The key relationships
📖 Formulae
Q = I Δt — charge (C) = current (A) × time (s)
n(e⁻) = Q ÷ F — F = 96 500 C·mol⁻¹ (the charge on one mole of electrons)
Number of electrons = Q ÷ e, with e = 1,6 × 10⁻¹⁹ C
The half-reaction tells you how many moles of electrons are needed per mole of product: Ag⁺ + e⁻ → Ag (1 : 1); Cu²⁺ + 2e⁻ → Cu (2 : 1); Al³⁺ + 3e⁻ → Al (3 : 1); 2Cl⁻ → Cl₂ + 2e⁻ (2 : 1).
These calculations appear in IEB papers. DBE papers test the same half-reactions, so the method is still useful for understanding.
Mass deposited
✏️ Worked example 1: copper plating
1A current of 2,0 A flows through CuSO₄(aq) for 30 minutes. Δt = 30 × 60 = 1 800 s.
2Q = IΔt = 2,0 × 1 800 = 3 600 C.
3n(e⁻) = Q/F = 3 600 ÷ 96 500 = 0,0373 mol.
4Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0,0373 ÷ 2 = 0,0187 mol.
5m = nM = 0,0187 × 63,5 = 1,18 g of copper deposited on the cathode.
Time or current needed
✏️ Worked example 2: working backwards
1How long must 3,0 A flow to deposit 5,0 g of copper?
2n(Cu) = 5,0 ÷ 63,5 = 0,0787 mol, so n(e⁻) = 2 × 0,0787 = 0,157 mol.
3Q = n(e⁻) × F = 0,157 × 96 500 = 15 200 C.
4Δt = Q/I = 15 200 ÷ 3,0 = 5 070 s (about 84 minutes).
✏️ Worked example 3: gas released
10,10 mol of electrons pass through concentrated NaCl(aq). At the anode, 2Cl⁻ → Cl₂ + 2e⁻.
2n(Cl₂) = 0,10 ÷ 2 = 0,050 mol. At STP, V = nVm = 0,050 × 22,4 = 1,12 dm³.
Checklist
- Convert minutes and hours to seconds first.
- Write the half-reaction and use its electron ratio.
- The same charge flows through every electrode in a series circuit, so 1 mol of e⁻ deposits 1 mol of Ag but only 0,5 mol of Cu.
- Reduction (metal deposits) happens at the cathode; oxidation at the anode.