Electrolysis, electrode reactions, industrial applications, electroplating — electricity meets chemistry! ⚡🔋
Electrolysis is the decomposition of an ionic compound (electrolyte) using electricity. It requires a direct current (DC) supply, two electrodes, and an electrolyte (molten or dissolved ionic compound).
| Component | Definition | Rule |
|---|---|---|
| Cathode | Negative electrode | Cations (+) attracted here → Reduction (gains electrons) |
| Anode | Positive electrode | Anions (−) attracted here → Oxidation (loses electrons) |
| Electrolyte | Ionic compound that conducts electricity in solution or when molten | Must contain free-moving ions |
Only Na⁺ and Cl⁻ ions present. Simple — no competition.
| Electrode | Ion discharged | Half-equation | Product |
|---|---|---|---|
| Cathode (−) | Na⁺ | Na⁺ + e⁻ → Na | Sodium metal (liquid) |
| Anode (+) | Cl⁻ | 2Cl⁻ → Cl₂ + 2e⁻ | Chlorine gas |
Brine contains Na⁺, Cl⁻, H⁺, and OH⁻ ions (from water). Competition occurs at each electrode.
| Electrode | Competing ions | Ion discharged & why | Product |
|---|---|---|---|
| Cathode (−) | Na⁺ and H⁺ | H⁺ discharged — lower in reactivity series (H₂ preferred over Na metal) | Hydrogen gas (H₂) |
| Anode (+) | Cl⁻ and OH⁻ | Cl⁻ discharged — higher concentration effect | Chlorine gas (Cl₂) |
| Solution left behind | Na⁺ + OH⁻ | Remaining ions form NaOH | Sodium hydroxide (NaOH) |
| Electrode | Ion discharged | Half-equation | Product |
|---|---|---|---|
| Cathode (−) | Cu²⁺ (preferred over H⁺ — Cu is lower in reactivity series) | Cu²⁺ + 2e⁻ → Cu | Pink copper deposits on cathode |
| Anode (+) | OH⁻ / H₂O (preferred over SO₄²⁻) | 2H₂O → O₂ + 4H⁺ + 4e⁻ | Oxygen gas |
| Electrode | Half-equation | What happens |
|---|---|---|
| Cathode (−) — object to be plated | Cu²⁺ + 2e⁻ → Cu | Copper deposits — cathode gets heavier |
| Anode (+) — pure copper | Cu → Cu²⁺ + 2e⁻ | Copper dissolves — anode gets lighter |
The Cu²⁺ concentration in solution stays constant — copper dissolves from anode at the same rate it deposits on the cathode!
When multiple ions compete, the one discharged depends on position in the reactivity series and concentration.
| At the Cathode (reduction — cations) | At the Anode (oxidation — anions) |
|---|---|
| Less reactive metal ions discharged first. Order: Cu²⁺ > H⁺ > Zn²⁺ > Na⁺ (Cu²⁺ most easily reduced) | OH⁻ / halide ions (Cl⁻, Br⁻, I⁻) discharged in preference to SO₄²⁻, NO₃⁻. At high conc. of Cl⁻ → Cl₂ At low conc. or no Cl⁻ → O₂ |
Aluminium cannot be extracted by reduction with carbon (too high reactivity) so electrolysis is used.
Electroplating deposits a thin layer of metal onto an object for protection, appearance, or conductivity.
| Component | What to use | Why |
|---|---|---|
| Cathode | Object to be plated | Metal ions deposit here (reduction) |
| Anode | Pure metal being plated (e.g. copper, silver) | Dissolves to replenish ions in solution |
| Electrolyte | Solution of the plating metal's salt | Provides ions; e.g. CuSO₄ for copper plating, AgNO₃ for silver plating |
First Law: The mass of substance deposited/released is proportional to the quantity of electricity (charge) passed.
Second Law: The mass deposited is inversely proportional to the charge of the ion (i.e. doubly-charged ions need twice the charge per mole).
