A solution of CuSO4 is electrolysed for 10 min with a current of 1.5 A. What is the mass of copper deposited at the cathode (molar mass of Cu = 63 g) (1F = 96487 C).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.2938 g
By Faraday’s first law, the mass deposited is proportional to the quantity of electricity passed. Copper needs 2 electrons per ion, so 2 F deposit 1 mol (63 g) of Cu.
\( Q = I \times t = 1.5\ \text{A} \times (10 \times 60)\ \text{s} = 900\ \text{C} \)
\( \ce{Cu^2+ + 2e- -> Cu} \)
\( \displaystyle \text{Mass of Cu} = \frac{63 \times 900}{2 \times 96487} = \frac{56700}{192974} = 0.2938\ \text{g} \)
Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Electrolytic Cells and Electrolysis – “Example 2.10 A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?”
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 20-06 S3 02-06 S3 31-05 S1 | – |