In the Kolbe electrolysis of sodium propanoat | vedclass

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Question

(C) The overall reaction for the Kolbe electrolysis of sodium propanoate is:$2 CH_3 CH_2 COO^{-} Na^{+} + 2 H_2 O \xrightarrow{	ext{electrolysis}} CH

Vedclass · Accepted Answer

$X=CH_3-CH_2-CH_2-CH_3$ at Anode; $Y=H_2$ at Cathode

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(C) The overall reaction for the Kolbe electrolysis of sodium propanoate is:$2 CH_3 CH_2 COO^{-} Na^{+} + 2 H_2 O \xrightarrow{	ext{electrolysis}} CH_3 CH_2-CH_2 CH_3 + 2 CO_2 + H_2 + 2 NaOH$At the anode (oxidation),the propanoate ions undergo decarboxylation to form butane $(X)$:$2 CH_3 CH_2 COO^{-} ightarrow CH_3 CH_2-CH_2 CH_3 + 2 CO_2 + 2 e^-$At the cathode (reduction),water is reduced to form hydrogen gas $(Y)$:$2 H_2 O + 2 e^- ightarrow 2 OH^{-} + H_2(Y)$Thus,$X$ is butane formed at the anode,and $Y$ is hydrogen gas formed at the cathode.

In the Kolbe electrolysis of sodium propanoate,the products $X$ and $Y$ are formed at respective electrodes. What are $X$ and $Y$?

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