What is the value of \(x\) and \(y\) in order to balance following redox reaction? \(\mathrm{xCuO}+\mathrm{yNH}_3 \longrightarrow \mathrm{xCu}+\mathrm{N}_2+\mathrm{xH}_2 \mathrm{O}\)

Reaction: \(x \mathrm{CuO}+y \mathrm{NH}_3 \rightarrow x \mathrm{Cu}+\mathrm{N}_2+x \mathrm{H}_2 \mathrm{O}\).
Step 1: Assign oxidation states:
- Cu in CuO: +2 , reduced to 0 in Cu .
- N in \(\mathrm{NH}_3:-3\), oxidized to 0 in \(\mathrm{N}_2\).
Step 2: Write half-reactions:
1. Reduction: \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}\),
- \(x \mathrm{CuO}\) contributes \(2 x\) electrons.
2. Oxidation: \(2 \mathrm{NH}_3 \rightarrow \mathrm{~N}_2+6 e^{-}\),
- Each \(y \mathrm{NH}_3\) contributes \(3 y\) electrons.
Step 3: Equate total electrons:
\(2 x=3 y \quad \Rightarrow \quad x=\frac{3 y}{2} .\)
Step 4: Trial values to balance: Let \(x=2: y=3\).
Balanced Equation:
\(2 \mathrm{CuO}+3 \mathrm{NH}_3 \rightarrow 2 \mathrm{Cu}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}\)
Answer: \(x=2, y=3\), Option 3.