Question

Two electrolytic cells containig $CuS{O}_4$ and$AgN{O}_3$ respectively are connected in series and a current is passed through them where $1mg$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is
[Atomic weight of copper and silver are 63.6 and 108 respectively.]

• A. $1.7mg$
• B. $3.5mg$
• C. $5.1mg$
• D. $6.8mg$

Answer 1

The correct option is B $3.5mg$

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

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

$E=\frac{\text{Atomic Mass}}{n-factor}$
$E_1=\frac{\text{63.5}}{2}\because C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$

$E_1=31.8$

$E_2=\frac{\text{108}}{1}\because A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$

$\frac{W_1}{W_2}=\frac{E_1}{E_2}$
Hence,
$\therefore \frac{1mg}{W_2}=\frac{31.8}{108}\Rightarrow W_2=\frac{108}{31.8}mg=3.4mg$