Question

When during electrolysis of a solution of $Ag{NO}_3$, 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on cathode will be:

• A. 1.08g
• B. 10.8g
• C. 21.6g
• D. 108g

Answer 1

Answer: (B)

10.8g

Here, no. of moles of electrons = $\frac{Q}{F}=\frac{9650}{96500}=0.1mole$

Now,

1 mole of $AgN{O}_3$ will produce one mole of monovalent silver ion on dissociation.

Therefore,

The number of electrons involved will also be one mole for one mole of $AgN{O}_3$.

Thus, 0.1 moles of silver will be produced by 0.1 mole of $AgN{O}_3$ .

Since,

Silver nitrate dissociates completely into $A{g}^{+}$ and $N{O}_3^{-}$ ion.

Hence,

Mass of silver = no. of moles x molar mass= 0.1 x 108 = 10.8 g

Hence,option B is correct answer.