In the cell Cd(s) | CdSO4(aq) || Hg2SO4(aq) | Hg(l) | Pt(s), what is the cathode half-reaction?

• A. Hg2SO4(s) + 2 e- ⟶ 2 Hg(l) + SO4^2- (aq)
• B. 2 Cd(l) + SO4^2- (aq) ⟶ CdSO4(s) + 2 e-
• C. CdSO4(s) + 2 e- ⟶ 2 Cd(l) + SO4^2- (aq)
• D. 2 Hg(l) + SO4^2- (aq) ⟶ Hg2SO4(s) + 2 e-

Answer: (A)

The cathode is where reduction happens, so the species on the right-hand side that can gain electrons is Hg2SO4(aq). The mercurous/mercuric species there contains Hg2+ that is reduced to metallic mercury. Two electrons are needed to reduce Hg2+ to two Hg atoms, and the sulfate ion simply remains in solution as SO4^2-. The correct half-reaction is Hg2SO4(aq) + 2 e- -> 2 Hg(l) + SO4^2- (aq). This fits the cell setup where the Cd electrode on the left undergoes oxidation, while the mercury-containing species on the right is reduced at the cathode.