\(30\, \ mL\) of \(NiSO_4\) is electrolysed using a current of \(360\, \ mA\) for \(30\) minutes. How much of the metal would have been plated out if the current efficiency is only \(60\%\)? (\(Ni = 58.7\, \ g/mol)\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.1182 g
By Faraday’s first law, the mass deposited at an electrode is proportional to the quantity of electricity actually used for the electrode reaction. With a current efficiency of 60%, only 60% of the charge passed deposits nickel.
\( Q = I \times t = 0.360\,\mathrm{A} \times (30 \times 60)\,\mathrm{s} = 648\,\mathrm{C} \)
\( Q_{\text{used}} = 0.60 \times 648 = 388.8\,\mathrm{C} \)
At the cathode, \( \ce{Ni^2+ + 2e^- -> Ni} \), so 2 × 96500 C deposit 1 mol (58.7 g) of nickel.
\( \displaystyle m = \frac{58.7 \times 388.8}{2 \times 96500} = 0.1182\,\mathrm{g} \)
The volume of the solution (30 mL) is not needed.
NCERT (2026-27 edition):
Faraday’s Laws of Electrolysis – …
