The cell is
and
and n = 2
∴
As K is very high, the reaction is favoured in the forward direction, so, Sn2+ can easily reduce Fe3+ ion to Fe2+ ion.
The cathode reaction is Cu2+ + 2e– → Cu
I = 1.5 amperes, t = 10 x 60 = 600 s
∴ Q = It = 1.5 x 600 = 900 C
The reaction states that 2 x 96500 C are required to deposit 63.5 g Cu at cathode
900 C will deposit
t = 600 s
Charge = current x time
= 1.5 A x 600s = 900 C
According to the reaction : Cu2+(aq) + 2e– → Cu(s)
We require 2F or 2 x 96487 C to deposit 1 mol or 63 g of Cu
For 900 C, the mass of Cu deposited =