WAEC Chemistry 2014 Paper

Concept:During electrolysis of aqueous $\mathrm{CuSO}_4$ using inert carbon electrodes, oxidation takes place at the anode, where $\mathrm{OH}^-$ ions are preferentially discharged over $\mathrm{SO}_4^{2-}$ ions.Explanation:In aqueous $\mathrm{CuSO}_4$, the possible anions are $\mathrm{OH}^-$ and $\mathrm{SO}_4^{2-}$. At the anode, oxidation occurs and anions lose electrons. Sulphate ions are very stable and difficult to oxidize. Therefore, water supplies $\mathrm{OH}^-$ ions, which easily lose electrons to be oxidized. The half-reaction at the anode is represented as:$$4\mathrm{OH}^-_{(aq)} \longrightarrow 2\mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)} + 4e^-$$This can also be written as:$$4\mathrm{OH}^-_{(aq)} - 4e^- \longrightarrow 2\mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)}$$Hence, oxygen gas is evolved at the carbon anode during electrolysis of aqueous copper sulfate.Answer:Option C is the correct anode reaction.