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(D) To balance the redox reaction,we use the ion-electron method.Step $1$: Write the half-reactions.Reduction: $MnO_4^- \rightarrow Mn^{2+}$Oxidation:

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$5$

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(D) To balance the redox reaction,we use the ion-electron method.Step $1$: Write the half-reactions.Reduction: $MnO_4^- \rightarrow Mn^{2+}$Oxidation: $C_2O_4^{2-} \rightarrow CO_2$Step $2$: Balance atoms and charges.Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$Step $3$: Equalize the number of electrons.Multiply reduction half-reaction by $2$ and oxidation half-reaction by $5$.$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$Step $4$: Add the half-reactions.$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$The stoichiometric coefficient of the oxalate ion $(C_2O_4^{2-})$ is $5$.

What is the stoichiometric coefficient of the oxalate ion after balancing the following redox reaction?
$MnO_4^- + C_2O_4^{2-} + H^+ \rightarrow Mn^{2+} + CO_2 + H_2O$

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What is the stoichiometric coefficient of the oxalate ion after balancing the following redox reaction?$MnO_4^- + C_2O_4^{2-} + H^+ \rightarrow Mn^{2+} + CO_2 + H_2O$