Animated Solution for Chemistry - s and p-Block Elements: Hydrogen peroxide reacts with iodine in basic medium to give

A Redox Adventure

Hydrogen peroxide ($H_2{O}_2$) is one of the most fascinating molecules in inorganic chemistry because of its chemical bipolarity. It can act as both an oxidizing agent and a reducing agent, depending entirely on the chemical environment and the species it is reacting with.

In this problem, we are asked to determine the product when hydrogen peroxide reacts with iodine ($I_2$) in a basic medium.

Let's look closely at the reactants. We have elemental iodine ($I_2$) and hydrogen peroxide ($H_2{O}_2$). The presence of hydroxide ions ($O{H}^{-}$) confirms that the reaction is taking place in a basic medium.

Halogens, like iodine, are generally good oxidizing agents, meaning they prefer to gain electrons and get reduced. Because iodine wants to be reduced, hydrogen peroxide is forced to step up and act as the reducing agent.

To truly understand the flow of electrons, we must assign oxidation states to our key players.

In elemental iodine ($I_2$), the oxidation state of iodine is exactly $0$.

In hydrogen peroxide ($H_2{O}_2$), oxygen is trapped in a peroxide linkage ($-O-O-$). Because each oxygen atom shares an electron equally with the other oxygen, it only gains a formal $-1$ charge from the less electronegative hydrogen atom. Thus, the oxidation state of oxygen in $H_2{O}_2$ is $-1$.

Since $H_2{O}_2$ is acting as the reducing agent, it will donate electrons to iodine. Iodine gains these electrons, and its oxidation state drops from $0$ to $-1$. This reduction transforms elemental iodine into the iodide ion ($I^{-}$).

Simultaneously, hydrogen peroxide must undergo oxidation. The oxygen atoms lose electrons, moving from a $-1$ oxidation state to a $0$ oxidation state. This results in the release of oxygen gas ($O_2$).

When we balance the half-reactions in the basic medium, we arrive at the final chemical equation:

The question specifically asks for the product formed from iodine. As we can see from our balanced equation, iodine is converted into the iodide ion ($I^{-}$). Therefore, the correct option is (c).

A Classic Trap

Here is a crucial catch that examiners love to test. What if the medium was acidic instead of basic?

In an acidic medium, hydrogen peroxide becomes a powerful oxidizing agent. If you were to react $H_2{O}_2$ with iodide ions ($I^{-}$) in the presence of $H^{+}$, it would oxidize the iodide back into elemental iodine ($I_2$).

Always pay strict attention to the medium provided in the question, as it completely flips the chemical behavior of hydrogen peroxide!