Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 u)
A
0 g
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B
63.5 g
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C
2 g
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D
127 g
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Solution
The correct option is B 63.5 g Given Q = 2F
Atomic mass of Cu = 63.5u
Valcency of the metal Z = 2
We have, CuSO4→Cu2++SO2−4 Cu(2+)+2e−→Cu
1mol 2mol 1mol = 63.5g
2F
Alternatively, W=ZQ=EF.2F=2E=2×63.52=63.5g