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

Options

A. `(8.96)`
B. `(4.48)`
C. `(6.72L)`
D. `2.24L`

Explanation

To solve the problem of determining the total volume of gases liberated when 0.2 Faraday of electricity is passed through an aqueous solution of potassium succinate, we can follow these steps:

### Step 1: Understand the Reaction
The oxidation of the succinate ion (C4H4O4^2-) produces ethylene (C2H4) and carbon dioxide (CO2) gases. The balanced reaction can be represented as:
\[ \text{C}_4\text{H}_4\text{O}_4^{2-} + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_4 + 2\text{CO}_2 + 2\text{H}^+ + 2\text{OH}^- \]

### Step 2: Calculate Moles of Electrons
According to Faraday's law, the amount of substance produced at an electrode is proportional to the charge passed. Given that 0.2 Faraday of electricity is passed, we can calculate the moles of electrons:
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