Question

During the electrolysis of an aqueous solution of $NaCl,10.3g$ of chlorine was liberated at STP. Calculate the volume of hydrogen that would be liberated at ${20}^{o}C$ and $740mmofHg$ pressure from acidulated water, when the same quantity of current as that used in the electrolysis of $NaCl$ was passed.

• A. $5521cc$
• B. $3249cc$
• C. $3582cc$
• D. $2843cc$

Answer 1

The correct option is C $3582cc$

Faradays second law of electrolysis:
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).

Thus,
$\frac{W_1}{W_2}=\frac{E_1}{E_2}=\frac{Z_1}{Z_2}$
$\therefore$
$\frac{\text{Weight of}{H}_2\text{liberated}}{\text{Equivalent weight of}{H}_2}=\frac{\text{Weight of}C{l}_2\text{liberated}}{\text{Equivalent weight of}C{l}_2}$

$\frac{\text{Weight of}{H}_2\text{liberated}}{1}=\frac{10.3}{35.5}$

$\text{Weight of}{H}_2\text{liberated}=\frac{10.3}{35.5}g$

As $2g$ of $H_2$ occupies $22400cc$ at STP

So $\frac{10.3}{35.5}g$ of $H_2$ occupies,

$\Rightarrow \frac{22400\times 10.3}{2\times 35.5}=3249cc\text{at STP}$

The volume of $H_2$ at ${20}^{o}C$ and $740mmofHg$ will be from

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

At STP,
$P_1=760mmHg$
$T_1=273K$
$V_1=3249cc$

$P_2=740mmHg$
$T_2=293K$
$V_2=?$
$V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

$V_2=\frac{760\times 3249\times 293}{273\times 740}$

$V_2=3582cc$