Identify reductant in following reaction. $\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+}…
Identify reductant in following reaction.
$\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
$\mathrm{H}^{+}$
$\mathrm{H}_2 \mathrm{O}$
$\mathrm{C}_2 \mathrm{O}_4^{2-}$
$\mathrm{MnO}_4^{-}$
Solution
$\stackrel{+3}{\mathrm{C}_2} \mathrm{O}_4^{2-}+\stackrel{+7}{\mathrm{MnO}_4^{-}}+\mathrm{H}^{+} \longrightarrow \stackrel{+2}{\mathrm{Mn}^{2+}} \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
The oxidation number of $\mathrm{C}$ increase from $+3\left(\mathrm{C}_2 \mathrm{O}_4^{2-}\right)$ to $+4\left(\mathrm{CO}_2\right)$ in above reaction.
$\therefore\left(\mathrm{C}_2 \mathrm{O}_4^{2-}\right)$ acts as a reductant.