Question

In the electrolysis of $H_{2}O$, $11.2$ litre of $H_2$ liberated at cathode at NTP. How much $O_2$ will be liberated at anode under the same conditions?

• A. $11.2$ litre
• B. $22.4$ litre
• C. $32$g
• D. $5.6$ litre

Answer 1

The correct option is D $5.6$ litre

$H_2{O}_L\to H_{2_{\left(g\right)}}+\frac{1}{2}{O}_{2_{\left(g\right)}}$

$\downarrow$ $\downarrow$ $\downarrow$

$1$ mole $1$ mole $1/2$ mole

As we know from $PV=nRT$

$P\to$ Pressure

$V\to$ Volume of gas

$n\to$ moles of gas

$R\to$ universal gas constant

$T\to$ Temperature

from $PV=nRT$

$V\propto n$

So,

$H_2{O}_L\to H_{2_{\left(g\right)}}+\frac{1}{2}{O}_{2_{\left(g\right)}}$

$Vo{l}^m\left(lt\right)\frac{Vo{l}^m}{2}\left(lt\right)$

Therefore, volume of $O_2$ litrated at anode

$=\frac{Vo{l}^m}{2}(\therefore Vo{l}^m=11.2lt)$

$=\frac{11.2}{2}=5.6lt$