1. Identify the electrode reactions

Potassium succinate gives succinate ion, which undergoes Kolbe electrolysis at the anode.

Succinate ion is: $^{-}OOC-CH_2-CH_2-COO^{-}$

On oxidation, it loses $2e^-$ overall and decomposes to give ethylene and carbon dioxide: $^{-}OOC-CH_2-CH_2-COO^{-} \rightarrow C_2H_4 + 2CO_2 + 2e^-$

So, at the anode, for every $2$ faradays passed:

• $1$ mol $C_2H_4$
• $2$ mol $CO_2$

Total anode gas = $3$ mol per $2F$

2. Cathode reaction

In aqueous solution, water is reduced: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

Thus, at the cathode, for every $2F$ passed:

• $1$ mol $H_2$

3. Total gas produced for 2F

For $2F$ electricity:

• Anode gas = $1 + 2 = 3$ mol
• Cathode gas = $1$ mol

Hence total gas = $3 + 1 = 4 \text{ mol}$

4. Gas produced for 0.2 F

Gas produced is proportional to charge passed.

For $2F \rightarrow 4$ mol gas

For $0.2F$: $\text{moles of gas} = 4 \times \frac{0.2}{2} = 0.4 \text{ mol}$

5. Convert moles to volume at STP

At STP, $1$ mol gas occupies $22.4$ L.

Therefore, $V = 0.4 \times 22.4 = 8.96 \text{ L}$

6. Check options

• A: $2.24$ L
• B: $4.48$ L
• C: $6.72$ L
• D: $8.96$ L

So the correct option is: $\boxed{D}$