Memory Aids

Part of Electrolysis of Aqueous Solutions · Section 9 of 13

Memory AidUnit: ElectrolysisGCSE

This memory aid covers Memory Aids 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 9 of 13 in this topic. Use it for quick recall, then test yourself straight afterwards so the memory aid becomes usable in an answer.

🧠 Memory Aids

For the CATHODE rule: "If the metal is ABOVE hydrogen, HYDROGEN forms — it takes its place!"

Think of it as hydrogen "jumping the queue" above reactive metals — because the metal is so reactive (so keen to stay ionic), hydrogen gets the electrons instead.

For the ANODE rule: "HALIDE = Halogen at the anode. No halide = no halogen = OXYGEN instead."

Remember the three halides: Cl⁻, Br⁻, I⁻ — if any of these are present in solution, a halogen gas forms at the anode.

Brine products mnemonic: "3 Hs from Brine — Hydrogen, cHlorine, Hydroxide (sodium)" — three useful products from one reaction!

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?

  • A. Na⁺, Cl⁻, H⁺ and OH⁻
  • B. Na⁺, Cl⁻, H₂O and OH⁻
  • C. Na⁺, Cl⁻ only
  • D. Na⁺, Cl⁻, H₂ and O²⁻
1 markfoundation

Describe the three products formed when concentrated brine is electrolysed, and state where each is produced.

3 marksstandard

Quick recall flashcards

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.
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.

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