Half-equations and ionic equationsEdexcel A-Level Chemistry: Revision notes

A half-equation shows either oxidation or reduction on its own, with the electrons written explicitly. Electrons are written on the right for oxidation (loss) and on the left for reduction (gain).

• Oxidation: Zn → Zn²⁺ + 2e⁻
• Reduction: Cu²⁺ + 2e⁻ → Cu

Every half-equation must balance for atoms and for charge. The overall redox reaction is the sum of one oxidation and one reduction half-equation, with the electrons cancelled.

For species containing oxygen in acidic solution, balance in this order:

1. Balance the atom being oxidised or reduced (e.g. 2Cr)
2. Balance oxygen by adding H₂O
3. Balance hydrogen by adding H⁺
4. Balance charge by adding electrons

Worked example: MnO₄⁻ → Mn²⁺. Add 4H₂O on the right (for 4 O). Add 8H⁺ on the left (for 8 H). Charge on left: −1 + 8 = +7; on the right: +2; so add 5e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

Dichromate: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

To build the overall ionic equation:

1. Write the oxidation and reduction half-equations
2. Multiply one or both so that the number of electrons is the same
3. Add them together and cancel the electrons
4. Cancel any species (such as H⁺ or H₂O) that appear on both sides

Example: MnO₄⁻ with Fe²⁺. Fe²⁺ → Fe³⁺ + e⁻ is multiplied by 5, then added to the manganate half-equation:

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

Simple example: Zn + Cu²⁺ → Zn²⁺ + Cu.

• Halogen displacement: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ (Cl₂ + 2e⁻ → 2Cl⁻ and 2Br⁻ → Br₂ + 2e⁻)
• Iron(III) and iodide: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
• Dichromate and iron(II): Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
• Hydrogen peroxide and manganate(VII): 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂

When the half-equation is not given, use the oxidation number change to decide how many electrons are involved, then check atoms and charge. Include state symbols when asked, with ions as (aq).