Electrolysis of Aqueous Solutions

This lesson covers the electrolysis of aqueous solutions as required by the AQA GCSE Combined Science Trilogy specification (8464). When an ionic compound is dissolved in water, there are additional ions from the water. You must be able to predict the products using simple rules and write half equations (higher tier).

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Why Are Aqueous Solutions Different?

When an ionic compound is dissolved in water, the water itself provides additional ions:

$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$H2​O(l)⇌H+(aq)+OH−(aq)

So an aqueous solution contains:

• Cations from the compound (e.g. Na⁺, Cu²⁺)
• Anions from the compound (e.g. Cl⁻, SO₄²⁻)
• H⁺ ions from water
• OH⁻ ions from water

At each electrode, there is competition between the ions, and rules determine which ion is discharged.

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Rules for Predicting Products

At the Cathode (Reduction — Positive Ions)

flowchart TD
    A["Which cation is<br/>present?"] --> B{"Is the metal LESS<br/>reactive than hydrogen?"}
    B -- "Yes<br/>(e.g. Cu²⁺, Ag⁺)" --> C["The METAL is<br/>deposited"]
    B -- "No<br/>(e.g. Na⁺, K⁺, Ca²⁺, Al³⁺)" --> D["HYDROGEN gas<br/>is produced<br/>(H⁺ ions are reduced<br/>instead)"]

    style C fill:#2e7d32,color:#fff
    style D fill:#1565c0,color:#fff

Rule: If the metal is more reactive than hydrogen, hydrogen gas is produced instead of the metal. If the metal is less reactive than hydrogen, the metal is deposited.

At the Anode (Oxidation — Negative Ions)

flowchart TD
    A["Which anion is<br/>present?"] --> B{"Is a HALIDE ion present?<br/>(Cl⁻, Br⁻, I⁻)"}
    B -- "Yes" --> C["The HALOGEN is<br/>produced<br/>(Cl₂, Br₂, or I₂)"]
    B -- "No<br/>(e.g. SO₄²⁻, NO₃⁻)" --> D["OXYGEN gas<br/>is produced<br/>(OH⁻ ions are oxidised<br/>instead)"]

    style C fill:#d32f2f,color:#fff
    style D fill:#e65100,color:#fff

Rule: If a halide ion (Cl⁻, Br⁻, I⁻) is present, the halogen is produced at the anode. If no halide is present, oxygen is produced (from the discharge of OH⁻ ions from water).

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Summary Table of Aqueous Electrolysis Products

Electrolyte	At Cathode	At Anode
Copper sulfate (CuSO₄)	Copper (Cu is less reactive than H)	Oxygen (no halide)
Sodium chloride (NaCl)	Hydrogen (Na is more reactive than H)	Chlorine (halide present)
Copper chloride (CuCl₂)	Copper (Cu is less reactive than H)	Chlorine (halide present)
Sodium sulfate (Na₂SO₄)	Hydrogen (Na is more reactive than H)	Oxygen (no halide)
Dilute sulfuric acid (H₂SO₄)	Hydrogen	Oxygen
Potassium bromide (KBr)	Hydrogen (K is more reactive than H)	Bromine (halide present)
Silver nitrate (AgNO₃)	Silver (Ag is less reactive than H)	Oxygen (no halide)

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Half Equations (Higher Tier)

At the Cathode

If hydrogen is produced: $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$2H+(aq)+2e−→H2​(g)

If a metal (e.g. copper) is deposited: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$Cu2+(aq)+2e−→Cu(s)

At the Anode

If a halogen (e.g. chlorine) is produced: $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$2Cl−(aq)→Cl2​(g)+2e−

If oxygen is produced: $4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^-$4OH−(aq)→O2​(g)+2H2​O(l)+4e−

Exam Tip: The half equation for oxygen production at the anode (4OH⁻ → O₂ + 2H₂O + 4e⁻) is often tested. Make sure you can write it correctly and check it is balanced for atoms, charges and electrons.

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Testing for the Products

Product	Gas Test
Hydrogen	Hold a burning splint near the gas — it gives a squeaky pop
Oxygen	Insert a glowing splint — it relights
Chlorine	Hold damp blue litmus paper near the gas — it turns red then white (bleaches)

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Electrolysis of Aqueous Copper Sulfate — Detailed Example

Products

• Cathode: Copper metal is deposited (brown/pink coating on the electrode).
• Anode: Oxygen gas is produced (bubbles).

Observations

• The blue colour of the solution fades as Cu²⁺ ions are removed from the solution and deposited as copper metal.
• Bubbles appear at the anode (oxygen).

Half Equations (Higher Tier)

Cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$Cu2+(aq)+2e−→Cu(s)

Anode: $4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^-$4OH−(aq)→O2​(g)+2H2​O(l)+4e−

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Electrolysis of Sodium Chloride Solution (Brine) — Detailed Example

Products

• Cathode: Hydrogen gas (Na is more reactive than H, so H⁺ is reduced instead).
• Anode: Chlorine gas (halide ion present).
• Solution: Sodium hydroxide (NaOH) remains in solution — used in industry.

Half Equations (Higher Tier)

Cathode: $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$2H+(aq)+2e−→H2​(g)

Anode: $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$2Cl−(aq)→Cl2​(g)+2e−

Industrial Importance

The electrolysis of brine is commercially very important. The three products are:

Product	Use
Chlorine	Water purification, making bleach, making PVC
Hydrogen	Making margarine, making ammonia (Haber process), fuel
Sodium hydroxide	Making soap, making paper, making bleach

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Common Mistakes

Mistake	Correction
Thinking a reactive metal (e.g. sodium) will be deposited from an aqueous solution	Reactive metals are NOT deposited — hydrogen is produced instead
Predicting oxygen at the anode when a halide is present	If Cl⁻, Br⁻ or I⁻ is present, the halogen is produced, not oxygen
Writing the wrong half equation for oxygen at the anode	It is 4OH⁻ → O₂ + 2H₂O + 4e⁻ (NOT 2O²⁻ → O₂ + 4e⁻ — there are no O²⁻ ions in aqueous solution)
Forgetting that water provides H⁺ and OH⁻ ions	These additional ions from water are what make aqueous electrolysis different from molten electrolysis

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Worked Example

Question: Predict the products when aqueous potassium iodide (KI) is electrolysed. Write half equations for each electrode.

Answer:

• At the cathode: Potassium is more reactive than hydrogen, so hydrogen gas is produced. Half equation: 2H⁺(aq) + 2e⁻ → H₂(g)
• At the anode: Iodide is a halide ion, so iodine is produced. Half equation: 2I⁻(aq) → I₂(aq) + 2e⁻

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Summary