Topic 4 of 6
Aqueous electrolysis: more than one possible ion
Water introduces competing products.
O-Level 6092 (2026) / SEC G3 K324 (2027)
Pure onlyAqueous electrolysis: more than one possible ion
Water introduces competing products.
In aqueous electrolysis, consider the solute ions and the water. With inert electrodes, a metal less reactive than hydrogen, such as copper, is usually deposited at the cathode. For a very reactive metal ion such as Na+, hydrogen forms instead. At the anode, sulfate is generally not discharged; oxygen forms from water/hydroxide. Halide concentration matters: concentrated chloride favours chlorine, while dilute chloride solution is treated here as giving oxygen.
- List species and electrode material
For aqueous potassium iodide with inert electrodes, consider K+, I- and water (H+ and OH-). Molten KI has no water competing.
- Choose the cathode product
Potassium is above hydrogen in the reactivity series. In the school aqueous model, hydrogen forms rather than potassium metal. For a metal below hydrogen, such as copper or silver, the metal is normally deposited.
- Choose the anode product
Halide discharge forms the corresponding halogen: bromide gives bromine; iodide gives iodine. For this iodide solution iodine forms; 2I-(aq) -> I2(aq) + 2e-. Chloride needs the dilute/concentrated comparison shown below. Sulfate is not discharged in these examples: oxygen forms from water/hydroxide.
- Check atoms, charge and observations
The iodide half-equation has two iodine atoms and total charge -2 on each side. The anode region becomes brown as iodine dissolves; hydrogen bubbles at the cathode. Do not describe every halogen product as a gas.
| Electrolyte | Cathode | Anode |
|---|---|---|
| Aqueous copper(II) sulfate | Copper | Oxygen |
| Dilute aqueous sodium chloride | Hydrogen | Oxygen |
| Concentrated aqueous sodium chloride | Hydrogen | Chlorine |
| Process | Half-equation |
|---|---|
| Copper deposition | Cu2+(aq) + 2e- -> Cu(s) |
| Hydrogen from water | 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) |
| Oxygen from hydroxide | 4OH-(aq) -> O2(g) + 2H2O(l) + 4e- |
| Chlorine from chloride | 2Cl-(aq) -> Cl2(g) + 2e- |
Worked example
Construct a half-equation from an unfamiliar ion charge
An aqueous solution contains Ag+ ions and silver is discharged at an inert cathode. A separate molten lead(II) bromide cell produces bromine at the anode. Construct the relevant half-equations.
- At the cathode an Ag+ ion gains one electron to make a neutral silver atom: Ag+(aq) + e- -> Ag(s).
- At the anode bromide ions lose electrons. Bromine is Br2, so begin 2Br- -> Br2, then add 2e- on the right to balance charge.
- For the molten-cell anode use Br-(l) and Br2(g); the molten electrolyte is hot enough for bromine to be a gas. Check both atoms and net charge.
Cathode in the silver solution: Ag+(aq) + e- -> Ag(s). Anode in molten lead(II) bromide: 2Br-(l) -> Br2(g) + 2e-. Each equation must balance atoms and charge.
Worked example
Compare dilute and concentrated brine
Explain why changing NaCl concentration can change one electrode product.
- Na+ is not deposited from either aqueous solution; hydrogen forms at the cathode.
- In dilute solution, oxygen forms at the inert anode.
- At high chloride concentration, chloride is preferentially discharged and chlorine forms.
Cathode product remains hydrogen; anode product changes. Always state aqueous/molten, concentration and electrode material.