For the redox reaction:\ $\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \l

Question

For the redox reaction:\ $\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$\ the correct coefficients of the reactants for the balanced reaction are

Professor PC Thomas Classes · Answer

(a) : To get the balanced reaction, we write two half reactions.\ (i) $\mathrm{C}_{2} \mathrm{O}_{4}^{2-} ightarrow 2 \mathrm{CO}_{2}+2 e^{-}$(Oxidation)\ (ii) $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} ightarrow 2 \mathrm{Mn}^{2+}$ $+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}$ (Reduction) \To get balanced net reaction multiply (i) by 5 and (ii) by 2 and add, we get\ $2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} ightarrow 2 \mathrm{Mn}^{2+}$ $+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}$\ From this balanced equation we get various coefficients as 2,5 and 16 for $\mathrm{MnO}_{4}^{-}, \mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{H}^{+}$respectively.\