The electrolysis of copper(ii) sulfate solution using inert electrodes
2 marks · HardWrite the balanced half-equation for the reaction that takes place at the anode during the electrolysis of copper(II) sulfate solution using inert electrodes.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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4OH^-(aq) → 2H2O(l) + O2(g) + 4e^- OR 2H2O(l) → O2(g) + 4H^+(aq) + 4e^-
Must show: correct ions and state symbols, correct balancing, electrons shown on correct side
Also accept: 2H2O(l) → O2(g) + 4H^+(aq) + 4e^-
Common mistakes: Using sulfate ions instead of hydroxide ions; Incorrect number of electrons; Omitting state symbols
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Marking points
M1 · 1 mark · Correct identification of hydroxide ions or water as the species being oxidised; includes electrons on the correct side; correct reactants.
A1 · 1 mark · Fully balanced and correct half-equation for oxygen production at the anode with appropriate state symbols and electrons.
Worked solution
At the anode, hydroxide ions from water lose electrons to form oxygen and water.
Alternative, using water as the reactant: