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:

We are asked to find the species that accompany the iron-cyanide complex when hydrogen peroxide reacts in two different media.

First, recall the dual nature of hydrogen peroxide:

• In acidic medium it behaves as an oxidising agent and is reduced to water. • In alkaline medium it behaves as a reducing agent and is oxidised to dioxygen.

Let us tackle the acidic case step by step.

In acid, hydrogen peroxide accepts electrons. We write the two half-reactions.

Oxidation half (for the complex): $$[Fe(CN)6]^{4- \;\longrightarrow\; [Fe(CN)6]^{3-} + e^-}$$

Reduction half (for H$$_2$$O$$_2$$). The standard form is stated first: $$H_{2}O_{2} + 2H^+ + 2e^- \;\longrightarrow\; 2H_{2}O$$

To equate electrons we multiply the iron half-reaction by 2:

$$2[Fe(CN)_6]^{4-} \longrightarrow 2[Fe(CN)_6]^{3-} + 2e^-$$

Adding the two halves gives

$$2[Fe(CN)_6]^{4-} + H_2O_2 + 2H^+ \longrightarrow 2[Fe(CN)_6]^{3-} + 2H_2O$$

We see clearly that the only additional product, apart from the ferricyanide ion, is water. Hence in acidic medium H$$_2$$O is formed.

Now we move to the alkaline case where hydrogen peroxide donates electrons. The standard oxidation half-reaction in base is

$$H_{2}O_{2} + 2OH^- \longrightarrow O_{2} + 2H_{2}O + 2e^-$$

The reduction half for the iron complex is simply the reverse of the earlier oxidation step:

$$[Fe(CN)6]^{3- + e^- \;\longrightarrow\; [Fe(CN)6]^{4-}}$$

Again we multiply this half-reaction by 2 to match electrons:

$$2[Fe(CN)_6]^{3-} + 2e^- \longrightarrow 2[Fe(CN)_6]^{4-}$$

Adding the two halves gives

$$2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^- \longrightarrow 2[Fe(CN)_6]^{4-} + O_2 + 2H_2O$$

Here the extra species produced, besides the ferrocyanide ion, are O$$_2$$ and H$$_2$$O. They appear on the product side together.

Collecting our results:

• Acidic medium → additional product: $$H_{2}O$$ • Alkaline medium → additional products: $$O_{2}$$ and $$H_{2}O$$

The option that lists “H$$_2$$O” for the first reaction and “(H$$_2$$O + O$$_2$$)” for the second is Option D.

Hence, the correct answer is Option D.