How much of current is required to pass through CuSO4 for 6 minute 26 seconds to deposit 0.3177gm of Cu?

Answer 1

2 F deposits 63.5 g

So, 0.3177 g is deposited by 1x10^-2 F = 10^-2 x 96500 C

= 965 C

6 min 26 sec is 386 secs.

So, current is 965/386 = 2.5 A