Electrolysis of Copper Sulphate: Platinum versus Copper Electrodes
A student electrolyses 500 mL of 0.50 M CuSO₄ with platinum electrodes, passing a steady current of 2.0 A for 90 minutes (F = 96 500 C mol⁻¹, Ar(Cu) = 63.5). The solution is stirred and held near 25 °C. The student observes that the blue colour fades only slightly, that a colourless gas collecting at the anode relights a glowing splint, and that pH paper dipped into the solution turns red. The student concludes that 'all the copper has been plated out and the acid left behind is sulphuric acid'. A second student repeats the experiment with copper electrodes under otherwise identical conditions and reports no gas at the anode, an unchanged blue colour, and a cathode mass gain exactly matched by the anode mass loss. Both students claim their results demonstrate 'the same reaction'.
1. Use the charge passed to test the first student's conclusion quantitatively, and account for the anode gas and the pH change.
Read the worked answer
Charge passed: Q = It = 2.0 × 5400 = 10 800 C, so moles of electrons = 10 800/96 500 = 0.112 mol.
Cathode: Cu²⁺ + 2e⁻ → Cu, so Cu deposited = 0.112/2 = 0.056 mol, mass = 0.056 × 63.5 = 3.55 g.
Initial copper: 0.500 L × 0.50 M = 0.250 mol. Fraction removed = 0.056/0.250 = 22%.
Anode: sulphate is not oxidised at platinum, so water is oxidised: 2H₂O → O₂ + 4H⁺ + 4e⁻. Moles O₂ = 0.112/4 = 0.028 mol, volume at STP = 0.028 × 22.4 = 0.63 L, which explains the glowing splint. Moles H⁺ = 0.112 in 0.500 L = 0.224 M, so pH ≈ 0.65 and the red pH paper is expected.
Judgement: the first student is wrong. Only 22% of the copper has been removed, so the solution remains distinctly blue, and the acid generated is not simply 'left behind' - it is produced by the anodic oxidation of water, with the sulphate acting only as a spectator.
2. Explain why the copper-electrode experiment gives different observations, and say which description of the cell reaction is correct in each case.
Read the worked answer
With a copper anode the metal itself is oxidised, because copper is a far better electron donor than water: Cu(s) → Cu²⁺ + 2e⁻.
Anode and cathode reactions are mirror images: 3.55 g of copper dissolves at the anode while 3.55 g deposits at the cathode, so the blue colour is unchanged, no oxygen forms, and no H⁺ is generated, so the pH stays near its starting value. The overall process is simply transfer of copper from anode to cathode - the basis of electrorefining - and the net cell reaction has no chemical change in the electrolyte.
With platinum the anode cannot dissolve, so water must be oxidised, and the overall reaction is the electrolytic decomposition of the electrolyte: copper is removed at the cathode while oxygen and acid are generated at the anode.
Judgement: the two students are observing genuinely different reactions, not 'the same reaction'. The electrode material, not the electrolyte, decides which anode reaction is possible; a claim about a cell must always specify the electrodes. This also explains why copper plating baths use copper anodes - to keep the bath composition constant.
3. Estimate how long it would take at 2.0 A to remove 95% of the copper with platinum electrodes, and state why the real time differs.
Read the worked answer
Copper to remove = 0.95 × 0.250 = 0.2375 mol.
Electrons required = 2 × 0.2375 = 0.475 mol; charge = 0.475 × 96 500 = 45 838 C.
Time = 45 838/2.0 = 22 919 s ≈ 6.4 hours.
Caveat: this assumes 100% current efficiency for copper deposition, which fails as the solution is stripped. The cathode potential shifts toward hydrogen evolution (2H⁺ + 2e⁻ → H₂) once [Cu²⁺] becomes small, so part of the current produces hydrogen rather than copper, the deposition efficiency falls below 100%, and the real time exceeds 6.4 hours. The pH also falls faster than the simple anode calculation predicts, because hydrogen evolution removes H⁺ at the cathode while the anode continues to generate it, and the extra acid must be tracked.
Judgement: 6.4 hours is a lower bound, valid only for the early, copper-rich part of the run. In practice the last few per cent of copper is the expensive part, and the experiment should be stopped when the blue colour has essentially gone rather than at a calculated time.