Animated Solution for Chemistry - Redox Reactions: Hydrogen peroxide in its reaction with ${\text{KIO}}_4$ and ${\text{NH}}_2\text{OH}$ respectively, is acting as a

The universe loves balance. If one atom loses an electron, another atom must gain it. This simultaneous dance of electron transfer is what we call a reduction-oxidation, or redox, reaction.

We use a mathematical bookkeeping tool called oxidation states to track where the electrons are going. An increase in oxidation state means an atom has lost electrons (Oxidation). A decrease in oxidation state means an atom has gained electrons (Reduction).

The chemical that causes oxidation by taking electrons is the oxidising agent (and it gets reduced in the process). Conversely, the chemical that causes reduction by giving away electrons is the reducing agent (and it gets oxidised).

It is a chemical chameleon.

Why? Because of the oxidation state of its oxygen atoms. In most compounds, like water (${\text{H}}_2\text{O}$) or carbon dioxide (${\text{CO}}_2$), oxygen loves to be in a $-2$ oxidation state. In elemental oxygen gas (${\text{O}}_2$), it is perfectly balanced at $0$.

But in peroxides, oxygen is stuck right in the middle at $-1$.

This intermediate state is the secret to its incredible versatility. It stands at a chemical crossroads. It can choose to gain electrons and go down to $-2$ (acting as an oxidising agent), or it can lose electrons and go up to $0$ (acting as a reducing agent). Its behavior entirely depends on the chemical personality of the molecule it is reacting with. It adapts to its environment.

The Encounter with Potassium Periodate Let's look at the first reaction where hydrogen peroxide meets potassium periodate (${\text{KIO}}_4$).

To understand what will happen, we must look at the oxidation state of iodine in ${\text{KIO}}_4$. Potassium is an alkali metal, always $+1$. Oxygen is $-2$. If we set up the equation for neutrality:

Iodine is in a $+7$ oxidation state. This is a crucial piece of information. Iodine belongs to Group 17 of the periodic table, meaning it has 7 valence electrons. It has already "lost" all 7 of its valence electrons to the highly electronegative oxygen atoms. It cannot lose any more electrons; $+7$ is its absolute maximum oxidation state.

Because it cannot be oxidised further, ${\text{KIO}}_4$ has only one option: it must gain electrons and get reduced. This makes it a powerful and strict oxidising agent.

When an unstoppable oxidising agent meets the versatile hydrogen peroxide, it forces the peroxide to act as a reducing agent. The iodine in ${\text{KIO}}_4$ gets reduced to a lower oxidation state (like $0$ in ${\text{I}}_2$ or $-1$ in ${\text{I}}^{-}$), and in the process, it aggressively strips electrons from the oxygen in ${\text{H}}_2{\text{O}}_2$.

The oxygen in ${\text{H}}_2{\text{O}}_2$ goes from $-1$ to $0$, releasing oxygen gas (${\text{O}}_2$). Conclusion for Reaction 1: Hydrogen peroxide acts as a reducing agent.

The Encounter with Hydroxylamine Now, let's shift our focus to the second reaction involving hydroxylamine (${\text{NH}}_2\text{OH}$).

First, we need to determine the oxidation state of nitrogen in ${\text{NH}}_2\text{OH}$. Hydrogen is generally $+1$ (except in metal hydrides) and oxygen is $-2$. Let's set up the equation:

Nitrogen is in a $-1$ oxidation state.

Now, let's look at the product side. The reaction produces dinitrogen trioxide (${\text{N}}_2{\text{O}}_3$). What is the oxidation state of nitrogen here?

Nitrogen has gone from $-1$ to $+3$. Its oxidation state has increased by 4 units, which means it has lost 4 electrons per nitrogen atom. Nitrogen has been oxidised.

If nitrogen is being oxidised, something must be taking those electrons. That "something" is our hydrogen peroxide. In this scenario, ${\text{H}}_2{\text{O}}_2$ accepts the electrons, and its oxygen atoms are reduced from $-1$ to $-2$, forming water (${\text{H}}_2\text{O}$). The weak O-O peroxide bond breaks, and the oxygen atoms happily accept electrons to reach their stable $-2$ state.

Conclusion for Reaction 2: Hydrogen peroxide acts as an oxidising agent.

By carefully calculating the oxidation states before and after the reactions, we have unmasked the role of hydrogen peroxide in both scenarios.

In the presence of a strong oxidising agent like ${\text{KIO}}_4$, it acts as a reducing agent. In the presence of a species that can be oxidised like ${\text{NH}}_2\text{OH}$, it acts as an oxidising agent.

Therefore, the correct sequence of its roles is reducing agent, oxidising agent, which perfectly matches option (A).

This problem is a beautiful reminder that in chemistry, context is everything. A molecule's behavior is defined by who it interacts with. Always remember, in redox chemistry, the oxidation states hold the ultimate truth! Never guess—always calculate.