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This lesson covers the electrolysis of aqueous solutions for Edexcel GCSE Chemistry (1CH0). Aqueous electrolysis is more complex than molten electrolysis because water introduces additional competing ions. You need to know the rules for predicting products at each electrode and be able to write half equations.
When an ionic compound is dissolved in water, there are more than two types of ion present. In addition to the ions from the dissolved compound, water itself produces a small number of ions:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
This means:
The product formed depends on which ion is preferentially discharged at each electrode.
The cathode attracts cations. The competing ions are the metal ion and H⁺ from water.
| Condition | Product at Cathode | Explanation |
|---|---|---|
| Metal is more reactive than hydrogen (e.g. Na, K, Ca, Mg, Al) | Hydrogen gas (H₂) | The metal ion stays in solution; H⁺ ions are discharged instead |
| Metal is less reactive than hydrogen (e.g. Cu, Ag, Au) | The metal | The metal ion is preferentially discharged |
In short: If the metal is reactive, you get hydrogen. If the metal is unreactive, you get the metal.
Half equation for hydrogen production: 2H⁺(aq) + 2e⁻ → H₂(g)
Half equation for copper deposition: Cu²⁺(aq) + 2e⁻ → Cu(s)
The anode attracts anions. The competing ions are the anion from the compound and OH⁻ from water.
| Condition | Product at Anode | Explanation |
|---|---|---|
| A halide ion is present (Cl⁻, Br⁻, I⁻) | The halogen (Cl₂, Br₂, I₂) | Halide ions are preferentially discharged |
| No halide ion present (e.g. sulfate SO₄²⁻, nitrate NO₃⁻) | Oxygen gas (O₂) | OH⁻ ions from water are discharged instead |
Half equation for chlorine production: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Half equation for oxygen production: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻
Exam Tip: The half equation for oxygen production from OH⁻ ions is the hardest one to remember: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻. Make sure you can write it and balance it. It is frequently tested.
Sodium chloride solution (brine) is one of the most important industrial electrolysis processes.
Ions present in solution:
Competing ions: Na⁺ and H⁺
Sodium is more reactive than hydrogen, so hydrogen gas is produced.
Half equation: 2H⁺(aq) + 2e⁻ → H₂(g)
Observation: Bubbles of colourless gas at the cathode; gas gives a squeaky pop with a lighted splint.
Competing ions: Cl⁻ and OH⁻
A halide ion (Cl⁻) is present, so chlorine gas is produced.
Half equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Observation: Bubbles of pale green/yellow gas at the anode; gas bleaches damp litmus paper (turns it white).
Na⁺ and OH⁻ ions remain in solution → this forms sodium hydroxide (NaOH), an important alkali used in industry.
| Product | Where | Test |
|---|---|---|
| Hydrogen | Cathode | Squeaky pop with lighted splint |
| Chlorine | Anode | Bleaches damp litmus paper white |
| Sodium hydroxide | Left in solution | pH test shows alkaline (pH 13–14) |
Exam Tip: You must know all three products of brine electrolysis and how to test for hydrogen and chlorine. This is a guaranteed exam topic.
Ions present in solution:
Competing ions: Cu²⁺ and H⁺
Copper is less reactive than hydrogen, so copper metal is produced.
Half equation: Cu²⁺(aq) + 2e⁻ → Cu(s)
Observation: Orange-brown/pink solid copper is deposited on the cathode.
Competing ions: SO₄²⁻ and OH⁻
Sulfate is not a halide ion, so oxygen gas is produced (from OH⁻).
Half equation: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻
Observation: Bubbles of colourless gas at the anode; gas relights a glowing splint.
As electrolysis proceeds, Cu²⁺ ions are removed from solution (deposited as copper at the cathode). The blue colour of the solution fades over time as the concentration of Cu²⁺ ions decreases. Eventually, the solution may become colourless if all the copper ions are deposited.
Ions present: Na⁺(aq), SO₄²⁻(aq), H⁺(aq), OH⁻(aq)
| Electrode | Competing Ions | Product | Reason |
|---|---|---|---|
| Cathode | Na⁺ vs H⁺ | Hydrogen (H₂) | Sodium is more reactive than hydrogen |
| Anode | SO₄²⁻ vs OH⁻ | Oxygen (O₂) | No halide present |
This is essentially the electrolysis of water — the sodium sulfate is just there to make the water conduct electricity.
Overall equation: 2H₂O(l) → 2H₂(g) + O₂(g)
When copper electrodes are used (instead of inert graphite or platinum electrodes), the results are different because the copper anode is reactive.
At the cathode: Copper ions from solution are deposited as copper metal (same as with inert electrodes).
Cu²⁺(aq) + 2e⁻ → Cu(s)
The cathode gains mass and becomes coated in pink/brown copper.
At the anode: Instead of OH⁻ ions being discharged, the copper anode itself dissolves, releasing Cu²⁺ ions into solution.
Cu(s) → Cu²⁺(aq) + 2e⁻
The anode loses mass and becomes smaller/thinner.
| Observation | Explanation |
|---|---|
| Cathode gains mass | Copper deposited from solution |
| Anode loses mass | Copper from the anode dissolves into solution |
| Blue colour of solution stays the same | Cu²⁺ ions are removed at cathode but replaced at anode |
| No gas produced at either electrode | Copper is deposited/dissolved, not gas |
For every Cu²⁺ ion removed from solution at the cathode, one Cu²⁺ ion is added to solution from the dissolving anode. The concentration of Cu²⁺ ions (and therefore the blue colour) remains constant.
Exam Tip: Be prepared to compare electrolysis with inert electrodes vs copper electrodes. With inert (graphite) electrodes, the solution fades and oxygen is produced at the anode. With copper electrodes, the anode dissolves, no gas is produced, and the solution colour stays constant. This comparison is a very popular exam question.
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