Ionic compounds can be decomposed when an electric current is passed through it. This process is called electrolysis.

For electrolysis to happen, the ionic compound must be able to conduct electricity. 

• In solid state, ionic compounds cannot conduct electricity because ions are fixed in the ionic lattice. 
• In molten or in aqueous solution, ionic compounds can conduct electricity because ions are free to move.

👇 For IGCSE Chemistry, you need to understand what happens during electrolysis 👇

1. Ionic compound dissociates into its ions when current is passed on between the two electrodes.
2. Positive ions (cation) are attracted to the negative electrode (cathode). Negative ions (anion) are attracted to the positive electrode (anode).
3. Cations receive electrons and get reduced at the cathode. Anions lose electrons and get oxidized at the anode.
4. Substances that are formed at each electrode differ based on whether the ionic compound is molten or in aqueous solution.

Electrolysis of Molten Ionic Compound

• At the cathode, the positively charged metal ion gets reduced to form metal.
• At the anode, negatively charged non-metal ion gets oxidized to form the non-metal element.

• At the cathode: Na⁺(l) + e^- → Na(s)
• At the anode: 2Cl^-(l) → Cl₂(g) + 2e^-

Electrolysis of Aqueous Ionic Compound

Electrolysis of aqueous ionic compounds can be complicated because water can provide H⁺ and OH^- ions in addition to the ions provided by the ionic compound.

• At the cathode: metal ions and H⁺ ions are attracted to the cathode. If the metal is above hydrogen in the reactivity series, hydrogen gas will be formed. If the metal is below hydrogen, solid metal will be formed.
• At the anode: non-metal ions and OH^- ions are attracted to the anode. If the non-metal ion is a simple halide ion, halogen will be produced. Otherwise, oxygen gas will be produced.

• At the cathode: 2H₂O(l) + 2e^- → 2OH^-(aq) + H₂(g)
• At the anode: 2Cl^-(aq) → Cl₂(g) + 2e^-

During the electrolysis of aqueous sodium sulfate:

• At the cathode: 2H⁺(aq) + 2e^- → H₂(g)
• At the anode: 4OH^-(l) → 2H₂O(l) + O₂(g) + 4e^- 

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