻ → Ag into the overall equation.
Solution
- The aluminium half releases 3 e⁻; the silver half accepts only 1 e⁻. Multiply the silver half by 3 so both involve 3 e⁻:
- Now add the two halves and cancel the 3e⁻:
- Check: atoms balance and the total charge is +3 on each side.
Final answer
Al + 3Ag⁺ → Al³⁺ + 3Ag.
How this is tested: Half-equations turn up in two ways.
- Paper 1A (MCQ): you are given two half-equations and must pick the correct combined overall equation (watch for the electrons cancelling). - Paper 2: 'deduce the oxidation half-equation' and then 'deduce the overall balanced equation' — the electrons and charges must both balance.
The markers award separate marks for the half-equation and for the correctly combined overall equation.
Score every mark: Include the electrons in each half-equation, check the charge balances, multiply so the electrons are equal, then cancel them. State symbols are usually not required unless the question asks.
IB-style question — magnesium and silver (a)
Magnesium metal reacts with silver(I) ions, Ag⁺, in solution. (a) Deduce the oxidation and reduction half-equations for this reaction. [2]
How to score the marks
- Mark 1 — oxidation. Magnesium loses 2 electrons (it is oxidised):
- Mark 2 — reduction. Each silver(I) ion gains 1 electron (it is reduced):
Final answer
Oxidation: Mg → Mg²⁺ + 2e⁻. Reduction: Ag⁺ + e⁻ → Ag.
IB-style question — magnesium and silver (b)
(b) Hence deduce the overall balanced equation for the reaction. [2]
How to score the marks
- Mark 1 — equalise the electrons. Magnesium releases 2 e⁻ but silver accepts only 1 e⁻, so multiply the silver half by 2:
- Mark 2 — add and cancel. Add the two halves and cancel the 2e⁻ on each side:
Final answer
Mg + 2Ag⁺ → Mg²⁺ + 2Ag (charge +2 on each side; electrons cancel).