Electrolysis in Aqueous Solutions
When ionic compounds are dissolved in water, water also breaks into H⁺ and OH⁻ ions. This means there are more ions present, so more possible reactions at the electrodes.
What is an aqueous solution?
- Aqueous means dissolved in water
- The solution contains ions from the ionic compound and from water:
- Water breaks down into H⁺ and OH⁻ ions
- So the electrolyte now contains:
- Positive ions: metal ions and H⁺
- Negative ions: non-metal ions and OH⁻
What happens at each electrode?
Cathode (negative electrode):
- Cations (positive ions) are attracted
- Either H⁺ or metal ions are discharged
- The less reactive one is discharged (gains electrons)
💡 If the metal is more reactive than hydrogen, hydrogen gas is produced instead
Anode (positive electrode):
- Anions (negative ions) are attracted
- Usually OH⁻ or halide ions (Cl⁻, Br⁻, I⁻)
- If halide ions are present → halogen forms
- If not → oxygen gas is formed from OH⁻
Example 1: Electrolysis of sodium chloride solution (NaCl (aq))
- Cathode: H⁺ is discharged (Na is more reactive) → hydrogen gas
- Anode: Cl⁻ is discharged → chlorine gas
Products: Hydrogen and chlorine gases
Example 2: Electrolysis of copper(II) sulfate solution (CuSO₄ (aq))
- Cathode: Cu²⁺ is discharged (less reactive than H⁺) → copper metal
- Anode: No halide ions, so OH⁻ is discharged → oxygen gas
Products: Copper and oxygen
General rules to remember
Electrode | Ions present | What gets discharged |
Cathode | Metal⁺, H⁺ | Least reactive (metal or H₂) |
Anode | Non-metal⁻, OH⁻ | Halide (if present), otherwise O₂ |
Questions
- What ions does water break into?
- Which gas forms if the metal is more reactive than hydrogen?
- What happens at the anode if no halide ions are present?
- In CuSO₄ (aq), what forms at the cathode?
- What is discharged at the anode if Cl⁻ is present?
Summary
- Aqueous solutions contain ions from the salt and water (H⁺ and OH⁻)
- At the cathode, the least reactive positive ion is discharged
- At the anode, halide ions are discharged if present; otherwise OH⁻ → oxygen gas
- Use the reactivity series to decide what forms
- Products depend on the ions present and their reactivity
