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Learn9701Paper 4 · A Level Structured QuestionsElectrochemistryElectrode potentials and cells
Electrochemistry · Subtopic 1 of 2

Electrode Potentials and Cells

A half-cell's standard electrode potential E⦵ is measured against the standard hydrogen electrode (SHE, 0.00 V) at 298 K, 1 mol dm⁻³ and 100 kPa. Half-equations are written as reductions; the more positive E⦵, the stronger the oxidising agent (left side). Ecell∘=E(positive, reduction)∘−E(negative, oxidation)∘E^\circ_\text{cell}=E^\circ_\text{(positive, reduction)}-E^\circ_\text{(negative, oxidation)}. A reaction is feasible if Ecell∘>0E^\circ_\text{cell}>0 (but may be slow, or conditions may be non-standard). Changing concentration shifts E (Nernst: E=E∘+0.059zlog⁡[ox][red]E=E^\circ+\frac{0.059}{z}\log\frac{[\text{ox}]}{[\text{red}]}). ΔG∘=−nFEcell∘\Delta G^\circ=-nFE^\circ_\text{cell}. Fuel cells and batteries apply this.

Cambridge 9701 Paper 4Edexcel YCH11 WCH15AQA 7405 Paper 1≈ 55 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Describe the SHE and how E⦵ is measured.
  • Calculate E⦵cell and predict feasibility.
  • Write cell diagrams.
  • Explain the effect of concentration; use the Nernst equation.
Pause and think

Why can copper metal not displace hydrogen from dilute acid, while zinc can?

Section 2 of 14

Core idea

Start hereA battery is a controlled redox reaction

Put zinc in copper sulfate and it just gets warm. Separate the two halves with a wire and a salt bridge, and the same reaction lights a bulb.

The big idea

Each half-cell has a reduction potential measured against hydrogen. The more positive half-cell is reduced, the other is oxidised, and E⦵cell is the difference.

1. Building a cell

Zinc–copper cell

Electrons flow through the wire; ions flow through the salt bridge.

E⦵cell and the overall equation
  1. Cu2++2e−⇌CuE∘=+0.34\mathrm{Cu^{2+}+2e^-\rightleftharpoons Cu}\quad E^\circ=+0.34
    More positive: goes forward (reduction).
  2. Zn2++2e−⇌ZnE∘=−0.76\mathrm{Zn^{2+}+2e^-\rightleftharpoons Zn}\quad E^\circ=-0.76
    Less positive: goes backward (oxidation).
  3. Ecell∘=0.34−(−0.76)=+1.10 VE^\circ_\text{cell}=0.34-(-0.76)=+1.10\ \text{V}
    Positive minus negative.
  4. Zn+Cu2+→Zn2++Cu\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}
    Add the half-equations; electrons cancel.
Worked exampleConcentration effect (Nernst)

Find E for Cu²⁺/Cu when [Cu²⁺] = 0.010 mol dm⁻³.

Goal
E in V.
Strategy
E=E∘+0.0592log⁡[Cu2+]E=E^\circ+\frac{0.059}{2}\log[\mathrm{Cu^{2+}}].
  1. log⁡0.010=−2\log0.010=-2
    Logarithm.
  2. E=0.34+0.0295×(−2)=+0.28 VE=0.34+0.0295\times(-2)=+0.28\ \text{V}
    Substitute.
Answer+0.28 V.
✓ CheckLower [Cu²⁺] → less positive ✓.
Exam tipDon't scale E⦵ If you multiply a half-equation by 2 to balance electrons, E⦵ stays the same.

Summary

  • SHE = 0.00 V.
  • E⦵cell = positive − negative.
  • Positive E⦵cell → feasible.
  • Nernst; ΔG = −nFE.
Key words
Half-cell
An element and its ions (or two ions) in equilibrium at an electrode.
Salt bridge
Allows ions to move between half-cells.
Standard conditions
298 K, 100 kPa, 1.00 mol dm⁻³.
HistoryVolta's pile. Alessandro Volta stacked zinc and copper discs separated by brine-soaked card in 1800, making the first battery.
Fun factFruit batteries. A lemon with a copper coin and a galvanised nail produces about 0.9 V, close to the Zn/Cu E⦵cell, reduced by internal resistance.
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