In the following balanced reaction, values of $X, Y$ and $Z$ respectively are
In the following balanced reaction,
values of $X, Y$ and $Z$ respectively are
-
2,5,16
-
$8,2,5$
-
$5,2,16$
-
$5,8,4$
Solution
$X \mathrm{MnO}_4^{-}+Y \mathrm{C}_2 \mathrm{O}_4^{-}+Z \mathrm{H}^{+} \leftrightharpoons$
$
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}
$
Asked in: JEE Main 2012 (12 May Online)
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