Explain what happens at each electrode when a voltaic cell is constructed using a magnesium half-cell (Mg²⁺/Mg, E° = −2.37 V) and a silver half-cell (Ag⁺/Ag, E° = +0.80 V), including identification of the anode and cathode and the direction of electron flow in the external circuit.

IB DP Chemistry Standard Level (2023 syllabus) — R3.2 Electron transfer reactions · Explain · 4 marks · View as Markdown

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

Model answer (4 marks)

Magnesium is the anode – oxidation occurs:
Mg(s) → Mg²⁺(aq) + 2e⁻
Silver is the cathode – reduction occurs:
Ag⁺(aq) + e⁻ → Ag(s)
Electrons leave the magnesium electrode, travel through the external circuit and enter the silver electrode.
The Mg electrode has the more negative E° (−2.37 V) so it is the stronger reducing agent and is oxidised, driving the spontaneous cell reaction.

Examiner tips

  • State the half‑reactions and identify anode/cathode
  • Show electron flow direction
  • Explain why Mg is the anode using E° values

Common mistakes

  • Confusing the direction of electron flow
  • Calling the silver electrode the anode
  • Ignoring the sign of the standard potentials

Mark scheme (4 marks)

  1. Magnesium electrode is the anode where oxidation occurs (Mg → Mg²⁺ + 2e⁻)
  2. Silver electrode is the cathode where reduction occurs (Ag⁺ + e⁻ → Ag)
  3. Electrons flow through the external circuit from the magnesium electrode (anode) to the silver electrode (cathode)
  4. The magnesium electrode has the more negative (lower) standard electrode potential, so it has the greater tendency to be oxidised / acts as the stronger reducing agent, driving the spontaneous reaction

Key terms in this question

anode · cathode · electron flow · voltaic cell

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