Half equations for electrolysis

Chemical changes

GCSE Chemistry (8462) · 4.4.3.4 & 4.4.3.5 · Higher

ORIGINAL

List the ions, decide each electrode's product by the rules, then balance atoms AND charge with electrons.

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Question

Sodium chloride solution is electrolysed using inert electrodes. Name the product at each electrode, write a half equation for each, and state which electrode reaction is an oxidation.

Every step worked, with the reasoning.

1
Ions present: $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ from the salt, $\mathrm{H^+}$ and $\mathrm{OH^-}$ from the water
In an aqueous solution, water ionises slightly, so four ions are attracted to the electrodes.
2
Cathode (negative electrode): hydrogen, $\mathrm{2H^+ + 2e^- \rightarrow H_2}$
Sodium is more reactive than hydrogen, so hydrogen is produced instead of sodium. Positive ions gain electrons at the cathode.
3
Anode (positive electrode): chlorine, $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$
The solution contains halide ions, so the halogen is produced rather than oxygen.
4
Check the charge: cathode $2(+1) + 2(-1) = 0$ on both sides; anode $-2$ on both sides
A half equation balances atoms and charge. Common error: writing $\mathrm{Cl^- \rightarrow Cl + e^-}$ — chlorine is diatomic, so two chloride ions make one molecule.
5
The anode reaction is the oxidation
Oxidation is loss of electrons: each $\mathrm{Cl^-}$ ion loses one. Gaining electrons at the cathode is reduction.

Answer: Cathode: hydrogen, $\mathrm{2H^+ + 2e^- \rightarrow H_2}$ (reduction). Anode: chlorine, $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$ (oxidation).

Question

Stuck on electrolysis half equations?