Animated Solution for Chemistry - Redox Reactions: In order to oxidise a mixture of one mole of each of ${\text{FeC}}_2{\text{O}}_4$, ${\text{Fe}}_2({\text{C}}_2{\text{O}}_4{)}_3$, ${\text{FeSO}}_4$ and ${\text{Fe}}_2({\text{SO}}_4{)}_3$ in acidic medium, the number of moles of ${\text{KMnO}}_4$ required is

Imagine a beaker containing a complex mixture of four different iron compounds: ${\text{FeC}}_2{\text{O}}_4$, ${\text{Fe}}_2({\text{C}}_2{\text{O}}_4{)}_3$, ${\text{FeSO}}_4$, and ${\text{Fe}}_2({\text{SO}}_4{)}_3$. We have exactly one mole of each. We are tasked with completely oxidizing this entire mixture using potassium permanganate (${\text{KMnO}}_4$) in an acidic medium.

At first glance, writing and balancing four separate redox reactions seems like a nightmare. But we don't have to! The Law of Chemical Equivalence is our ultimate shortcut. It states that in any redox reaction, the total equivalents of the oxidizing agent must perfectly match the total equivalents of the reducing agents.

Mathematically, this is expressed as:

Let's start with our oxidizing agent, ${\text{KMnO}}_4$. The medium is crucial here. In an acidic medium, the permanganate ion (${\text{MnO}}_4^{-}$) undergoes a massive transformation. The manganese atom drops from a $+7$ oxidation state all the way down to a $+2$ state.

Because it gains 5 electrons per molecule, the n-factor for ${\text{KMnO}}_4$ in an acidic medium is exactly $5$.

Now, we must carefully evaluate each compound in our mixture to determine its n-factor. The n-factor for a salt acting as a reducing agent is the total number of electrons lost per molecule.

1. Ferrous Oxalate (${\text{FeC}}_2{\text{O}}_4$) Here, both the cation and the anion are eager to oxidize. The ${\text{Fe}}^{2+}$ ion oxidizes to ${\text{Fe}}^{3+}$ (losing $1$ electron). Simultaneously, the oxalate ion (${\text{C}}_2{\text{O}}_4^{2-}$) oxidizes to two molecules of ${\text{CO}}_2$ (losing $2$ electrons). Therefore, the total n-factor is $1+2=3$. Since we have $1$ mole, it contributes $3$ equivalents.

2. Ferric Oxalate (${\text{Fe}}_2({\text{C}}_2{\text{O}}_4{)}_3$) Notice the iron here is ${\text{Fe}}^{3+}$. It is already in its highest stable oxidation state, so it will not oxidize further. However, the compound contains three oxalate ions. Each oxalate ion loses $2$ electrons, so three of them will lose a total of $6$ electrons. Therefore, the n-factor is $6$. It contributes $6$ equivalents.

3. Ferrous Sulfate (${\text{FeSO}}_4$) The sulfate ion (${\text{SO}}_4^{2-}$) is fully oxidized and inert here. Only the ${\text{Fe}}^{2+}$ ion oxidizes to ${\text{Fe}}^{3+}$, losing $1$ electron. Therefore, the n-factor is $1$. It contributes $1$ equivalent.

4. Ferric Sulfate (${\text{Fe}}_2({\text{SO}}_4{)}_3$) This is the catch! Both the ${\text{Fe}}^{3+}$ ion and the ${\text{SO}}_4^{2-}$ ion are already in their maximum oxidation states. Permanganate cannot extract any more electrons from them. Therefore, no oxidation occurs, the n-factor is $0$, and it contributes $0$ equivalents.

Now, we sum up the total reducing equivalents from our mixture:

According to the Law of Equivalence, the equivalents of ${\text{KMnO}}_4$ must also be $10$. Let $n$ be the required moles of ${\text{KMnO}}_4$.

It takes exactly 2 moles of potassium permanganate to completely oxidize the mixture. The elegance of the equivalence concept turns a complex multi-reaction problem into a simple addition and division exercise!