Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of $CuSO_{4}$.

Since, 63.5 g of Cu is deposited by = 96500C

.’. 1.27 g of Cu is deposited by

= 96500/63.5 x l 2? =1930C

As, Q = i X t [Where, Q = charge, i = current, t = time]

=> t = Q/i

Here, i = 2A, Q = 1930 C

.-. t =1930/2 = 965 s