The products obtained at cathode and anode on electrolysis of aqueous sodium succinate are :

anode : ethene +

C O_{2}

, cathode :

H_{2}

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anode : ethyne +

C O_{2}

, cathode :

H_{2}

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anode : ethyne +

H_{2}

, cathode : Na

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anode : ethyne +

C O_{2}

, cathode : Na

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Solution

The correct option is A anode : ethene + $C O_{2}$ , cathode : $H_{2}$
At anode: ${C H}_{2} - C O O N a | ⟶ {C H}_{2} + 2 {C O}_{2} + {2 N a}^{+} + {2 e}^{-} {C H}_{2} C O O N a ∥ {C H}_{2}$

At cathode: ${2 H}^{+} + 2 e ⟶ H_{2}$

So, A option is correct.

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Aqueous $C u S O_{4}$ is electrolyzed using $A g$ as anode and $C u$ as cathode, then product of electrolysis at anode and cathode are, respectively

H_{2}

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Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

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