Animated Solution for Chemistry - Redox Reactions: In dilute aqueous ${\text{H}}_2{\text{SO}}_4$, the complex diaquodioxalatoferrate(II) is oxidized by ${\text{MnO}}_4^{-}$. For this reaction, the ratio of the rate of change of $[{\text{H}}^{+}]$ to the rate of change of $[{\text{MnO}}_4^{-}]$ is -

Permanganate vs. The Complex

When you see a redox problem in JEE Advanced, you should immediately be on high alert for hidden traps. In this problem, we are asked to find the ratio of the rate of change of ${\text{H}}^{+}$ to ${\text{MnO}}_4^{-}$ during the oxidation of a bulky coordination complex: diaquodioxalatoferrate(II).

Let's decode the complex first. The name tells us we have a central iron atom in the $+2$ oxidation state, surrounded by two water molecules (diaquo) and two oxalate ions (dioxalato). The formula is $[\text{Fe}({\text{H}}_2\text{O}{)}_2({\text{C}}_2{\text{O}}_4{)}_2{]}^{2-}$.

The oxidizing agent is our old friend, the permanganate ion (${\text{MnO}}_4^{-}$). In an acidic medium (provided by the dilute ${\text{H}}_2{\text{SO}}_4$), permanganate undergoes a classic 5-electron reduction. The manganese atom drops from a $+7$ oxidation state down to $+2$:

This half-reaction is straightforward and should be second nature to you. The real magic happens with the reducing agent.

The Hidden Trap

Here is where many students make a fatal error. They see the ${\text{Fe}}^{2+}$ and immediately write down the oxidation of iron to ${\text{Fe}}^{3+}$, accounting for a loss of $1e^{-}$. But look closely at the ligands!

The complex contains two oxalate ions (${\text{C}}_2{\text{O}}_4^{2-}$). Oxalate is a well-known reducing agent that readily oxidizes to carbon dioxide (${\text{CO}}_2$). Each oxalate ion loses $2e^{-}$ in the process. Since our complex has two oxalate ligands, they will collectively lose $4e^{-}$.

Therefore, the total oxidation of the complex involves both the central metal ion and the ligands:

- ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+1e^{-}$ - $2{\text{C}}_2{\text{O}}_4^{2-}\to 4{\text{CO}}_2+4e^{-}$

The complete oxidation half-reaction is:

Notice the beautiful symmetry here? The complex loses exactly $5e^{-}$, and the permanganate ion requires exactly $5e^{-}$. This means they react in a perfect $1:1$ molar ratio. We can simply add the two half-reactions together to get the overall balanced chemical equation:

The final step bridges stoichiometry with chemical kinetics. The question asks for the ratio of the rate of change of $[{\text{H}}^{+}]$ to the rate of change of $[{\text{MnO}}_4^{-}]$.

Recall that the rate of disappearance of a reactant is directly proportional to its stoichiometric coefficient in the balanced equation. For a general reaction $aA+bB\to \text{Products}$, the rate is defined as:

Applying this to our balanced equation:

Rearranging this to find the ratio gives:

The ratio is exactly 8. This problem is a brilliant reminder to always inspect every part of a molecule—both the metal and the ligands—when balancing redox reactions!