0.1 Faraday of electricity was passed through a solution ...
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
6.4g
3.2g
Correct answer is E
CuSO4 → Cu2+ + SO42+
Cu2+ + 2e → Cu
1 Faraday will deposit 642gms of copper
0.1 faraday will deposit 642×110=6420
= 3210 = 3.2gms
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