Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the volume of gases (at both cathode and anode) at STP (1atm and 273 K) is:

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6.72 L

8.96 L

4.48 L

2.24 L

answer is A.

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Detailed Solution

The reactions occurring are:

Anode

Cathode $2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 {OH}^{-} (aq)$

On passing 0.2 F of electricity, 0.1mol of $C_{2} H_{4}, 0.2 mol$ of ${CO}_{2}$ and 0.1mol of $H_{2}$ will be released. The volume of gases will be $(22.4 L {mol}^{- 1}) (0.4 mol) = 8.96 L$

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