A given quantity of electricity was passed through electrolytic cells connected in series containing solution of Silver trioxonitrate v, Copper ii, tetraoxosulphate vi and Sodium Chloride respectively. If
10.5g of Copper are deposited in the second electrolytic cell. Calculate (a)the mass of silver deposited in the first
cell (b)the volume of Chlorine liberated in the third cell at 18°C and 740mmHg pressure [Ag= 108g/mol, Cu =
63.6,/mol, IF-95600C, Molar volume of gas S. T.P =22.4dm

a)35.8 g

b) 24.5 L

Explanation:

Now we have to remember Faraday's second law of electrolysis; the mass of substances deposited during electrolysis when the same quantity of electricity is passed through them is directly proportional to the chemical equivalents.

E1/E2 =m1/m2

The chemical equivalent is mass/valency

E1= chemical equivalent of silver= 108/1 =108

E2= chemical equivalent of Cu= 63.6/2= 31.8

m1= mass of Ag =x

m2= mass of Cu= 10.5g

108/31.8 = x/10.5

108 × 10.5 = 31.8 × x

x= 108.5 × 10.5/31.8

x= 35.8g

ii) 2Cl^-(aq) + 2e ---> Cl2(g)

Given that;

V1= 22.4 L

T1= 273 K

P1= 760 mmHg

T2 = 18 + 273= 291 K

P2= 740mmHg

V2?

P1V1/T1 = P2V2/T2

V2= P1 V1 T2/P2T1

V2= 760 × 22.4 × 291/740 × 273

V2 = 4953984/202020

V2= 24.5 L