Hydrogen peroxide oxidises [Fe(CN)6]4- to [Fe(CN)6]3- in acidic medium but reduces [Fe(CN)6]3- to [F

Q: Why does $H_2O_2$ act differently in acidic and alkaline mediums?

The redox potential of $H_2O_2$ depends heavily on the pH. In acidic conditions, the presence of $H^+$ ions favors its reduction to $H_2O$, making it a strong oxidizing agent. In alkaline conditions, the presence of $OH^-$ ions facilitates the removal of protons from $H_2O_2$, making it easier to oxidize to $O_2$.

Q: How do I remember the oxidation state of oxygen in peroxides?

In most compounds, oxygen is -2. But in peroxides, there is an O-O single bond. Since the two oxygen atoms share electrons equally, each oxygen only gains one electron from the attached hydrogen (or metal), resulting in a -1 oxidation state.

Q: Is water always a product when $H_2O_2$ reacts?

Yes, in almost all its redox reactions in aqueous solutions, whether it acts as an oxidizing or reducing agent, $H_2O$ is formed either as the main reduced product or as a byproduct of balancing hydrogen and oxygen atoms.

Q: What is the oxidation state of Iron in $[Fe(CN)_6]^{4-}$ and $[Fe(CN)_6]^{3-}$?

Cyanide ($CN^-$) has a charge of -1. For $[Fe(CN)_6]^{4-}$, let Fe be $x$. $x + 6(-1) = -4 \implies x = +2$. For $[Fe(CN)_6]^{3-}$, $x + 6(-1) = -3 \implies x = +3$.

Q: Can $H_2O_2$ act as a reducing agent in an acidic medium?

Yes, it can! If it reacts with a much stronger oxidizing agent like $KMnO_4$ in an acidic medium, $H_2O_2$ will act as a reducing agent and release $O_2$ gas. The medium alone doesn't dictate the role; the relative strength of the other reactant matters too.

Question

Hydrogen peroxide oxidises $[\mathrm{Fe(CN)_6}]^{4-}$ to $[\mathrm{Fe(CN)_6}]^{3-}$ in acidic medium but reduces $[\mathrm{Fe(CN)_6}]^{3-}$ to $[\mathrm{Fe(CN)_6}]^{4-}$ in alkaline medium. The other products formed are, respectively.

Options:

(A) $(\mathrm{H_2O} + \mathrm{O_2})$ and $\mathrm{H_2O}$

(B) $(\mathrm{H_2O} + \mathrm{O_2})$ and $(\mathrm{H_2O} + \mathrm{OH^-})$

(C) $\mathrm{H_2O}$ and $(\mathrm{H_2O} + \mathrm{O_2})$

(D) $\mathrm{H_2O}$ and $(\mathrm{H_2O} + \mathrm{OH^-})$

Sigma Percentile · Accepted Answer

### The Chameleon Molecule: Hydrogen Peroxide

Hydrogen peroxide ($H_2O_2$) is one of the most fascinating molecules in chemistry because of its dual nature.

It can act as both an oxidizing agent and a reducing agent. This chameleon-like behavior stems from the oxidation state of oxygen in peroxides, which is $-1$.

Since $-1$ is an intermediate state, oxygen can either gain electrons to reach $-2$ (acting as an oxidant) or lose electrons to reach $0$ (acting as a reductant).

### The Acidic Medium: The Electron Thief

In an acidic medium, $H_2O_2$ typically acts as a strong oxidizing agent.

When it reacts with the ferrocyanide ion, $[Fe(CN)_6]^{4-}$, it steals electrons. The iron in ferrocyanide is in a $+2$ oxidation state. By losing an electron, it gets oxidized to ferricyanide, $[Fe(CN)_6]^{3-}$, where iron is in a $+3$ state.

But what happens to the hydrogen peroxide?

Since it is acting as an oxidizing agent, it gets reduced. The oxygen atoms go from a $-1$ oxidation state to a $-2$ oxidation state, forming water ($H_2O$).

The balanced half-reaction is:
$$H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$$

Thus, the only other product formed in the acidic medium is **water**.

### The Alkaline Medium: The Generous Donor

Now, let's flip the script. In an alkaline (basic) medium, the presence of $OH^-$ ions changes the dynamics.

Here, $H_2O_2$ acts as a reducing agent when reacting with the ferricyanide ion, $[Fe(CN)_6]^{3-}$. It generously donates electrons, reducing the iron from a $+3$ state back to a $+2$ state, forming ferrocyanide, $[Fe(CN)_6]^{4-}$.

As a reducing agent, hydrogen peroxide itself must get oxidized.

The oxygen atoms lose electrons, moving from a $-1$ oxidation state to $0$, which means they form oxygen gas ($O_2$). To balance the reaction in a basic medium, water is also produced.

The balanced half-reaction is:
$$H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$$

So, in the alkaline medium, the other products formed are **water and oxygen gas**.

### The Final Verdict

By analyzing both scenarios, we can clearly see the products formed alongside the iron complexes.

In the acidic medium, we get $H_2O$. In the alkaline medium, we get $H_2O$ and $O_2$.

This perfectly aligns with option (c). Understanding the intermediate oxidation state of peroxides is the master key to unlocking these types of redox problems!

Animated Solution for Chemistry - s and p-Block Elements: Hydrogen peroxide oxidises $[Fe (CN)_{6}]^{4 -}$ to $[Fe (CN)_{6}]^{3 -}$ in acidic medium but reduces $[Fe (CN)_{6}]^{3 -}$ to $[Fe (CN)_{6}]^{4 -}$ in alkaline medium. The other products formed are, respectively.

Select Answer:

Visualized Solution

\text{Dual Nature of } \mathrm{H_2O_2}

\text{Acidic Medium: Oxidizing Agent}

\text{Reaction in Acidic Medium}

\text{Alkaline Medium: Reducing Agent}

\text{Reaction in Alkaline Medium}

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Hydrogen, Hydrides and Water

The Chameleon Molecule

The Acidic Medium

The Alkaline Medium

The Final Verdict

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