Describe the Kolbe electrolysis reaction of $C_{2}H_{5}COO^{-}Na^{+}$ at the anode and cathode.

(N/A) In an aqueous solution of $C_{2}H_{5}COO^{-}Na^{+}$,the salt dissociates into $C_{2}H_{5}COO^{-}$ and $Na^{+}$ ions.
At the Anode (Oxidation):
The propanoate ion $(C_{2}H_{5}COO^{-})$ migrates to the anode and undergoes oxidation by losing an electron to form an ethyl free radical and carbon dioxide gas.
$2C_{2}H_{5}COO^{-} \rightarrow 2C_{2}H_{5}^{\bullet} + 2CO_{2} + 2e^{-}$
The two ethyl free radicals then combine to form butane:
$2C_{2}H_{5}^{\bullet} \rightarrow C_{4}H_{10}$ (Butane)
Overall reaction at the anode: $2C_{2}H_{5}COO^{-} \rightarrow C_{4}H_{10} + 2CO_{2} + 2e^{-}$
At the Cathode (Reduction):
Water molecules are reduced at the cathode to produce hydrogen gas and hydroxide ions:
$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$