Redox Reactions - Result Question 13
####13. For the redox reaction
[2018]
$MnO_4^{-}+C_2 O_4^{2-}+H^{+} \to Mn^{2+}+CO_2+H_2 O$
The correct coefficients of the reactants for the balanced equation are
\begin{tabular}{llll} & $MnO_4^{-}$ & $C_2 O_4^{2-}$ & $H^{+}$ \\
(a) & 16 & 5 & 2 \\
(b) & 2 & 5 & 16 \\
(c) & 5 & 16 & 2 \\
(d) & 2 & 16 & 5 \end{tabular}
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Solution:
- (b) $MnO_4^{-} \longrightarrow Mn^{2+} ; 5 e^{-}$gain
Multiplying (i) by 2 and (ii) by 5 to balance $e^{-}$
$ 2 MnO_4^{-}+5 C_2 O_4^{2-} \longrightarrow 2 Mn^{2+}+10 CO_2 $
On balancing charge;
$ \begin{aligned} & 2 MnO_4^{-}+5 C_2 O_4^{2-}+ 16 H^{+} \longrightarrow \\ & 2 Mn^{2+}+10 CO_2+8 H_2 O \end{aligned} $
