A student constructs an electrochemical cell using a standard manganese(II)/manganese half-cell and a standard tin(II)/tin half-cell connected by a salt bridge. The standard electrode potentials are: Mn²⁺(aq) + 2e⁻ ⇌ Mn(s), E° = −1.18 V and Sn²⁺(aq) + 2e⁻ ⇌ Sn(s), E° = −0.14 V. Explain how the standard electrode potentials are used to predict the overall cell potential, identify which half-cell acts as the negative electrode, and describe the role of the salt bridge in this cell.

Edexcel A-Level Chemistry (9CH0) — 14.1 Electrode potentials and electrochemical cells · Explain · 5 marks · View as Markdown

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

Standard electrode potentials: Mn²⁺(aq)/Mn(s), E° = −1.18 V; Sn²⁺(aq)/Sn(s), E° = −0.14 V.

Model answer (5 marks)

The overall cell potential is found by subtracting the less positive electrode potential from the more positive one:
E°cell = E°(Sn²⁺/Sn) – E°(Mn²⁺/Mn) = (‑0.14) – (‑1.18) = +1.04 V.
The manganese half‑cell has the more negative potential (‑1.18 V) and therefore acts as the negative electrode (anode). At this electrode oxidation occurs: Mn(s) → Mn²⁺(aq) + 2e⁻.
The tin half‑cell is the positive electrode (cathode) where reduction takes place: Sn²⁺(aq) + 2e⁻ → Sn(s).
The salt bridge completes the circuit by allowing ions to flow between the two half‑cells. It maintains electrical neutrality in each half‑cell by balancing the charge build‑up as electrons are transferred.

Examiner tips

  • Show the calculation of E°cell with correct sign convention. Identify the electrode with the more negative potential as the anode. Explain oxidation/reduction at each electrode. Mention the salt bridge’s role in completing the circuit and maintaining neutrality.

Common mistakes

  • Using the wrong sign for the cell potential (e.g. +0.14 V instead of +1.04 V). Failing to state which electrode is the anode and which is the cathode. Ignoring the role of the salt bridge or describing it as a passive component only.

Mark scheme (5 marks)

  1. The overall cell potential (e.m.f.) is calculated by subtracting the less positive (more negative) electrode potential from the more positive electrode potential, giving E°cell = −0.14 − (−1.18) = +1.04 V
  2. The manganese half-cell acts as the negative electrode because it has the more negative standard electrode potential (−1.18 V)
  3. Oxidation occurs at the manganese electrode (negative electrode); manganese is oxidised to Mn²⁺ ions
  4. The salt bridge completes the electrical circuit by allowing ions to flow between the two half-cells
  5. The salt bridge maintains electrical neutrality in each half-cell by balancing the build-up of charge as ions are produced or consumed

Key terms in this question

standard electrode potential · negative electrode · salt bridge

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