For the redox reactionMnO_{4}^{-} + C_{2}O_{4 | vedclass

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(B) The reduction half-reaction is: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_{2}O$ ... $(I)$The oxidation half-reaction is: $C_{2}O

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$2 \quad 5 \quad 16$

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(B) The reduction half-reaction is: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_{2}O$ ... $(I)$The oxidation half-reaction is: $C_{2}O_{4}^{2-} \longrightarrow 2CO_{2} + 2e^{-}$ ... $(II)$To balance the electrons,multiply $(I)$ by $2$ and $(II)$ by $5$:$2MnO_{4}^{-} + 16H^{+} + 10e^{-} \longrightarrow 2Mn^{2+} + 8H_{2}O$$5C_{2}O_{4}^{2-} \longrightarrow 10CO_{2} + 10e^{-}$Adding these two equations gives the balanced redox reaction:$2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+} \longrightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$Thus,the coefficients for $MnO_{4}^{-}$,$C_{2}O_{4}^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

For the redox reaction
$MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+} \longrightarrow Mn^{2+} + CO_{2} + H_{2}O$
the correct coefficients of the reactants for the balanced equation are
$MnO_{4}^{-} \quad C_{2}O_{4}^{2-} \quad H^{+}$

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For the redox reaction$MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+} \longrightarrow Mn^{2+} + CO_{2} + H_{2}O$the correct coefficients of the reactants for the balanced equation are$MnO_{4}^{-} \quad C_{2}O_{4}^{2-} \quad H^{+}$