3.16 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 was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Hint: Faraday’s Second Laws of Electrolysis

Step 1:
According to the reaction:
Ag+(aq)+e-Ag(s)
                          108 g
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by=96487×1.45108C
= 1295.43 C

Step 2:
 
Given,
Current=1.5 A
Time=1295.431.5s
= 863.6 s
= 864 s
= 14.40 min

Step 3:
 
Again,
Cu2+(aq)+2e-Cu(s)
                              63.5 g
i.e., 2 x 96487 C of charge deposit=63.5 g of Cu
Therefore, 1295.43 C of charge will deposit=65.4×1295.432×96487g
= 0.439 g of Zn