Q = I × t charge (coulombs) = current (A) × time (s)moles of e⁻ = Q / 96500 (Faraday constant F = 96500 C/mol)moles of substance = moles of e⁻ / n (n = charge on ion)mass = moles × Mᵣ
Select an electrolyte and watch ions migrate to the correct electrode, products appear, and half-equations update live.
An interactive reference chart. Click any cell to get the full explanation for that combination.
👆 Click a cell to see the explanation
Click to flip!
Problem: A steady current of 3.0 A is passed through a solution of silver nitrate (AgNO₃) for 1608 seconds using platinum electrodes. Calculate the mass of silver deposited at the cathode. (Aᵣ: Ag = 108, F = 96500 C/mol)
Use Q = I × t. Current = 3.0 A, time = 1608 s.
Use n(e⁻) = Q / F = 4824 / 96500.
Half-equation: Ag⁺ + e⁻ → Ag. Ag⁺ has a +1 charge so 1 mole of e⁻ deposits 1 mole of Ag. Moles of Ag = ?
Use m = n × Mᵣ. n(Ag) = 0.05 mol, Mᵣ(Ag) = 108 g/mol.
1 (a) Cathode: hydrogen (H₂) ✓ | Anode: chlorine (Cl₂) ✓
2 (b) Cathode: 2H⁺ + 2e⁻ → H₂ ✓ | Anode: 2Cl⁻ → Cl₂ + 2e⁻ ✓
3 (c) Sodium hydroxide (NaOH) left in solution ✓
4 H₂: used as a fuel / making margarine / rocket propellant ✓
5 Cl₂: used to disinfect water supplies / making PVC plastic / bleach production ✓
6 NaOH: soap manufacture / paper production / drain cleaners ✓ [Award any 2 for 2 marks]
1 (a) The metal spoon (object to be plated) ✓
2 (b) Pure copper (the plating metal) ✓
3 (c) The copper anode is oxidised and dissolves into solution as Cu²⁺ ions: Cu → Cu²⁺ + 2e⁻. This replenishes the Cu²⁺ used at the cathode, keeping the electrolyte concentration constant. ✓✓
4 (d) Cu²⁺(aq) + 2e⁻ → Cu(s) ✓
1 (a) Cathode: 2H⁺ + 2e⁻ → H₂ ✓
2 Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ ✓
3 (b) To produce 2 mol H₂: 2 × (2e⁻) = 4 mol e⁻ used ✓
4 At the anode, 4e⁻ released produces 1 mol O₂ ✓
5 So 2 mol H₂ : 1 mol O₂. The volume ratio H₂:O₂ = 2:1 — this matches the electrolysis of water. ✓
1 (a) Cathode: lead (Pb) ✓ | Anode: bromine (Br₂) ✓
2 (b) Q = I × t = 5.0 × 1930 = 9650 C ✓
3 n(e⁻) = 9650 / 96500 = 0.1 mol ✓
4 Pb²⁺ + 2e⁻ → Pb: n(Pb) = 0.1/2 = 0.05 mol ✓
5 m(Pb) = 0.05 × 207 = 10.35 g ✓
Cathode (−) = Reduction
Anode (+) = Oxidation
RED CAT | AN OX
Cathode: H₂ (2H⁺+2e⁻→H₂)
Anode: Cl₂ (2Cl⁻→Cl₂+2e⁻)
Solution: NaOH
Q = I × t
n(e⁻) = Q / 96500
n = n(e⁻) / ion charge
m = n × Mᵣ
Cathode = object (gains mass)
Anode = plating metal (loses mass)
Electrolyte = metal salt solution
Cathode: Cu²⁺+2e⁻→Cu (pink)
Anode: O₂ from H₂O
Solution loses Cu²⁺, paler
Al₂O₃ dissolved in cryolite
Cathode: Al³⁺+3e⁻→Al
Anode burns away as CO₂