Ratio of Rate of Change of Proton to Permanganate in Diaquodioxalatoferrate Oxidation

To find the ratio of the rate of change of $[\text{H}^+]$ to the rate of change of $[\text{MnO}_4^-]$, we first write and balance the redox reaction occurring in dilute acidic medium ($\text{H}_2\text{SO}_4$).

1. Oxidation Half-Reaction

The diaquodioxalatoferrate(II) complex has the formula: $[\text{Fe}(\text{C}_2\text{O}_4)_2(\text{H}_2\text{O})_2]^{2-}$

In this complex ion:

• The iron center is in the $+2$ oxidation state and is oxidized to $\text{Fe}^{3+}$: $\text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + e^-$
• The two oxalate ligands ($\text{C}_2\text{O}_4^{2-}$) are oxidized to carbon dioxide ($\text{CO}_2$): $2\text{C}_2\text{O}_4^{2-} \longrightarrow 4\text{CO}_2 + 4e^-$

Combining these processes, the total number of electrons lost per complex ion is: $n = 1 + 4 = 5\,e^-$

The balanced oxidation half-reaction is: $[\text{Fe}(\text{C}_2\text{O}_4)_2(\text{H}_2\text{O})_2]^{2-} \longrightarrow \text{Fe}^{3+} + 4\text{CO}_2 + 2\text{H}_2\text{O} + 5e^-$

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2. Reduction Half-Reaction

In an acidic medium, the permanganate ion ($\text{MnO}_4^-$) is reduced to manganese(II) ($\text{Mn}^{2+}$): $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

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3. Overall Balanced Redox Reaction

Adding the two half-reactions together (since both involve $5e^-$): $[\text{Fe}(\text{C}_2\text{O}_4)_2(\text{H}_2\text{O})_2]^{2-} + \text{MnO}_4^- + 8\text{H}^+ \longrightarrow \text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{CO}_2 + 6\text{H}_2\text{O}$

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4. Ratio of the Rates of Change

From the stoichiometry of the overall balanced chemical equation, the rate of reaction is defined as: $\text{Rate} = -\frac{1}{8}\frac{d[\text{H}^+]}{dt} = -\frac{1}{1}\frac{d[\text{MnO}_4^-]}{dt}$

Thus, the ratio of the rate of change (disappearance) of $[\text{H}^+]$ to that of $[\text{MnO}_4^-]$ is: $\frac{-\dfrac{d[\text{H}^+]}{dt}}{-\dfrac{d[\text{MnO}_4^-]}{dt}} = \frac{8}{1} = 8$