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Hydrocarbons

Section: Chemistry

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Question : 3 of 23

Marks: +1, -0

Consider the following

Statement-I : Kolbe's electrolysis of sodium propionate gives n-hexane as product

Statement-II : In Kolbe's process

\mathrm{CO}_2

is liberated at anode and

\mathrm{H}_2

is liberated at cathode

Correct answer is

[AP EAPCET 22 May 2025 S1]

Both statement-I and statement-II are correct
Both statement-I and statement-II are not correct
Statement-I is correct, but statement-II is not correct
Statement-I is not correct, but statement-II is correct

Solution:

Statement I is wrong, but Statement II is right.

Statement I should say: In Kolbe's electrolysis of sodium propionate, the product is butane, not n-hexane.

The chemical reaction is:

\;2\mathrm{CH}_3\mathrm{CH}_2\mathrm{COONa}+2\mathrm{H}_2\mathrm{O}\rightarrow

\;\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3+2\mathrm{CO}_2+2\mathrm{NaOH}+\mathrm{H}_2\uparrow

Butane

In Statement II, it is correct that in Kolbe's process, carbon dioxide (

\mathrm{CO}_2

) is released at the anode and hydrogen gas

(\mathrm{H}_2)

is released at the cathode.

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