For the redox reaction MnO_4^{-} + C_2O_4^{2- | vedclass

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(A) To balance the redox reaction,we use the half-reaction method:$1$. Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$$2$. Reduction h

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$2, 5, 16$

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(A) To balance the redox reaction,we use the half-reaction method:$1$. Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$$2$. Reduction half-reaction: $MnO_4^{-} + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$$3$. To balance the electrons,multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$$2MnO_4^{-} + 16H^{+} + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$4$. Adding both half-reactions gives the balanced equation:$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Thus,the coefficients for $MnO_4^{-}$,$C_2O_4^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

For the redox reaction $MnO_4^{-} + C_2O_4^{2-} + H^{+} \rightarrow Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced reaction are respectively $MnO_4^{-}$,$C_2O_4^{2-}$ and $H^{+}$:

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