Explain what happens at each electrode when dilute sulfuric acid is electrolysed using inert electrodes.

Edexcel GCSE Chemistry (1CH0) — 3.2 Electrolytic processes · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

At the cathode (negative electrode) the positively charged hydrogen ions (H⁺) are reduced, gaining electrons to form hydrogen gas:
H⁺ + e⁻ → ½ H₂(g)

At the anode (positive electrode) the hydroxide ions (OH⁻) are oxidised, losing electrons to produce oxygen gas:
4 OH⁻ → O₂(g) + 2 H₂O + 4 e⁻

Examiner tips

  • Use the correct electrode names (cathode = negative, anode = positive).
  • Show the half‑reactions with electrons to demonstrate reduction/oxidation.
  • Mention the gases produced (H₂ at cathode, O₂ at anode).

Common mistakes

  • Confusing the cathode and anode or reversing the reactions.
  • Failing to include the electron transfer in the half‑reactions.

Mark scheme (4 marks)

  1. Hydrogen is produced at the cathode (negative electrode)
  2. At the cathode, positively charged hydrogen ions (H⁺) gain electrons / reduction occurs
  3. Oxygen is produced at the anode (positive electrode)
  4. At the anode, negatively charged hydroxide ions (OH⁻) lose electrons / oxidation occurs

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