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

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Solution

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,
$\frac{W_{1}}{W_{2}} = \frac{E_{1}}{E_{2}} = \frac{Z_{1}}{Z_{2}}$
$∴$
$\frac{Weight of H_{2} liberated}{Equivalent weight of H_{2}} = \frac{Weight of C l_{2} liberated}{Equivalent weight of C l_{2}}$

$\frac{Weight of H_{2} liberated}{1} = \frac{10.3}{35.5 \times 2}$

$Weight of H_{2} liberated = \frac{10.3}{35.5} g$

As $2 g$ of $H_{2}$ occupies $22400 c c$ at STP

So $\frac{10.3}{35.5} g$ of $H_{2}$ occupies,

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

The volume of $H_{2}$ at $20^{o} C$ and $740 m m o f H g$ will be from

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

At STP,
$P_{1} = 760 m m H g$
$T_{1} = 273 K$
$V_{1} = 3249 c c$

$P_{2} = 740 m m H g$
$T_{2} = 293 K$
$V_{2} = ?$
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$

$V_{2} = \frac{760 \times 3249 \times 293}{273 \times 740}$

$V_{2} = 3582 c c$

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