Question

Calculate the time taken to deposit 1.27 g at copper at cathode when a current of 2 amp. is passed through the solution of CuSO₄.

(Atomic weight of Cu = 63.5 gmol^-1)

Answer 1

From Faraday’s law of electrolysis

W = Zit = `"E"/"F"` it

where W = Deposited amount

E = Equivalent weight

F = Faraday constant = 96500

i = Electricity

⇒ `1.27 = 63.5/(96500 xx 2) xx "t"`

⇒ t = 1930 sec

Thus, 1930 sec is required to deposit 1.27 g of copper at the cathode.