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  4. How many minutes would be required to deposit copper in 500 mL of 0.25 N CuSO4 by a current of 75 milliampere?

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Q. How many minutes would be required to deposit copper in 500 mL of 0.25 N $CuSO_{4}$ by a current of 75 milliampere?

Electrochemistry

Solution:

$V=500 mL =0.5 L$
Equivalents = 0.25 × 0.5 = 0.125
Current =$75\times10^{-3} $ ampere; Q = It
$0.125=\frac{75\times10^{-3}\times t}{96500} \Rightarrow t=2680 \, min$