In the Kolbe electrolysis of sodium propanoate, the products $\mathrm{X}$ and $\mathrm{Y}$ are formed at respected electrodes. What are $\mathrm{X}$ and $\mathrm{Y}$ ?

• A

  $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ at Cathode; $\mathrm{Y}=\mathrm{H}_2$ at Anode
• B

  $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3$ at Cathode; $\mathrm{Y}=\mathrm{H}_2$ at Anode
• Correct answer
  C

  $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ at Anode; $\mathrm{Y}=\mathrm{H}_2$ at Cathode
• D

  $\mathrm{X}=\mathrm{CH}_3-\mathrm{CH}_3$ at Anode; $\mathrm{Y}=\mathrm{H}_2$ at Cathode

Solution

$2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CO} \overline{\mathrm{O}^2} \mathrm{Na} \xrightarrow[2 \mathrm{H}_2 \mathrm{O}]{\text { electrolysis }}$ $$ \mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{CH}_3+2 \mathrm{CO}_2+\mathrm{H}_2+2 \mathrm{NaOH} $$ At anode (oxidation):
At cathode (reduction): $$ \begin{aligned} & 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}+2 \mathrm{H}^{+} \\ & 2 \mathrm{H}^{+} \rightarrow \mathrm{H}_2(\mathrm{Y}) \\ & 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{NaOH} \end{aligned} $$

Hence, the correct answer is option: 'C'.