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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]

A.

64.0g

B.

32.0g

C.

16.0g

D.

6.4g

E.

3.2g

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Correct answer is E

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

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