Question

A current of $i$ ampere was passed for $t$ sec, through three calls $P,Q$ and $T$ connected in series. These contain respectively silver nitrate, mercuric nitrate, and mercurous nitrate. At the cathode of the cell $P,0.216g$ of $Ag$ was deposited. The weights of mercury deposited in the cathode of $Q$ and $R$ respectively are: (at wt. of $Hg=200.59$)

• A. $0.4012$ and $0.8024g$
• B. $0.4012$ and $0.2006g$
• C. $0.2006$ and $0.4012g$
• D. $0.1003$ and $0.2006g$

Answer 1

Answer: (C)

$0.2006$ and $0.4012g$

Cells are connected in series therefore same amount of current will Pass through them:

Faraday's second law:

$\frac{W_1}{W_2}=\frac{E_1}{E_2}$

(a) For Mercuric nitrate $\left(H{g}^{2+}\right)$

$\frac{W_1}{\left(W_2\right)Ag}=\frac{E_1}{\left(E_2\right)Ag}$

$W_1=\frac{200.59}{2\times 108}\times 0.216$

$=0.2006g$

(b) For mercurous nitrate $\left(H{g}^{+}\right)$

$\frac{W_1}{\left(W_2\right)Ag}=\frac{E_1}{\left(E_2\right)Ag}$ $(\therefore E_{Ag}=108)$

$W_1=\frac{200.59}{108}\times 0.216$

$=0.4012g$