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Practise 16 exam-style questions on Representation of reactions at electrodes as half equations (HT only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Write a half equation for the formation of potassium from potassium ions at the cathode.
K⁺ + e⁻ → K
Potassium is in Group 1, so its ion has a 1+ charge and needs to gain one electron to become an atom. The charge is 0 on both sides.
K⁺ + e⁻ → K
allow K+ + e- → K; ignore state symbols
Explain how you can tell that this half equation is balanced:
2Br⁻ → Br₂ + 2e⁻
There are two bromine atoms on the left (in 2Br⁻) and two on the right (in Br₂), so the atoms balance. The total charge on the left is 2 × (−1) = −2, and on the right it is 0 + 2 × (−1) = −2, so the charges also balance.
A half equation must balance both atoms and charge. Electrons carry the charge, so counting them is how you make the charges equal.
there are two bromine atoms on each side
allow the atoms balance
the total charge is the same on both sides (2−)
allow the charges balance; do not accept 'the electrons balance' alone
In the electrolysis of aluminium oxide the half equations are:
Al³⁺ + 3e⁻ → Al
2O²⁻ → O₂ + 4e⁻
Use the half equations to explain why 4 moles of aluminium are produced for every 3 moles of oxygen.
The electrons released at the anode are the same electrons that are used at the cathode, so the numbers must be equal. Making 4 mol of aluminium needs 4 × 3 = 12 mol of electrons. Each mole of O₂ formed releases 4 mol of electrons, so 12 mol of electrons are released when 3 mol of O₂ forms. So 4 mol of Al are made for every 3 mol of O₂.
To combine half equations, multiply them so that the electrons are equal: 4Al³⁺ + 12e⁻ → 4Al and 6O²⁻ → 3O₂ + 12e⁻. This gives the overall ratio 4Al : 3O₂, matching 2Al₂O₃ → 4Al + 3O₂.
the number of electrons lost at the anode must equal the number gained at the cathode
4 mol of Al needs 4 × 3 = 12 mol of electrons
12 mol of electrons are released when 3 mol of O₂ forms (3 × 4 = 12)
The table shows four half equations, A to D, for reactions at electrodes during electrolysis.
For each half equation, state whether it happens at the cathode or at the anode, and whether it is oxidation or reduction.
A: cathode, reduction
B: anode, oxidation
C: anode, oxidation
D: cathode, reduction
Half equations with electrons on the left are reductions and happen at the cathode, where positive ions gain electrons. Those with electrons on the right are oxidations and happen at the anode, where negative ions lose electrons.
A: cathode, reduction
both needed for each mark
B: anode, oxidation
C: anode, oxidation
D: cathode, reduction
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