0.1 Faraday of electricity was passed through a solution of copper (ll) sulphate. The maximum weight of copper deposited on the cathode would be [Cu = 64]
- 64.0g
- 32.0g
- 16.0g
- 6.4 g
-
3.2g
The correct answer is: E
Explanation
CuSO4 \(\to\) Cu2+ + SO42+Cu2+ + 2e \(\to\) Cu
1 Faraday will deposit \(\frac{64}{2}\)gms of copper
0.1 faraday will deposit \(\frac{64}{2} \times \frac{1}{10} = \frac{64}{20}\)
= \(\frac{32}{10}\) = 3.2gms