Consider the given sequence of reactions.
\(\text{C}_2\text{H}_6 + \frac{3}{2}\text{O}_2 \xrightarrow[\Delta]{(\text{CH}_3\text{COO})_2\text{Mn}} \text{X} \xrightarrow{\text{Na}} \text{Y}\)
Electrolysis of aqueous solution of Y gives gases P and Q at anode. P and Q are respectively.
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Kolbe's Electrolysis of sodium salt of carboxylic acid (\(R\text{COONa}\)) always yields the alkane \(R-R\) and \(\text{CO}_2\) at the anode.
Step 1: Identify X:
The reaction of Ethane (\(\text{C}_2\text{H}_6\)) with oxygen in the presence of Manganese Acetate \([(\text{CH}_3\text{COO})_2\text{Mn}]\) catalyst is a method of controlled oxidation of alkanes.
\(2\text{C}_2\text{H}_6 + 3\text{O}_2 \xrightarrow{\text{Mn-acetate}} 2\text{CH}_3\text{COOH} + 2\text{H}_2\text{O}\).
So, X is Ethanoic Acid (Acetic Acid), \(\text{CH}_3\text{COOH}\).
Step 2: Identify Y:
Reaction of X (\(\text{CH}_3\text{COOH}\)) with Na metal:
\(2\text{CH}_3\text{COOH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2 \uparrow\).
So, Y is Sodium Acetate, \(\text{CH}_3\text{COONa}\).
Step 3: Electrolysis of Y (Kolbe's Electrolysis):
Electrolysis of aqueous sodium acetate:
At Anode (Oxidation): Acetate ions oxidize.
\(2\text{CH}_3\text{COO}^- \rightarrow 2\text{CH}_3\text{COO}^\bullet + 2e^-\)
\(2\text{CH}_3\text{COO}^\bullet \rightarrow 2\dot{\text{C}}\text{H}_3 + 2\text{CO}_2 \uparrow\)
\(\dot{\text{C}}\text{H}_3 + \dot{\text{C}}\text{H}_3 \rightarrow \text{C}_2\text{H}_6 \uparrow\) (Ethane).
So, gases produced at the anode are Ethane (\(\text{C}_2\text{H}_6\)) and Carbon Dioxide (\(\text{CO}_2\)).
Note: Hydrogen gas (\(\text{H}_2\)) is produced at the Cathode.
Final Answer: P and Q are \(\text{C}_2\text{H}_6\) and \(\text{CO}_2\).
