Question

A current of 1.5A was passed through two electrolytes arranged

in series for 10 minutes and 2.16 g of silver deposited on the cathode

of the first cell. (Atomic wt. of Ag=108 ,Ni =59) .Calculate

a) The quantity of electricity that passes through the two cells.

b) The amount of Nickel deposited on the cathode of the second cell

containing Nickel(II) nitrate, Ni(NO3)2 ?​

Answer 1

0.59 g

Step-by-step explanation:

The reaction at the first cathode;

Ag^+(aq) + e -----> Ag

if Q = It = 1.5 * 10 * 60 = 900C

According to Faraday's second law of electolysis; if we pass the same quantity of electricity through different electrolytes, the mass of substances deposited on each cathode is proportional to the equivalent weights.

Equivalent weight E = Atomic mass/valency

For Ag = 108/1 = 108

For Ni = 59/2 = 29.5

Hence

Let m1 = mass of Ag deposited =2.16

Let m2 = mass of Ni deposited

Let E1 = equivalent weight of Ag

Let E2 = equivalent weight of Ni

m1/m2 = E1/E2

2.16/m2 = 108/29.5

m2 = 2.16 * 29.5/108

m2 = 0.59 g