In the Kolbe electrolysis of sodium propanoate, the products $\mathrm{X}$ and $\mathrm{Y}$ are formed at…
- $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ at Cathode; $\mathrm{Y}=\mathrm{H}_2$ at Anode
- $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3$ at Cathode; $\mathrm{Y}=\mathrm{H}_2$ at Anode
- $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ at Anode; $\mathrm{Y}=\mathrm{H}_2$ at Cathode
- $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_3$ at Anode; $\mathrm{Y}=\mathrm{H}_2$ at Cathode
Solution

At cathode (reduction): $ \begin{aligned} & 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}+2 \mathrm{H}^{+} \\ & 2 \mathrm{H}^{+} \rightarrow \mathrm{H}_2(\mathrm{Y}) \\ & 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{NaOH} \end{aligned} $
Asked in: AP EAMCET 2023 (15 May Shift 1)