In the electrolysis of an aqueous solution of copper sulphate using copper electrodes, the mass of the cathode increases by 2 g. What happens at the anode?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 g of copper goes into the solution as Cu2+ ions
With copper electrodes, copper is reduced at the cathode and the copper anode itself is oxidised:
Cathode: \( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
Anode: \( \displaystyle \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \)The same charge passes through both electrodes, and each reaction uses 2 electrons per Cu atom. So the number of Cu atoms that dissolve at the anode equals the number deposited at the cathode. The cathode gains 2 g, so 2 g of copper leaves the anode as \( \ce{Cu^2+} \) ions.
Why the other options are incorrect
Option (1) 0.2 mol of Cu2+ ions go into the solution — Incorrect. 2 g of copper is \( \frac{2}{63.5}\approx0.031 \) mol, not 0.2 mol.
Option (2) 560 mL of O2 is liberated — Incorrect. Oxygen would come from water only at an inert anode. Here the copper anode is oxidised instead, so no gas is given off.
Option (3) There is no loss in mass — Incorrect. The anode dissolves, so it loses 2 g, the same mass the cathode gains.
NCERT (2026-27 edition):
2.5 Electrolytic Cells and Electrolysis – “… One of the simplest electrolytic cell consists of two copper strips dipping in an aqueous solution of copper sulphate. … Thus copper is dissolved (oxidised) at anode and deposited (reduced) at cathode.”