Statement I: In the electrolysis of an aqueous CuSO4 solution with copper electrodes, copper is dissolved at the anode and deposited at the cathode.
Statement II: In an electrolytic cell, oxidation takes place at the anode and reduction takes place at the cathode.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
In an electrolytic cell an external voltage drives the chemical change. With two copper electrodes dipping in aqueous \( \ce{CuSO4} \), the electrode reactions are:
Cathode: \( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
Anode: \( \displaystyle \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \)
Statement I: In the electrolysis of an aqueous CuSO4 solution with copper electrodes, copper is dissolved at the anode and deposited at the cathode. — True. The copper anode goes into solution as \( \ce{Cu^2+} \) ions, and \( \ce{Cu^2+} \) ions from the solution are deposited as copper metal on the cathode.
Statement II: In an electrolytic cell, oxidation takes place at the anode and reduction takes place at the cathode. — True. The anode is the electrode where electrons are lost (oxidation) and the cathode is where they are gained (reduction).
Why Statement II explains Statement I? At the anode copper atoms lose two electrons each, which is oxidation, so the anode dissolves. At the cathode \( \ce{Cu^2+} \) ions gain two electrons each, which is reduction, so copper is deposited. Statement I is this general rule applied to copper electrodes.
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.”