How to predict electrolysis products
How to decide what forms at the anode and cathode, half equations that balance, and why molten and aqueous compounds give different answers.
What the examiner is testing
Whether you can predict the products before writing any equation, and whether your half equations balance for both atoms and charge. Marks are given separately for the product and for the equation, so a correct prediction with a broken equation still scores.
The two electrodes
| Electrode | Charge | Attracts | Process |
|---|---|---|---|
| Cathode | Negative | Cations (+) | Reduction — gain of electrons |
| Anode | Positive | Anions (−) | Oxidation — loss of electrons |
OIL RIG: Oxidation Is Loss, Reduction Is Gain. Combined with "reduction happens at the cathode", that fixes everything else.
Molten or aqueous — decide this first
Molten compounds contain only the ions of the compound. The metal forms at the cathode and the non-metal at the anode. No exceptions to consider.
Aqueous solutions also contain \( \text{H}^+ \) and \( \text{OH}^- \) from the water, so there is competition:
- At the cathode: the less reactive of the metal and hydrogen is discharged. Metals below hydrogen in the reactivity series (copper, silver) are deposited. Metals above it (sodium, potassium, calcium, magnesium, aluminium) stay in solution and hydrogen is released.
- At the anode: halide ions are discharged in preference to \( \text{OH}^- \), giving chlorine, bromine or iodine. If no halide is present, oxygen is released.
Worked example
Predict the products of electrolysing aqueous copper(II) sulfate with inert electrodes, and write the half equations.
Step 1 — list the ions present. \( \text{Cu}^{2+} \), \( \text{SO}_4^{2-} \), and from the water \( \text{H}^+ \) and \( \text{OH}^- \).
Step 2 — cathode. The contest is copper against hydrogen. Copper is below hydrogen in the reactivity series, so copper is discharged:
$$ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} $$
A pink-brown coating forms on the cathode.
Step 3 — anode. No halide is present, so hydroxide is discharged and oxygen is released:
$$ 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^- $$
Bubbles of a colourless gas that relights a glowing splint.
Step 4 — check the balance. Cathode: charge is \( +2 - 2 = 0 \) on the left and 0 on the right. Anode: \( -4 \) on the left, and \( -4 \) from the electrons on the right. Both balance.
Observation to quote: the blue colour of the solution fades as \( \text{Cu}^{2+} \) ions are removed. Examiners ask for this specifically.
The three mistakes that lose marks
1. Confusing the electrode names. The cathode is negative and attracts positive ions. Reduction happens there.
2. Forgetting the water in an aqueous solution. This is the difference between predicting sodium and predicting hydrogen.
3. Half equations that do not balance for charge. Count the charge on both sides including the electrons, every time.
30-second recap
Molten or aqueous first. Cathode negative, gains electrons, reduction. Anode positive, loses electrons, oxidation. In solution, the less reactive species wins at the cathode, and halides beat hydroxide at the anode. Balance atoms and charge before moving on.