| Electrode | Condition | Product | Half equation |
|---|---|---|---|
| CATHODE (−) | Metal above hydrogen in reactivity series (K, Na, Li, Ca, Mg, Al) | Hydrogen gas forms | 2H⁺ + 2e⁻ → H₂ |
| CATHODE (−) | Metal below hydrogen in reactivity series (Cu, Ag, Au, Pt) | Metal deposits | Cu²⁺ + 2e⁻ → Cu |
| ANODE (+) | Halide ion present (Cl⁻, Br⁻, I⁻) | Halogen gas forms | 2Cl⁻ → Cl₂ + 2e⁻ |
| ANODE (+) | No halide present (SO₄²⁻, NO₃⁻, etc.) | Oxygen gas forms | 4OH⁻ → O₂ + 2H₂O + 4e⁻ |
This comparison covers The Two Decision Rules — Learn These! within Electrolysis of Aqueous Solutions for GCSE Chemistry. Revise Electrolysis of Aqueous Solutions in Electrolysis for GCSE Chemistry with 21 exam-style questions and 15 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 4 of 13 in this topic. Use this comparison to connect the idea to the wider topic before moving on to questions and flashcards.
Practice questions for Electrolysis of Aqueous Solutions
When sodium chloride (NaCl) is dissolved in water, which four types of ion are present in the solution?
Describe the three products formed when concentrated brine is electrolysed, and state where each is produced.
Quick recall flashcards
State the cathode rule for aqueous electrolysis.
If the metal is MORE reactive than hydrogen (above H in reactivity series) → hydrogen gas forms.
If the metal is LESS reactive than hydrogen (below H) → the metal deposits.
Examples: Na above H → H₂ forms. Cu below H → Cu deposits.
State the anode rule for aqueous electrolysis.
If a halide ion (Cl⁻, Br⁻, or I⁻) is present → the halogen gas forms.
If NO halide is present → oxygen gas forms from the OH⁻ ions.
Example: Cl⁻ present → Cl₂ forms. SO₄²⁻ present (no halide) → O₂ forms.