A half-equation describes one electrode. At the cathode, positive ions gain electrons (reduction). At the anode, negative ions lose electrons (oxidation).
Two half-equations, with equal numbers of electrons, add up to the overall equation.
This lesson builds on predicting molten products and on balancing skills from formulas and equations.
How do I write a half-equation?
Work in four small steps.
- Write the ion and the product. Pb²⁺ becomes Pb. Br⁻ becomes Br₂.
- Balance the atoms. Two Br⁻ are needed for one Br₂.
- Add electrons to balance the charge. Put them on the left if gained, on the right if lost.
- Check the charges. The total charge must be the same on both sides.
Remember the memory aid OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain. The anode is where oxidation occurs, the cathode where reduction occurs.
Worked example
Molten lead(II) bromide is electrolysed. Write the half-equation at each electrode and combine them.
Cathode: Pb²⁺ → Pb. The left side has charge +2 and the right side has 0, so add two electrons to the left:
Pb²⁺ + 2e⁻ → Pb
Charge check: +2 + (−2) = 0 on the left, 0 on the right.
Anode: Br⁻ → Br₂. Balance atoms first: 2Br⁻ → Br₂. The left has charge −2 and the right has 0, so two electrons appear on the right:
2Br⁻ → Br₂ + 2e⁻
Charge check: −2 on the left, 0 + (−2) = −2 on the right.
Combine: each half-equation involves two electrons, so they cancel when added:
Pb²⁺ + 2Br⁻ → Pb + Br₂
That is PbBr₂ → Pb + Br₂, matching the result from the previous lesson.
When the electron numbers differ
For molten aluminium oxide, the cathode is Al³⁺ + 3e⁻ → Al and the anode is 2O²⁻ → O₂ + 4e⁻. The electron numbers are 3 and 4, so scale both to 12: 4Al³⁺ + 12e⁻ → 4Al and 6O²⁻ → 3O₂ + 12e⁻.
Adding gives 4Al³⁺ + 6O²⁻ → 4Al + 3O₂, which is 2Al₂O₃ → 4Al + 3O₂. Count: 4 Al and 6 O on each side.
The mistake to watch for
A typical slip is putting the electrons on the wrong side, or leaving them out.
Mistaken answer: Pb²⁺ → Pb + 2e⁻
The charge is +2 on the left but −2 on the right, so it does not balance. It also shows the lead ion losing electrons, the opposite of what happens.
The correction is to ask “is this ion gaining or losing electrons?” A positive ion becoming a neutral atom must gain electrons, so they belong on the left.
Check yourself
1. Write the half-equation for Zn²⁺ becoming zinc at the cathode.
Show answer
Zn²⁺ + 2e⁻ → Zn. Charge: 0 on both sides.
2. Write the half-equation for chloride ions forming chlorine at the anode.
Show answer
2Cl⁻ → Cl₂ + 2e⁻. Charge: −2 on the left, −2 on the right.
3. Combine the half-equations Na⁺ + e⁻ → Na and 2Cl⁻ → Cl₂ + 2e⁻ into an overall equation.
Show answer
Multiply the first by 2: 2Na⁺ + 2e⁻ → 2Na. Add to the second; the electrons cancel: 2Na⁺ + 2Cl⁻ → 2Na + Cl₂, which is 2NaCl → 2Na + Cl₂.
Where this leads next
Next, compare aqueous and molten cases within scope, where water adds extra ions to consider. The equation balance trainer is a good way to check your atom counts.
Students who find half-equations mechanical often understand them after a teacher asks “what is this ion doing?” at each step. Our teachers work this way in online one-to-one Chemistry tuition.