Assertion (A): Sodium acetate on Kolbe's elec | vedclass

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(C) In Kolbe's electrolysis,sodium acetate $(CH_3COONa)$ undergoes electrolysis to produce ethane $(C_2H_6)$. The dissociation is: $CH_3COONa \rightle

Vedclass · Accepted Answer

$A$ is true,$R$ is false

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(C) In Kolbe's electrolysis,sodium acetate $(CH_3COONa)$ undergoes electrolysis to produce ethane $(C_2H_6)$. The dissociation is: $CH_3COONa \rightleftharpoons CH_3COO^{-} + Na^{+}$. At the anode,the acetate ion loses an electron to form an acetoxy radical,which then decarboxylates to form a methyl free radical $(CH_3^{\bullet})$. Two methyl radicals combine to form ethane. At the cathode,water is reduced to produce hydrogen gas $(H_2)$ and hydroxide ions $(OH^{-})$. Therefore,the methyl free radical is formed at the anode,not the cathode. Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

Assertion $(A):$ Sodium acetate on Kolbe's electrolysis gives ethane.
Reason $(R):$ Methyl free radical is formed at cathode.

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Assertion $(A):$ Sodium acetate on Kolbe's electrolysis gives ethane. Reason $(R):$ Methyl free radical is formed at cathode.