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