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]
-
A.
64.0g -
B.
32.0g -
C.
16.0g -
D.
6.4 g -
E.
3.2g
Correct Answer: Option 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