When \(CuSO_4\) solution is electrolysed using platinum electrodes, which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Copper is liberated at cathode and oxygen at anode
Platinum electrodes are inert: they only supply or remove electrons. In aqueous \( \ce{CuSO4} \):
Cathode (reduction): Cu2+(aq) + 2e− → Cu(s) (E° = +0.34 V, preferred over reduction of water)
Anode (oxidation): 2H2O(l) → O2(g) + 4H+(aq) + 4e−
The sulphate ion is not oxidised (water is oxidised more easily), so copper is deposited at the cathode and oxygen is evolved at the anode; the solution becomes acidic (\( \ce{H2SO4} \)).
Why the other options are incorrect
Option (1) Copper at cathode and sulphur at anode — Incorrect. Sulphur in \( \ce{SO4^{2-}} \) is already +6; it cannot give elemental sulphur by oxidation.
Option (2) Sulphur at cathode and copper at anode — Incorrect. \( \ce{Cu^{2+}} \) is reduced, so copper appears at the cathode; it cannot be liberated at the anode, where oxidation occurs.
Option (3) Oxygen at cathode and copper at anode — Incorrect. Oxygen comes from oxidation and copper from reduction; the electrodes are interchanged.
NCERT (2026-27 edition):
Products of Electrolysis – “Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons.”