Which of the following equation depicts the oxidising nature of H2O2?
(a) KIO4 + H2O2 - KIO3 + H2O

Q: Why is the oxidation state of oxygen in $H_2O_2$ -1 and not -2 like in water?

In peroxides, there is a unique O-O single bond. Because the two oxygen atoms share electrons equally between themselves, and each oxygen is bonded to one less electronegative hydrogen atom, the assigned oxidation state is -1 for each oxygen atom, rather than the usual -2.

Q: Can $H_2O_2$ act as both an oxidising and reducing agent?

Yes, absolutely! Since the oxidation state of oxygen in $H_2O_2$ is -1, it sits in an intermediate state. It can either decrease to -2 (acting as an oxidising agent) or increase to 0 (acting as a reducing agent).

Q: How do I quickly identify if $H_2O_2$ is acting as a reducing agent in a reaction?

A very handy trick is to look at the products. If $O_2$ gas is evolved, it means the oxygen in $H_2O_2$ went from -1 to 0. This is an oxidation process, which means $H_2O_2$ is acting as a reducing agent.

Q: In option (a), why is $KIO_4$ the oxidising agent and not $H_2O_2$?

In $KIO_4$, Iodine is in its maximum possible oxidation state of +7. It cannot lose any more electrons; it can only gain them (be reduced). Therefore, it is forced to act as an oxidising agent, which in turn forces $H_2O_2$ to act as the reducing agent.

Q: Does the medium (acidic or basic) change whether $H_2O_2$ acts as an oxidising or reducing agent?

$H_2O_2$ can act as both an oxidising and reducing agent in both acidic and basic mediums. The specific role it plays depends entirely on the standard reduction potentials of the other chemical species it is reacting with, not just the pH of the medium.

Question

Which of the following equation depicts the oxidising nature of $H_2O_2$?

Options:

(A) $KIO_4 + H_2O_2 \longrightarrow KIO_3 + H_2O + O_2$

(B) $I_2 + H_2O_2 + 2OH^- \longrightarrow 2I^- + 2H_2O + O_2$

(C) $2I^- + H_2O_2 + 2H^+ \longrightarrow I_2 + 2H_2O$

(D) $Cl_2 + H_2O_2 \longrightarrow 2HCl + O_2$

Sigma Percentile · Accepted Answer

### Unveiling the Dual Nature of Hydrogen Peroxide

Hydrogen peroxide ($H_2O_2$) is one of the most fascinating and versatile molecules in chemistry. It is like a chemical chameleon, capable of acting as both an **oxidising agent** and a **reducing agent** depending entirely on the company it keeps. But how does it decide which role to play? The secret lies in the oxidation state of its oxygen atoms.

### The Secret is in the Oxidation State

In most compounds, like water ($H_2O$), oxygen happily sits at an oxidation state of $-2$. In elemental oxygen gas ($O_2$), it is at $0$. However, in peroxides like $H_2O_2$, oxygen is in an intermediate oxidation state of **$-1$**.

Because $-1$ is right in the middle, oxygen in $H_2O_2$ has two choices:
1. It can gain an electron and go down to $-2$ (Reduction). When it does this, it acts as an **oxidising agent**.
2. It can lose an electron and go up to $0$ (Oxidation). When it does this, it acts as a **reducing agent**.

To solve our problem, we need to find the reaction where $H_2O_2$ is acting as an oxidising agent, which means we are looking for the reaction where the oxygen in $H_2O_2$ is **reduced** from $-1$ to $-2$.

### Analyzing the Suspects

Let's put our detective hats on and analyze the given options one by one.

**Option (a):** $KIO_4 + H_2O_2 \longrightarrow KIO_3 + H_2O + O_2$
Here, Iodine in $KIO_4$ is in its maximum oxidation state of $+7$. It is desperate for electrons and acts as a strong oxidising agent, forcing itself to be reduced to $+5$ in $KIO_3$. Consequently, $H_2O_2$ is forced to act as a reducing agent, getting oxidised to $O_2$ gas (oxidation state $0$).

**Option (b):** $I_2 + H_2O_2 + 2OH^- \longrightarrow 2I^- + 2H_2O + O_2$
In this basic medium reaction, elemental Iodine ($I_2$, oxidation state $0$) is reduced to Iodide ions ($I^-$, oxidation state $-1$). Since Iodine is reduced, $H_2O_2$ must be the one getting oxidised. Again, we see $O_2$ gas being produced, confirming $H_2O_2$ is acting as a reducing agent.

**Option (d):** $Cl_2 + H_2O_2 \longrightarrow 2HCl + O_2$
Similar to option (b), Chlorine gas ($Cl_2$, oxidation state $0$) is reduced to Chloride ions in $HCl$ (oxidation state $-1$). $H_2O_2$ is once again oxidised to $O_2$ gas, playing the role of a reducing agent.

### The Culprit Revealed

Now, let's look at **Option (c):** $2I^- + H_2O_2 + 2H^+ \longrightarrow I_2 + 2H_2O$

Let's track the electrons carefully. The Iodide ion ($I^-$) starts with an oxidation state of $-1$. On the product side, it becomes elemental Iodine ($I_2$) with an oxidation state of $0$. The oxidation number has increased from $-1$ to $0$, meaning Iodide has been **oxidised**.

Who caused this oxidation? Hydrogen peroxide! Let's verify by checking its own oxidation state. The oxygen in $H_2O_2$ starts at $-1$ and ends up in $H_2O$ with an oxidation state of $-2$. The oxidation number has decreased, meaning $H_2O_2$ has been **reduced**.

Because $H_2O_2$ oxidises $I^-$ to $I_2$ while reducing itself to $H_2O$, it is unequivocally acting as an **oxidising agent** in this reaction.

**Pro Tip:** A quick trick to identify when $H_2O_2$ is acting as a reducing agent is to look for the evolution of $O_2$ gas on the product side. If you see $O_2$, $H_2O_2$ has been oxidised! In our correct option (c), there is no $O_2$ gas, only water, confirming its role as an oxidising agent.

Animated Solution for Chemistry - Redox Reactions: Which of the following equation depicts the oxidising nature of $H_{2} O_{2}$ ?

Select Answer:

Visualized Solution

\text{Decoding the Question}

\text{Oxidation State of Oxygen in } H_2O_2

\text{Evaluating Option (a)}

\text{Evaluating Option (b)}

\text{Evaluating Option (c)}

\text{Oxidation of Iodide}

\text{Reduction of } H_2O_2

\text{Conclusion}

The Sigma Insight: Oxidation and Reduction

Unveiling the Dual Nature of Hydrogen Peroxide

The Secret is in the Oxidation State

Analyzing the Suspects

The Culprit Revealed

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