Animated Solution for Chemistry - Redox Reactions: In the chemical reaction between stoichiometric quantities of ${\text{KMnO}}_4$ and $\text{KI}$ in weakly basic solution, what is the number of moles of ${\text{I}}_2$ released for 4 moles of ${\text{KMnO}}_4$ consumed?

Let's dive into this fascinating redox reaction. We are dealing with the reaction between potassium permanganate (${\text{KMnO}}_4$) and potassium iodide ($\text{KI}$) in a weakly basic solution.

Before we jump into the math, there is a crucial conceptual catch here. In a strictly chemical sense, when ${\text{KMnO}}_4$ reacts with $\text{KI}$ in a faintly alkaline or neutral medium, the iodide ion (${\text{I}}^{-}$) is typically oxidized to iodate (${\text{IO}}_3^{-}$). However, the question explicitly asks for the number of moles of ${\text{I}}_2$ released. In competitive exams like JEE Advanced, you must always follow the explicit constraints and assumptions provided in the question stem. Therefore, we will proceed by assuming that ${\text{I}}_2$ is the final oxidized product of iodine.

To solve this using the concept of equivalents, we first need to determine the n-factor (valency factor) for both the oxidizing and reducing agents.

Let's look at the oxidizing agent, ${\text{KMnO}}_4$. In a weakly basic medium, the permanganate ion (${\text{MnO}}_4^{-}$) is reduced to manganese dioxide (${\text{MnO}}_2$). The oxidation state of manganese changes from $+7$ to $+4$.

This represents a gain of 3 electrons per molecule of ${\text{KMnO}}_4$. Thus, the n-factor for ${\text{KMnO}}_4$ is $3$.

Next, let's examine the reducing agent, which produces ${\text{I}}_2$. The iodide ion (${\text{I}}^{-}$) has an oxidation state of $-1$, and it oxidizes to elemental iodine (${\text{I}}_2$) with an oxidation state of $0$.

Since we are calculating the n-factor with respect to the product ${\text{I}}_2$, we must account for both iodine atoms in the molecule. The total change in oxidation state is $2\times (0-(-1))=2$. Therefore, the n-factor for ${\text{I}}_2$ is $2$.

Law of Equivalence

The most elegant way to solve stoichiometry problems without balancing complex redox equations is by using the Law of Chemical Equivalence. This law states that in any chemical reaction, the number of equivalents of the oxidizing agent consumed must exactly equal the number of equivalents of the reducing agent produced.

We know that the number of equivalents is simply the product of the number of moles and the n-factor. Let's set up our master equation:

Now, it's just a matter of plugging in the values. The problem gives us $4$ moles of ${\text{KMnO}}_4$.

Dividing both sides by 2, we get:

Exactly $6$ moles of ${\text{I}}_2$ are released. By mastering the concept of n-factors and equivalents, you can bypass tedious equation balancing and arrive at the answer swiftly and accurately!