In electrolysis, electricity forces a redox reaction. To predict the products, you decide which cation is discharged at the cathode and which anion, or electrode, reacts at the anode.
This lesson is part of SPM Chemistry redox equilibrium. It uses the skills from writing half-equations.
What three questions decide the product?
For an aqueous solution, list every ion present, including H⁺ and OH⁻ from water. Then ask:
- Position in the electrochemical series: the cation lower in the series is discharged at the cathode.
- Concentration: when anions compete, the more concentrated ion is preferentially discharged.
- Type of electrode: carbon and platinum are inert, while copper can dissolve at the anode.
Worked example 1: copper(II) sulfate with carbon electrodes
Ions present: Cu²⁺, H⁺, SO₄²⁻ and OH⁻.
Cathode: Cu²⁺ is lower in the series than H⁺, so it is discharged. Cu²⁺ + 2e⁻ → Cu. A brown deposit forms.
Anode: sulfate ions are not discharged in this situation, so OH⁻ is discharged. 4OH⁻ → 2H₂O + O₂ + 4e⁻. Oxygen gas bubbles off.
Worked example 2: sodium chloride, two concentrations
Ions present: Na⁺, H⁺, Cl⁻ and OH⁻.
Cathode (both): H⁺ is lower than Na⁺, so hydrogen forms. 2H⁺ + 2e⁻ → H₂.
Anode, concentrated solution: Cl⁻ is at high concentration, so chlorine forms. 2Cl⁻ → Cl₂ + 2e⁻.
Anode, dilute solution: OH⁻ is preferentially discharged, so oxygen forms. 4OH⁻ → 2H₂O + O₂ + 4e⁻.
Worked example 3: copper electrodes in copper(II) sulfate
The copper anode is not inert. It loses electrons: Cu → Cu²⁺ + 2e⁻. The anode becomes thinner.
The cathode still gains copper: Cu²⁺ + 2e⁻ → Cu. The cathode becomes thicker, and the blue colour of the solution stays about the same.
The mistake that costs marks
The common slip is predicting sodium at the cathode for aqueous sodium chloride, because sodium ions are the only metal ions listed. Sodium is higher than hydrogen in the series, so hydrogen ions are discharged.
| Electrolyte | Wrong cathode | Right cathode |
|---|---|---|
| Aqueous NaCl | Sodium | Hydrogen gas |
| Aqueous CuSO₄ | Hydrogen | Copper |
Always write the full list of ions before choosing, including the ions from water.
Check yourself
Predict the products and write half-equations for the electrolysis of 0.5 mol/dm³ copper(II) sulfate solution using copper electrodes.
Answer
Cathode: Cu²⁺ + 2e⁻ → Cu, so the cathode gains a brown deposit.
Anode: Cu → Cu²⁺ + 2e⁻, so the copper anode dissolves and becomes thinner.
The concentration of Cu²⁺ in the solution stays about the same, because copper ions removed at the cathode are replaced at the anode.
What to study next
The same half-equations appear in cells, where the direction of the process is reversed. Continue with explaining simple chemical cells, then use the redox practice set.
The mole and stoichiometry steps tool supports the amount calculations. If you want a teacher to test your predictions on electrolytes you have not seen, see online one-to-one Chemistry tuition.