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Step-by-Step Solution
Step 1: Overall Reaction
During the electrolysis of potassium succinate solution, the succinate ion undergoes oxidation at the anode, and water is reduced at the cathode. The overall simplified reaction can be written as:
$2 CH_{3}COOK + 2 H_{2}O \xrightarrow[\text{Electrolysis}]{} CH_{3} - CH_{3} + 2 CO_{2} + H_{2} + 2 KOH$
Step 2: Anode (Oxidation) Reaction
At the anode, the succinate ion is oxidized to ethane (in the reaction scheme given). The process can be represented as:
Step 3: Cathode (Reduction) Reaction
Water is reduced at the cathode to produce hydrogen gas:
$2 H_{2}O + 2 e^{-} \longrightarrow 2 OH^{-} + 2H^{\bullet}$
$2H^{\bullet} \longrightarrow H_{2}$
Step 4: Calculation of Moles of Gaseous Products
From 0.2 Faraday of charge passed, the moles of gaseous products formed (ethane, carbon dioxide, and hydrogen) are determined by the respective electron requirements. A simplified approach gives:
• Moles of C2H6 produced = $ \frac{0.2}{2} = 0.1$
• Moles of CO2 produced = $ \frac{0.2}{1} = 0.2$
• Moles of H2 produced = $ \frac{0.2}{2} = 0.1$
Therefore, total moles of gases, $n = 0.1 + 0.2 + 0.1 = 0.4$.
Step 5: Volume of Gases at STP
Using the ideal gas equation $V = \frac{nRT}{p}$ at STP ($T = 273\,K$ and $p = 1\,\text{atm}$, $R = 0.0821\,L\,atm\,mol^{-1}K^{-1}$):
$V = 0.4 \times 0.0821 \times 273 \approx 8.96 \text{ L}$
Hence, the total volume of the gases produced at STP is 8.96 L.
