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Learn9701Paper 4 · A Level Structured QuestionsElectrochemistryElectrolysis calculations
Electrochemistry · Subtopic 2 of 2

Electrolysis Calculations

Charge Q=ItQ=It (coulombs = amps × seconds). The Faraday constant F=96 500F=96\,500 C mol⁻¹ is the charge on 1 mol of electrons, and F=LeF=Le (Avogadro constant × electron charge). Moles of electrons =Q/F=Q/F; then use the half-equation: Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu} needs 2 mol e⁻ per mol Cu. Products in aqueous electrolysis are predicted from E⦵: at the cathode, the ion with the more positive E⦵ is reduced (Cu²⁺, Ag⁺ before H₂O); at the anode, concentrated halide gives halogen, otherwise O₂ forms. The Avogadro constant can be found by electrolysis: L=F/eL=F/e.

Cambridge 9701 Paper 4≈ 40 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Use Q=ItQ=It and F=LeF=Le.
  • Calculate masses and volumes of electrolysis products.
  • Predict products using E⦵ and concentration.
  • Describe an experiment to find LL.
Pause and think

The same current flows for the same time through CuSO₄ and AgNO₃ in series. Why is a greater number of moles of silver deposited?

Section 2 of 14

Core idea

Start hereCounting electrons by the mole

An ammeter and a stopwatch are enough to count electrons. Weigh the metal they deposit and you can even work out Avogadro's constant.

The big idea

Charge Q = It. Divide by F to get moles of electrons, then use the half-equation to find moles of product.

1. The calculation chain

From current to product

Always go through moles of electrons.

Worked exampleCopper plating

2.00 A for 965 s through CuSO₄(aq). Mass of Cu?

Goal
Mass in g.
Strategy
Charge → electrons → copper.
  1. Q=2.00×965=1930 CQ=2.00\times965=1930\ \text{C}
    Q = It.
  2. n(e−)=193096 500=0.0200n(e^-)=\frac{1930}{96\,500}=0.0200
    Divide by F.
  3. n(Cu)=0.0100; m=0.0100×63.5=0.635 gn(\mathrm{Cu})=0.0100;\ m=0.0100\times63.5=0.635\ \text{g}
    Cu²⁺ + 2e⁻ → Cu.
Answer0.635 g.
✓ CheckTwo electrons per copper ✓.
Avogadro constant from electrolysis
  1. n(e−)=2×Δm63.5n(e^-)=2\times\frac{\Delta m}{63.5}
    From the mass of copper deposited.
  2. electrons=Qe\text{electrons}=\frac{Q}{e}
    Total charge divided by the charge on one electron.
  3. L=Q/en(e−)L=\frac{Q/e}{n(e^-)}
    Particles per mole.

Summary

  • Q = It (seconds).
  • n(e⁻) = Q/F.
  • Use half-equation ratios.
  • F = Le.
Key words
Faraday constant
Charge on 1 mol of electrons, 96 500 C mol⁻¹.
Coulomb
Unit of charge: 1 A for 1 s.
HistoryFaraday's laws, 1834. Michael Faraday found that the mass deposited is proportional to the charge passed, decades before the electron was discovered.
Fun factAluminium is electricity in solid form. Making 1 kg of aluminium takes about 3 × 10⁸ C of charge, which is why smelters are built next to cheap hydroelectric power.
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