4.15 Half equations at the electrodes

Writing and balancing half equations for the reactions at the cathode and anode during electrolysis, and explaining that reduction happens at the cathode and oxidation at the anode.

Higher only AQA Edexcel OCR 1 min read

Electrons at the electrodes

Key terms

Half equation: an equation showing the electrons gained or lost by one substance at an electrode.

At the cathode (−), positive ions gain electrons. This is reduction.
At the anode (+), negative ions lose electrons. This is oxidation.

Remember OIL RIG: oxidation is loss, reduction is gain.

Electrolysis of copper chloride solution with electrons flowing from the supply to the cathode, where copper ions gain electrons, and from the anode, where chloride ions lose electrons, with a list of common half equations

Writing a half equation

Write the ion on one side and the product on the other.
Balance the atoms (for example, two chloride ions make one Cl₂ molecule).
Add electrons so that the charges balance on both sides.

Example: chlorine at the anode

\mathrm{Cl^-} → Cl₂
balance atoms: 2\mathrm{Cl^-} → Cl₂
charges: left 2−, right 0, so add 2 electrons on the right: 2\mathrm{Cl^-} → Cl₂ + 2\mathrm{e^-}

Example: copper at the cathode

\mathrm{Cu^{2+}} → Cu: left 2+, right 0, so add 2 electrons on the left: \mathrm{Cu^{2+}} + 2\mathrm{e^-} → Cu

Common half equations

Half equations to learn
Electrode	Half equation	Product
Cathode	\mathrm{Cu^{2+}} + 2\mathrm{e^-} → Cu	copper
Cathode	2\mathrm{H^+} + 2\mathrm{e^-} → H₂	hydrogen
Cathode	\mathrm{Al^{3+}} + 3\mathrm{e^-} → Al	aluminium (molten)
Anode	2\mathrm{Cl^-} → Cl₂ + 2\mathrm{e^-}	chlorine
Anode	4\mathrm{OH^-} → O₂ + 2H₂O + 4\mathrm{e^-}	oxygen (from solutions)
Anode	2\mathrm{O^{2-}} → O₂ + 4\mathrm{e^-}	oxygen (molten oxides)

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5 exam questions · 8 marks · with mark schemes and model answers

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