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

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(A) The balanced redox reaction is obtained by balancing the half-reactions:Reduction half: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Oxidati

Vedclass · Accepted Answer

$2$ - $5$ - $16$

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(A) The balanced redox reaction is obtained by balancing the half-reactions:Reduction half: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$Oxidation half: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$To balance the electrons,multiply the reduction half by $2$ and the oxidation half by $5$:$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$Adding these gives the balanced equation:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$The coefficients for the reactants $MnO_4^-$,$C_2O_4^{2-}$,and $H^+$ are $2$,$5$,and $16$ respectively.

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

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