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In the following balanced reaction,

values of $X, Y$ and $Z$ respectively are
Options:

values of $X, Y$ and $Z$ respectively are
Solution:
1406 Upvotes
Verified Answer
The correct answer is:
2,5,16
2,5,16
$X \mathrm{MnO}_4^{-}+Y \mathrm{C}_2 \mathrm{O}_4^{-}+Z \mathrm{H}^{+} \leftrightharpoons$
$$
X \mathrm{Mn}^{++}+2 Y \mathrm{CO}_2+\frac{Z}{2} \mathrm{H}_2 \mathrm{O}
$$
First half reaction
$$
\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{++}
$$
On balancing
$$
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{++}+4 \mathrm{H}_2 \mathrm{O}
$$
Second half reaction
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2
$$
On balancing
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}
$$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$$
\begin{aligned}
& 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{--}+16 \mathrm{H}^{+} \longrightarrow \\
& 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
$$
X \mathrm{Mn}^{++}+2 Y \mathrm{CO}_2+\frac{Z}{2} \mathrm{H}_2 \mathrm{O}
$$
First half reaction
$$
\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{++}
$$
On balancing
$$
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{++}+4 \mathrm{H}_2 \mathrm{O}
$$
Second half reaction
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2
$$
On balancing
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}
$$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$$
\begin{aligned}
& 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{--}+16 \mathrm{H}^{+} \longrightarrow \\
& 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
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