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A current of 2 A was passed for 1 h through a solution of ${CuSO}_{4} . 0 .237 g of {Cu}^{2 +}$ ions was discharged at cathode. The current efficiency is

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10%

42.2%

40.01%

26.1%

answer is C.

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Detailed Solution

$W = \frac{63 .5}{2 \times 96500} \times 2 \times 60 \times 60 = 2 .37 g$

% of efficiency $= \frac{0 .237}{2 .37} \times 100 = 10 %$

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