Ionic Half Equations
Section: 4. Electrochemistry | Syllabus: Cambridge IGCSE Chemistry (0620)
Writing Half Equations
A half equation shows the reaction at just one electrode - the gain or loss of electrons by an ion.
- Cathode (reduction): electrons appear on the left - a positive ion gains electrons: Ion⁺ + e⁻ → Atom
- Anode (oxidation): electrons appear on the right - a negative ion loses electrons: Ion⁻ → Atom + e⁻
- Every half equation must be balanced for both atoms and charge
Method
- Identify the ion being discharged and its product
- Balance the atoms (add coefficients if the product is diatomic, e.g. Cl₂)
- Add electrons to balance the charge
- Check atoms and charge are balanced on both sides, then add state symbols
Worked Example: Copper at the Cathode
- Question: write the half equation for the discharge of copper ions at the cathode. [2]
- Answer: ion = Cu²⁺, product = Cu (1 atom, already balanced); charge on the left is +2, so add 2e⁻: Cu²⁺(aq) + 2e⁻ → Cu(s) [2]
Worked Example: Chlorine at the Anode
- Question: write the half equation for the formation of chlorine gas at the anode. [2]
- Answer: ion = Cl⁻, product = Cl₂ (diatomic, so 2 Cl⁻ needed); charge on the left is -2, so add 2e⁻ on the right: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ [2]
Worked Example: Oxygen from Hydroxide at the Anode
- Question: write the half equation for the formation of oxygen gas from hydroxide ions at the anode. [2]
- Answer: 4 OH⁻ are needed to form 1 O₂ and 2 H₂O; charge on the left is -4, so add 4e⁻ on the right: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ [2]
Common Half Equations - Quick Reference
| At cathode (reduction) | Half equation |
|---|---|
| Hydrogen | 2H⁺(aq) + 2e⁻ → H₂(g) |
| Copper | Cu²⁺(aq) + 2e⁻ → Cu(s) |
| Silver | Ag⁺(aq) + e⁻ → Ag(s) |
| Lead (molten) | Pb²⁺(l) + 2e⁻ → Pb(l) |
| Aluminium (molten) | Al³⁺(l) + 3e⁻ → Al(l) |
| Sodium (molten) | Na⁺(l) + e⁻ → Na(l) |
| At anode (oxidation) | Half equation |
|---|---|
| Chlorine | 2Cl⁻(aq) → Cl₂(g) + 2e⁻ |
| Bromine | 2Br⁻(aq) → Br₂(l) + 2e⁻ |
| Iodine | 2I⁻(aq) → I₂(s) + 2e⁻ |
| Oxygen (from OH⁻, aqueous) | 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ |
| Oxygen (from O²⁻, molten) | 2O²⁻(l) → O₂(g) + 4e⁻ |
Electrons released at the anode must equal electrons gained at the cathode when half equations are combined
Combining Half Equations
To find the overall equation, scale each half equation so the electron counts match, then add them together (the electrons cancel).
Worked Example: Molten Lead Bromide
- Cathode: Pb²⁺ + 2e⁻ → Pb; Anode: 2Br⁻ → Br₂ + 2e⁻ - both already involve 2 electrons, so add directly:
- Overall: Pb²⁺ + 2Br⁻ → Pb + Br₂ (i.e. PbBr₂ → Pb + Br₂)
Worked Example: When Electron Counts Don't Match
- Question: combine the half equations for molten aluminium oxide electrolysis into an overall equation. Cathode: Al³⁺ + 3e⁻ → Al; Anode: 2O²⁻ → O₂ + 4e⁻. [3]
- Answer: the lowest common multiple of 3 and 4 is 12; multiply the cathode equation by 4: 4Al³⁺ + 12e⁻ → 4Al [1]; multiply the anode equation by 3: 6O²⁻ → 3O₂ + 12e⁻ [1]; the 12 electrons now cancel: 4Al³⁺ + 6O²⁻ → 4Al + 3O₂ (i.e. 2Al₂O₃ → 4Al + 3O₂) [1]
Common Mistakes
- ❌ Putting electrons on the wrong side
- ✓ reduction (cathode) has electrons on the left, as a reactant; oxidation (anode) has electrons on the right, as a product
- ❌ Forgetting that Cl₂, Br₂, I₂, H₂, and O₂ are diatomic
- ✓ balance the ion first for a diatomic product, e.g. 2Cl⁻ (not Cl⁻) → Cl₂ + 2e⁻
- ❌ Using different electron numbers on each side when combining half equations
- ✓ scale both half equations to a common multiple of electrons first, so they cancel exactly when added
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