Q. 163.9( 40 Votes )

Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Given -

I = 1.5 A

W = 1.45 g of Ag

t = ?

n = 1

Answer :

Equivalent weight is Ag, EAg = = 180


Equivalent weight is Cu, ECu = = 31.75


Equivalent weight is Zn, EZn= = 32.5


Using Faraday’s second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,


Equation 1




WCu = 0.426 g




WZn = 0.436 g


To find the time of current flow, using Faraday’s first law of electrolysis we get,


M = Z ×I ×t Equation 2


Z = , Equation 2 becomes,


M =


t =


t = 864 seconds.


The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g


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