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9701 · 24.2

Standard electrode potentials E°, standard cell potentials Ecell and the Nernst equation — practice questions

Practice and worked examples for 9701 Standard electrode potentials E°, standard cell potentials Ecell and the Nernst equation. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the standard cell potential for a cell constructed from zinc and copper half-cells. Predict the spontaneous direction of the reaction and write the overall cell equation.

Given: Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E° = -0.76 V Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = +0.34 V

Show solution outline
  1. Identify cathode and anode: The Cu²⁺/Cu half-cell has a more positive E° (+0.34 V) than the Zn²⁺/Zn half-cell (-0.76 V). Therefore, the copper half-cell is the cathode (reduction) and the zinc half-cell is the anode (oxidation).
  2. Calculate E°cell: E°cell = E°(cathode) - E°(anode) E°cell = (+0.34 V) - (-0.76 V) E°cell = +1.10 V
  3. Determine reaction direction: Since E°cell is positive, the reaction is spontaneous as described. Reduction occurs at the copper electrode, and oxidation occurs at the zinc electrode.
  4. Write half-equations and overall equation: Cathode (Reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) Anode (Oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Worked example 2

Consider the reaction between acidified permanganate(VII) ions and iron(II) ions. Calculate the standard cell potential and the standard Gibbs free energy change at 298 K.

Given: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l) E° = +1.51 V Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E° = +0.77 V

Show solution outline
  1. Identify cathode and anode: The MnO₄⁻/Mn²⁺ half-cell has the more positive E° (+1.51 V), so it is the cathode. The Fe³⁺/Fe²⁺ half-cell is the anode, and its reaction will be reversed (Fe²⁺ → Fe³⁺ + e⁻).
  2. Calculate E°cell: E°cell = E°(cathode) - E°(anode) E°cell = (+1.51 V) - (+0.77 V) E°cell = +0.74 V
  3. Balance electrons and find 'n': To combine the half-equations, the iron reaction must be multiplied by 5 to balance the 5 electrons in the permanganate reaction. 5Fe²⁺(aq) → 5Fe³⁺(aq) + 5e⁻ This means n = 5 moles of electrons are transferred.
  4. Calculate ΔG°: ΔG° = -nFE°cell ΔG° = -(5) * (96500 C mol⁻¹) * (+0.74 V) ΔG° = -357100 J mol⁻¹ ΔG° = -357 kJ mol⁻¹ (to 3 s.f.) Since E°cell is positive and ΔG° is negative, the reaction is spontaneous under standard conditions.
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