How to write half-equations for electrolysis

Half-equations show what happens to ions at each electrode during electrolysis, and SPM examiners ask for them again and again in the redox equilibrium chapter. Students lose marks not because the chemistry is deep, but because they leave electrons out, put them on the wrong side, or forget to balance the charge. This guide gives you a reliable method so every half-equation you write is correct, cathode and anode, molten or aqueous.

First, what a half-equation is

Electrolysis is a redox process split across two electrodes. At the cathode (the negative electrode) positive ions gain electrons, that is reduction. At the anode (the positive electrode) negative ions lose electrons, that is oxidation. A half-equation is just the equation for one of those two events. The old memory hook still works: OIL RIG, Oxidation Is Loss, Reduction Is Gain, of electrons. So cathode half-equations have electrons on the left (they are added), and anode half-equations have electrons on the right (they are removed).

This guide assumes you already know which ion is discharged at each electrode. Deciding that is a separate skill, the selective discharge rules, which we cover in our companion guide on predicting the products of electrolysis.

The four-step method

Use the same routine every time:

1. Write the ion and the product, with the ion on the correct side. At the cathode the metal ion or H⁺ becomes a neutral atom or molecule. At the anode the non-metal ion becomes a neutral atom or molecule.
2. Balance the atoms first, ignoring charge for a moment.
3. Add electrons (e⁻) to make the total charge equal on both sides. Count the charge on the left and on the right, and add just enough electrons to balance them.
4. Check the side. Cathode = reduction = electrons on the left. Anode = oxidation = electrons on the right. If your electrons landed on the wrong side, you have made an error somewhere.

Cathode examples

A simple metal ion. Lead(II) ions are reduced to lead metal:

Pb²⁺ + 2e⁻ → Pb

The lead ion carries a 2+ charge, so it needs two electrons to become a neutral atom. The charge on the left is (2+) + (2−) = 0, matching the neutral Pb on the right.

A singly charged ion. Sodium ions need only one electron each:

Na⁺ + e⁻ → Na

Hydrogen from an aqueous solution. When hydrogen is discharged, two H⁺ ions combine and gain two electrons to form one H₂ molecule:

2H⁺ + 2e⁻ → H₂

Notice the 2 in front of H⁺, you need two hydrogen ions to build one molecule, and therefore two electrons. You can see this in action in the electrolysis of aqueous solutions.

Anode examples

At the anode the electrons come off, so they sit on the right.

A halide ion. Bromide ions are oxidised to bromine:

2Br⁻ → Br₂ + 2e⁻

Two bromide ions each give up one electron, so two electrons are released as they pair into a Br₂ molecule. Chloride behaves the same way: 2Cl⁻ → Cl₂ + 2e⁻.

Hydroxide ions giving oxygen. When oxygen is produced in an aqueous solution, hydroxide ions are discharged:

4OH⁻ → 2H₂O + O₂ + 4e⁻

This one looks intimidating but follows the same rules: four hydroxide ions carry a total charge of 4−, so four electrons must be released to leave neutral water and oxygen on the right. Balance the O and H atoms first (4 O and 4 H on the left; 2 H₂O gives 4 H and 2 O, plus O₂ gives 2 more O), then add the four electrons to balance the charge.

A dissolving metal anode. If the anode is copper rather than inert carbon, the copper itself is oxidised:

Cu → Cu²⁺ + 2e⁻

The metal atom loses two electrons and enters the solution as an ion. This is the reaction behind electroplating and copper purification.

Combining half-equations into the overall equation

To get the full ionic equation, make the electrons cancel. Take molten lead(II) bromide:

• Cathode: Pb²⁺ + 2e⁻ → Pb
• Anode: 2Br⁻ → Br₂ + 2e⁻

Both involve two electrons, so simply add them and cancel the 2e⁻ on each side:

Pb²⁺ + 2Br⁻ → Pb + Br₂

You can trace this whole process in the electrolysis of molten compounds. When the electron numbers differ, multiply one or both half-equations up until they match before adding, exactly as you would when finding a lowest common multiple.

The mistakes that cost marks

Four slips account for most lost marks. Leaving electrons out entirely, a half-equation without an e⁻ is always wrong. Putting them on the wrong side, remember reduction (cathode) adds them on the left. Unbalanced charge, always tally the charges on each side after adding electrons. And forgetting the coefficient, such as writing H⁺ + e⁻ → H instead of 2H⁺ + 2e⁻ → H₂, since hydrogen and the halogens exist as diatomic molecules.

Write out the cathode and anode half-equations for lead bromide, sodium chloride solution and copper(II) sulfate solution until the pattern is automatic. If you would like a teacher to check your charges and electron counts line by line and correct the exact step you slip on, our online one-to-one lessons with our experienced SPM Chemistry teachers are built for that, from RM50 an hour with a paid one-hour trial lesson.