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Q.

A current of 3 ampere was passed for 2 hour through a solution of $C u S O_{4} . 3 g$ of $C u^{2 +}$ ions were discharged at cathode. Then percentage of current efficiency. (at wt. of Cu = 63.5) is ___(round off the number)

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answer is 42.

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Detailed Solution

$w_{C u} = \frac{E . i . t}{96500}$
$3 = \frac{63.5 \times i \times 2 \times 60 \times 60}{2 \times 96500} \Rightarrow i = 1.266$ ampere
Current efficiency = $\frac{c u r r e n t p a s s e d a c t u a l l y}{T o t a l c u r r e n t p a s s e d \exp e r i m e n t a l y} \times 100$
$= \frac{1.266}{3} \times 100 = 42.2 %$

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