Question

When during electrolysis of a solution of $AgN{{O}_{3}}$ , 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on cathode will be:
A. 10.8 g
B. 21.6
C. 108 g
D. 1.08 g

Answer 1

Hint: The relation between the number of moles of electrons and the charge is as follows.
Number of moles of electrons = $\dfrac{Q}{F}$
\[\text{Number of moles of electrons =}\dfrac{\text{Q}}{\text{F}}\]
Here Q = charge
F = Faraday

Complete answer:
- In the question it is asked to calculate the mass of the silver deposited at cathode when 6950 coulombs of charge which is passed through the electroplating bath contains silver nitrate solution.
- Initially we have to calculate the number of moles of electrons generated when 9650 coulombs charge is passed and it is as follows.
Number of moles of electrons = $\dfrac{Q}{F}$
Here Q = charge = 9650
F = Faraday = 96500
\[\text{Number of moles of electrons =}\dfrac{\text{Q}}{\text{F}}=\dfrac{9650}{96500}=0.1mole\]
- We know that one mole of silver nitrate is going to produce one mole of monovalent silver ion when silver nitrate is dissociated.
- Therefore, the number of electrons produced will be 1 mole.
- So, 0.1 moles of silver are going to be produced by 0.1 moles of silver nitrate.
- Therefore, mass of the silver = (number of moles) (molar mass) = (0.1) (108) g.
- So, the silver which is deposited as per the data given in the question is 10.8 g.

Thus, the correct option is A.

Note:
We should know the number of moles of electrons generated when 9650 coulombs of charge pass through the bath. Later we should know how many silver ions are going to be produced when one mole of silver nitrate is going to be dissociated.