Example 1: Core Knowledge
Question 1
A student uses this apparatus to investigate the electrolysis of copper(II) sulfate solution.
2H2O → + +
Final answer
2H2O → 4H+ + O2 + 4e−
Mark scheme points
- M1 Includes O2 and e− on the product side.
- M2 Full half-equation correct: 2H2O → 4H+ + O2 + 4e−.
Explanation
At the positive electrode, water is oxidised to oxygen.
- First, make sure the products include oxygen gas, O2, and electrons, e−. That secures the first marking point.
- Then balance the equation fully. Starting from 2H2O gives 4 hydrogen atoms and 2 oxygen atoms on the left.
- Put O2 on the right for the 2 oxygen atoms, and 4H+ for the 4 hydrogen atoms.
- Now balance charge: the right side has +4 from 4H+, so add 4e− to make the total charge 0 on both sides.
This gives the complete balanced half-equation:
2H2O → 4H+ + O2 + 4e−
Common mistakes
- Examiners reported that this was poorly answered: only a small minority gained both marks. A very common error was writing 2H2 instead of 4H+, which may balance atoms but does not balance charge.
- Writing O instead of O2. Oxygen is diatomic.
- Forgetting the electrons, or putting them on the left-hand side.
- Balancing atoms but not balancing charge. Half-equations must balance both.
- Using 4OH− → 2H2O + O2 + 4e− does not complete the given equation and only gains partial credit here.