During the electrolysis of dilute sulfuric acid using inert platinum electrodes, describe what is produced at each electrode and explain why the volume of gas collected at one electrode is greater than at the other.

Cambridge International IGCSE Chemistry (0620) — 4.1 Electrolysis · Describe and 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) hydrogen gas is produced. At the anode (positive electrode) oxygen gas is produced.

The volume of hydrogen collected is twice the volume of oxygen. This is because the electrolysis of water (or dilute H₂SO₄) involves the reduction of H⁺ (or H₂O) to H₂ at the cathode, giving two H atoms per H₂ molecule, while the oxidation at the anode produces one O₂ molecule from two O atoms. Thus for every one O₂ molecule formed, two H₂ molecules are formed, giving a 2:1 volume ratio.

Examiner tips

  • Use the correct electrode names (cathode = negative, anode = positive).
  • State the gases produced at each electrode before giving the ratio.
  • Explain the 2:1 ratio in terms of the stoichiometry of water splitting (2 H₂ per O₂).

Common mistakes

  • Confusing cathode and anode, or saying oxygen is produced at the cathode.
  • Failing to mention the 2:1 volume ratio or giving the wrong ratio.
  • Not explaining the reason (stoichiometry of water splitting).

Mark scheme (4 marks)

  1. Hydrogen gas is produced at the cathode (negative electrode)
  2. Oxygen gas is produced at the anode (positive electrode)
  3. The volume of hydrogen is greater / twice the volume of oxygen
  4. Because water molecules / hydrogen ions and oxide/hydroxide ions split in a 2:1 ratio, so twice as many hydrogen molecules are formed for every oxygen molecule / water splits as H₂O giving two H for every one O

Key terms in this question

electrolysis

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