Writing electrode half-equations (Higher tier)
A half-equation shows what happens at one electrode: which ion is discharged and how many electrons it gains or loses. The two half-equations together describe the full electrolysis.
Rules for writing half-equations
- Write the species being discharged on the LHS.
- Write the product on the RHS.
- Balance atoms.
- Add electrons to balance charge.
- Reduction (cathode): electrons on the LHS.
- Oxidation (anode): electrons on the RHS.
Cathode reactions (reduction)
- Pb²⁺ + 2e⁻ → Pb (molten PbBr₂)
- 2H⁺ + 2e⁻ → H₂ (aqueous, when metal more reactive than H)
- Cu²⁺ + 2e⁻ → Cu (aqueous CuSO₄)
- Al³⁺ + 3e⁻ → Al (molten Al₂O₃ in cryolite)
Anode reactions (oxidation)
- 2Br⁻ → Br₂ + 2e⁻ (molten PbBr₂)
- 2Cl⁻ → Cl₂ + 2e⁻ (NaCl(aq))
- 4OH⁻ → O₂ + 2H₂O + 4e⁻ (sulfate solutions; OH⁻ from water)
- 2O²⁻ → O₂ + 4e⁻ (molten Al₂O₃)
Combining half-equations
To get the overall ionic equation, multiply each half-equation so electrons cancel, then add.
Example: NaCl(aq).
- Cathode: 2H⁺ + 2e⁻ → H₂
- Anode: 2Cl⁻ → Cl₂ + 2e⁻
Add: 2H⁺ + 2Cl⁻ → H₂ + Cl₂. The 2e⁻ on each side cancel.
✦Worked example— Example with different electron numbers
CuSO₄(aq) electrolysis with inert electrodes:
- Cathode: Cu²⁺ + 2e⁻ → Cu (×2 → 2Cu²⁺ + 4e⁻ → 2Cu)
- Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻
Add: 2Cu²⁺ + 4OH⁻ → 2Cu + O₂ + 2H₂O. (Sulfate is spectator.)
Predicting half-equations from solutions
Step 1: identify all ions present (the salt's + H⁺ + OH⁻ from water). Step 2: apply discharge rules (C4.10). Step 3: write each half-equation, ensuring atoms and charge balance.
⚠Common mistakes
- Electrons on wrong side. Cathode: gain (LHS). Anode: lose (RHS).
- Charge unbalanced. Always check the total charge on each side after adding electrons.
- Missing the coefficient on the halide. 2Cl⁻ → Cl₂ + 2e⁻ (need TWO chlorides for ONE Cl₂).
- Forgetting OH⁻ in sulfate solutions — write 4OH⁻ → O₂ + 2H₂O + 4e⁻.
Links
Builds on C4.10, C4.11. Used in C5 (electrochemical cells). Foundational for A-level redox.
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