Chemistry - Electrochemistry Question with Solution | TestHub

At cathode :

$${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to {\mathrm{OH}}^{-}+\frac{1}{2}{\mathrm{H}}_2\uparrow \text{ since }{\mathrm{E}}_{\mathrm{red}}^{\mathrm{O}}\left({\mathrm{H}}_2\mathrm{O}\right)>{\mathrm{E}}_{\mathrm{red}}^{\mathrm{O}}\left({\mathrm{Na}}^{+}\right)$$

At anode :

At cathode :

$1e^{-}=1F=\frac{1}{2}mol{H}_2=\frac{22.4}{2}=11.2L{H}_2$

At anode :

$\begin{array}{rlrlr}2e^{-}& =2F=\left(2molC{O}_2+1mol{C}_2{H}_6\right)& & & \end{array}$

$2F=3molgases$

$1F=\frac{3}{2}molgases=\frac{3}{2}\times 22.4=36.6L$

Total volume of gases $=11.2+36.6=44.8\mathrm{L}$