In the following balanced reaction, $\mathrm{XMnO}_{4}^{-}+\mathrm{Y} \mathrm{C}_{2}…
- $2,5,16$
- $8,2,5$
- $5,2,16$
- $5,8,4$
Solution
First half reaction $\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{+2} \ldots$(i)
On balancing
$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}$...(ii)
Second half reaction
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 2 \mathrm{CO}_{2}$...(iii)
On balancing
$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{e}^{-}$...(iv)
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{+2}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}$
Asked in: JEE-TOPICTESTS-CHEMISTRY