Consider the following:Statement-I : Kolbe's | vedclass

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(D) In Kolbe's electrolysis,the reaction is $2CH_3CH_2COONa + 2H_2O \xrightarrow{	ext{electrolysis}} CH_3CH_2CH_2CH_2CH_3 + 2CO_2 + H_2 + 2NaOH$.Stat

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Statement-$I$ is not correct,but statement-$II$ is correct.

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(D) In Kolbe's electrolysis,the reaction is $2CH_3CH_2COONa + 2H_2O \xrightarrow{	ext{electrolysis}} CH_3CH_2CH_2CH_2CH_3 + 2CO_2 + H_2 + 2NaOH$.Statement-$I$: Sodium propionate $(CH_3CH_2COONa)$ undergoes electrolysis to form $n$-butane $(CH_3CH_2CH_2CH_3)$,not $n$-hexane. Thus,statement-$I$ is incorrect.Statement-$II$: During the electrolysis of sodium salts of carboxylic acids,$CO_2$ is evolved at the anode and $H_2$ is evolved at the cathode. Thus,statement-$II$ is correct.

Consider the following:
Statement-$I$ : Kolbe's electrolysis of sodium propionate gives $n$-hexane as product.
Statement-$II$ : In Kolbe's process,$CO_2$ is liberated at the anode and $H_2$ is liberated at the cathode.
The correct answer is:

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Consider the following:Statement-$I$ : Kolbe's electrolysis of sodium propionate gives $n$-hexane as product.Statement-$II$ : In Kolbe's process,$CO_2$ is liberated at the anode and $H_2$ is liberated at the cathode.The correct answer is: