Half-Equation – IGCSE Chemistry Definition
IGCSE Chemistry definition of half-equation: an equation showing the gain or loss of electrons at one electrode. Covers writing and balancing half-equations for electrolysis.
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Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
A half-equation shows what happens at a single electrode during electrolysis — specifically, the gain or loss of electrons by the ions. Writing and balancing half-equations is a key Supplement skill on the 0620 syllabus. Examiners award marks separately for the correct formula, the correct number of electrons, and correct balancing.
The 0620 definition
A half-equation is an equation that shows the transfer of electrons at one electrode during electrolysis. Electrons are shown as e⁻.
Half-equations at the cathode (reduction)
Cations gain electrons. Electrons appear on the left side of the equation.
| Reaction | Half-equation |
|---|---|
| Copper deposited | Cu²⁺ + 2e⁻ → Cu |
| Lead deposited | Pb²⁺ + 2e⁻ → Pb |
| Silver deposited | Ag⁺ + e⁻ → Ag |
| Hydrogen gas produced | 2H⁺ + 2e⁻ → H₂ |
| Aluminium deposited | Al³⁺ + 3e⁻ → Al |
Half-equations at the anode (oxidation)
Anions lose electrons. Electrons appear on the right side of the equation.
| Reaction | Half-equation |
|---|---|
| Chlorine gas produced | 2Cl⁻ → Cl₂ + 2e⁻ |
| Bromine produced | 2Br⁻ → Br₂ + 2e⁻ |
| Oxygen gas produced | 4OH⁻ → 2H₂O + O₂ + 4e⁻ |
How to write a half-equation
- Write the ion on one side and the product on the other
- Balance the atoms (you may need to double the ions for diatomic products like Cl₂)
- Add electrons to balance the charge
- Check: total charge on the left = total charge on the right
Example: chloride ions at the anode
- Start: Cl⁻ → Cl₂ (chlorine is diatomic, so need 2 Cl⁻)
- Balance atoms: 2Cl⁻ → Cl₂
- Balance charge: left side = 2 x (-1) = -2. Right side = 0. Add 2e⁻ to the right.
- Final: 2Cl⁻ → Cl₂ + 2e⁻
Worked exam question
During the electrolysis of dilute sulfuric acid: (a) Write the half-equation at the cathode. (1) (b) Write the half-equation at the anode. (2) (c) State which half-equation represents oxidation. (1)
Mark scheme
(a) 2H⁺ + 2e⁻ → H₂ [1]
(b) 4OH⁻ → 2H₂O + O₂ + 4e⁻ [1 for correct formula/products, 1 for balance]
(c) The anode half-equation [1] — because electrons are lost / appear on the right / oxidation is loss of electrons
Common exam mistakes
- Putting electrons on the wrong side. At the cathode, electrons are gained (left side). At the anode, electrons are lost (right side).
- Forgetting to make gaseous products diatomic. Chlorine is Cl₂ not Cl; hydrogen is H₂ not H; oxygen is O₂ not O.
- Not balancing the charge. If you have 2Cl⁻ on the left (charge = -2) and Cl₂ on the right (charge = 0), you need 2e⁻ on the right to balance.
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