Question

Copper and silver voltameter are connected in series and an electric current is passed through them for sometime. It is observed that $14g$ of copper is deposited. Then, calculate the mass of silver deposited.
Given:
$\text{Electrochemical equivalent of copper}=3.2\times {10}^{-7}g{C}^{-1}$
$\text{Electrochemical equivalent of silver}=1.2\times {10}^{-6}g{C}^{-1}$

• A. $35.5g$
• B. $12.6g$
• C. $52.5g$
• D. $7.25g$

Answer 1

The correct option is C $52.5g$

If equal amount of charge (Q) is passed through two different solutions, the amount of substance deposited/liberated (W) is proportional to their chemical equivalent weights (E).

For copper cell,
By Faradays first law,
$W_1=Z_1\times Q_1$
where,
$W_1=\text{Amount of copper deposited}{Z}_1=\text{Electrochemical equivalent of copper}{Q}_1=\text{Charge passed through the copper cell}$

For Silver cell,
By Faradays first law,
$W_2=Z_2\times Q_2$
where,
$W_2=\text{Amount of silver deposited}{Z}_2=\text{Electrochemical equivalent of silver}{Q}_2=\text{Charge passed through the silver cell}$

Since, charge passed is same for both the cells,
$Q_1=Q_2$

So,

$\frac{W_1}{W_2}=\frac{Z_1}{Z_2}$

$Z_1=\frac{E_1}{96500}{Z}_2=\frac{E_2}{96500}$

$\therefore$

$\frac{W_1}{W_2}=\frac{E_1}{E_2}$
Also,
$\frac{W_1}{W_2}=\frac{E_1}{E_2}=\frac{Z_1}{Z_2}$

$E_1=\text{chemical equivalent weight of copper}{E}_2=\text{chemical equivalent weight of silver}$

So,

$\frac{W_1}{W_2}=\frac{Z_1}{Z_2}$

$\frac{14}{W_2}=\frac{3.2\times {10}^{-7}}{1.2\times {10}^{-6}}$

$W_2=52.5g$

$\text{Hence, amount of Ag deposited is 52.5 g}$