Question

2.68 ampere current is passed for 1 hour through the solutions of Hg2 (CIO4)2, CuSO4 and AgNO3
connected in series. How many moles of metal in each case will be deposited on cathode?

Answer 1

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