Animated Solution for Chemistry - Redox Reactions: Which of the following chemical reactions depicts the oxidising behaviour of ${\text{H}}_2{\text{SO}}_4$?

The concept of an oxidising agent is one of the most fundamental ideas in redox chemistry. But what exactly does it mean to be an "oxidising agent"?

Imagine a chemical reaction as a bustling marketplace where electrons are the currency. An oxidising agent is like a very aggressive debt collector. It forces other molecules to hand over their electrons (oxidising them), and by taking those electrons for itself, it gets reduced.

Therefore, the hallmark of an oxidising agent is that its own oxidation state must decrease during the reaction.

In our problem, we are looking for a reaction where sulfuric acid (${\text{H}}_2{\text{SO}}_4$) acts as this electron-hungry debt collector.

First, let's determine the oxidation state of sulfur in sulfuric acid. Hydrogen is typically $+1$ and oxygen is $-2$. Setting up the equation for the neutral molecule:

Sulfur is sitting at its maximum possible oxidation state of $+6$. It cannot lose any more electrons; it can only gain them. For sulfuric acid to act as an oxidising agent, we must find a reaction where the oxidation state of sulfur drops below $+6$ in the products.

Let's carefully examine the first option provided:

We need to track the flow of electrons by assigning oxidation states to the key players before and after the reaction.

On the reactant side, iodine in hydrogen iodide ($\text{HI}$) has an oxidation state of $-1$. On the product side, it appears as elemental iodine (${\text{I}}_2$), which has an oxidation state of $0$.

Iodine's oxidation state increased, meaning it lost electrons. It underwent oxidation.

Now, let's look at our prime suspect, sulfur. We already know it starts at $+6$ in ${\text{H}}_2{\text{SO}}_4$. What happens to it in the product, sulfur dioxide (${\text{SO}}_2$)?

In ${\text{SO}}_2$, oxygen is $-2$.

Sulfur's oxidation state went from $+6$ down to $+4$.

This decrease means sulfur gained electrons. It underwent reduction. Because sulfuric acid got reduced, it is the species responsible for oxidising the hydrogen iodide. Thus, it perfectly depicts its behaviour as an oxidising agent.

If we quickly glance at the other options, such as the formation of ${\text{CaSO}}_4$ or ${\text{NaHSO}}_4$, sulfur remains stubbornly at $+6$. Those are merely acid-base or displacement reactions, devoid of any electron transfer. Always follow the electrons, and the oxidising agent will reveal itself!