For how long must a steady current of 2.0 A be passed through an aqueous solution of copper sulphate to deposit 1.27 g of copper at the cathode? (Molar mass of Cu = 63.5 g mol−1; F = 96500 C mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1930 s
By Faraday’s first law, the amount of copper deposited is proportional to the quantity of electricity passed, Q = I × t. At the cathode each Cu2+ ion takes two electrons:
\( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
So 1 mol of copper (63.5 g) needs 2 mol of electrons, that is 2 × 96500 C.
\( \displaystyle n(\ce{Cu}) = \frac{1.27\ \mathrm{g}}{63.5\ \mathrm{g\ mol^{-1}}} = 0.020\ \mathrm{mol} \)
\( \displaystyle Q = 0.020 \times 2 \times 96500\ \mathrm{C} = 3860\ \mathrm{C} \)
\( \displaystyle t = \frac{Q}{I} = \frac{3860\ \mathrm{C}}{2.0\ \mathrm{A}} = 1930\ \mathrm{s} \)
This is 32 minutes and 10 seconds.
NCERT (2026-27 edition):
Faraday’s Laws of Electrolysis – “(i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).”