Explain the formation of alkanes by Kolbe's electrolysis method with an example. Alternatively,describe the preparation of alkanes from carboxylic acids via electrolysis.

(N/A) An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives an alkane containing an even number of carbon atoms.
Example: Electrolysis of sodium acetate yields ethane.
$2 CH_{3}COO^{-}Na^{+} + 2 H_{2}O \xrightarrow{\text{Electrolysis}} CH_{3}-CH_{3} + 2 CO_{2} + H_{2} + 2 NaOH$
Mechanism:
$a$. In aqueous solution,the salt dissociates:
$2 CH_{3}COO^{-}Na^{+} \rightleftharpoons 2 CH_{3}COO^{-}_{(aq)} + 2 Na^{+}_{(aq)}$
$b$. At the Anode (Oxidation):
$(i)$ $2 CH_{3}COO^{-} \rightarrow 2 CH_{3}COO^{\bullet} + 2 e^{-}$
$(ii)$ $2 CH_{3}COO^{\bullet} \rightarrow 2 \dot{C}H_{3} + 2 CO_{2} \uparrow$
$(iii)$ $\dot{C}H_{3} + \dot{C}H_{3} \rightarrow CH_{3}-CH_{3} \uparrow$ (Ethane)
$c$. At the Cathode (Reduction):
Water is reduced instead of $Na^{+}$ ions:
$2 H_{2}O + 2 e^{-} \rightarrow 2 OH^{-} + H_{2} \uparrow$
Overall reaction: $2 CH_{3}COO^{-}Na^{+} + 2 H_{2}O \rightarrow CH_{3}-CH_{3} + 2 CO_{2} + H_{2} + 2 NaOH$