Half-equations and balancing redox equationsAQA A-Level Chemistry: Revision notes

A half-equation shows either oxidation or reduction on its own, with electrons written explicitly.

• Oxidation: electrons appear on the right, e.g. Fe²⁺ → Fe³⁺ + e⁻ and 2Br⁻ → Br₂ + 2e⁻
• Reduction: electrons appear on the left, e.g. Cl₂ + 2e⁻ → 2Cl⁻ and Cu²⁺ + 2e⁻ → Cu

They must balance in both atoms and charge.

For a species that gains or loses oxygen in acidic solution:

1. Balance the element being oxidised or reduced
2. Balance O by adding H₂O
3. Balance H by adding H⁺
4. Balance charge by adding e⁻

Worked example. MnO₄⁻ → Mn²⁺. Add 4H₂O on the right, 8H⁺ on the left. Charge on left: −1 + 8 = +7; right: +2. Add 5e⁻ on the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

• MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
• Fe²⁺ → Fe³⁺ + e⁻
• H₂O₂ → O₂ + 2H⁺ + 2e⁻ (hydrogen peroxide as a reducing agent)
• Cl₂ + 2e⁻ → 2Cl⁻ and 2I⁻ → I₂ + 2e⁻
• Zn → Zn²⁺ + 2e⁻

You can build others using the method above, as long as the species and the medium are given.

To write the overall equation:

1. Multiply the half-equations so that the electrons lost equal electrons gained
2. Add them together
3. Cancel the electrons and any species that appear on both sides (H⁺, H₂O)

Worked example. Fe²⁺ with MnO₄⁻: multiply the Fe half-equation by 5. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

For H₂O₂ with MnO₄⁻, multiply by 2 and 5 (10 electrons): 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂

The balanced overall equation gives the mole ratio for titration calculations.

Worked example. 22.40 cm³ of 0.0200 mol dm⁻³ MnO₄⁻ reacts with 25.0 cm³ of Fe²⁺ solution.

• Amount MnO₄⁻ = 0.02240 × 0.0200 = 4.48 × 10⁻⁴ mol
• Ratio 1 : 5, so Fe²⁺ = 2.24 × 10⁻³ mol
• Concentration = 2.24 × 10⁻³ ÷ 0.0250 = 0.0896 mol dm⁻³