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  4. A current of 0.25 A is passed through CuSO 4 solution placed in voltameter for 45 minutes. The amount of Cu deposited on cathode is (at. weight of Cu =63.6)

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Q. A current of $0.25 \,A$ is passed through $CuSO _{4}$ solution placed in voltameter for $45 $ minutes. The amount of $Cu$ deposited on cathode is (at. weight of $Cu =63.6)$

Electrochemistry

Solution:

$ I =0.25 A , t=45 \times 60\,s $

$Z =\frac{63.5}{96500 \times 2}$

$w =\frac{63.5}{2 \times 96500} \times 0.25 \times 45 \times 60=0.22\,g $