MnO_4^- + C_2O_4^{2-} + H^+ to Mn^{2+} + CO_2 | vedclass

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(A) Let us balance the reaction using the oxidation number method.$Step (i)$: Write the unbalanced equation:$MnO_4^- + C_2O_4^{2-} + H^+ 	o Mn^{2+} +

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$2, 5, 16$

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(A) Let us balance the reaction using the oxidation number method.$Step (i)$: Write the unbalanced equation:$MnO_4^- + C_2O_4^{2-} + H^+ 	o Mn^{2+} + CO_2 + H_2O$$Step (ii)$: Identify oxidation number changes:$Mn$ changes from $+7$ to $+2$ (decrease of $5$ per atom).$C$ changes from $+3$ to $+4$ (increase of $1$ per atom).$Step (iii)$: Equalize the change in oxidation number:Total increase in $C = 2 	imes 1 = 2$.Total decrease in $Mn = 1 	imes 5 = 5$.To balance,multiply $MnO_4^-$ by $2$ and $C_2O_4^{2-}$ by $5$:$2MnO_4^- + 5C_2O_4^{2-} + H^+ 	o 2Mn^{2+} + 10CO_2 + H_2O$$Step (iv)$: Balance $H$ and $O$ atoms:There are $16$ oxygen atoms on the left (excluding $H_2O$) and $20$ on the right. Balancing $H^+$ and $H_2O$ gives:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ 	o 2Mn^{2+} + 10CO_2 + 8H_2O$The coefficients for $MnO_4^-$,$C_2O_4^{2-}$,and $H^+$ are $2, 5, 16$ respectively.

$MnO_4^- + C_2O_4^{2-} + H^+ \to Mn^{2+} + CO_2 + H_2O$ :
What are the correct stoichiometric coefficients for the balanced redox reaction?

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$MnO_4^- + C_2O_4^{2-} + H^+ \to Mn^{2+} + CO_2 + H_2O$ :What are the correct stoichiometric coefficients for the balanced redox reaction?