Three Electrolytic Cells A, b, c Containing Solutions of Znso4, Agn Exercise Chapter 3 Electrochemistry Chemistry

Question 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?

Answer: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