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

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

Solution:

Ag has +1 charge in AgNO₃

So charge transfer, n = 1

F = 96487 Coulombs

Molar mass of Ag = 108 g

Use formula required charge  = nF

So required charge for 1 mol or 108 g of Ag = Coulombs.

 Charge required for 1.45 g of Ag                  = 96487 Coulombs × 1.45g/108g

                                                                = 1295.43 Coulombs

Given that current I = 1.5 A

Use formula Charge = I × t

Time t  = charge / I

           = 1295.43 Coulombs/ 1.5 A

          = 863.6 s

Divide by 60 to convert in minute

= 864/60 min

= 14.40 min

Charge in Cu in CuSo4 is +2

Use formula required charge for 1 mol = nF

So charge required for 1 mol or 63.5 g  of Cu = 2 ×  96487 = 192974 Coulombs

192974 Coulombs of charge deposit = 63.5 g of Cu

1295.43 Coulombs of charge will deposit = 63.5 g × 1295.43/192974

        = 0.426 g of Cu

Similarly for Zn

Zn has charge in ZnSO₄  = +2

Use formula required charge for 1 mol = nF

So charge required for 1 mol or 63.5 g  of Zn = 2 ×  96487 = 192974 Coulombs

192974 Coulombs of charge deposit = 63.5 g of Zn

2 × 96487 C of charge deposit         = 65.4 g of Zn

1295.43 Coulombs of charge will deposit = 65.4 g × 1295.43/192974

        = 0.439 g of Zn