Question

(a) ${H_2 O_2 + O_3 -> H_2O + 2O_2}$ (b) ${H_2 O_2 + Ag_2O -> 2Ag + H_2O + O_2}$ Role of hydrogen peroxide in the above reactions is respectively -

• A. Oxidizing in (a) and reducing in (b)
• B. Reducing in (a) and oxidizing in (b)
• C. Reducing in (a) and (b)
• D. Oxidizing in (a) and (b)

Answer 1

The Correct Option is C

To determine the role of hydrogen peroxide (\(H_2O_2\)) in each of the given reactions, we need to understand the concepts of oxidation and reduction.

Oxidation: It is the process of losing electrons or increasing oxidation state by a molecule, atom, or ion.

Reduction: It is the process of gaining electrons or decreasing oxidation state.

1. Consider reaction (a): {H_2O_2 + O_3 \rightarrow H_2O + 2O_2}
   • In this reaction, the oxidation states of the elements need to be analyzed:
     • Hydrogen in \(H_2O_2\) is in +1 oxidation state.
     • Oxygen in \(H_2O_2\) is in -1 oxidation state, and in \(O_3\), it is 0.
     • In \(H_2O\), oxygen is in -2 oxidation state, and in \(O_2\), it is 0.
   • The transformation shows that \(H_2O_2\) is losing oxygen atoms (as \(O_2\)) while creating \(H_2O\). This essentially means that hydrogen peroxide is losing its own oxygen atoms, thus acting as a reducing agent (it reduces itself).
2. Consider reaction (b): {H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2}
   • Again, analyze the oxidation states:
     • In \(Ag_2O\), silver (Ag) is in +1 oxidation state, and oxygen is in -2.
     • In \(2Ag\), silver is in 0 oxidation state (indicating a reduction of silver from +1 to 0).
   • Similarly, for \(H_2O_2\):
     • The oxygen in \(H_2O_2\) (with oxidation state -1) changes to molecular oxygen (\(O_2\), with oxidation state 0), indicating it is itself being oxidized.
     • Since \(H_2O_2\) helps in reducing \(Ag_2O\) to \(Ag\), it is acting as a reducing agent here as well.

By analyzing both reactions, we conclude that hydrogen peroxide (\(H_2O_2\)) behaves as a reducing agent in both (a) and (b).

Thus, the correct answer is Reducing in (a) and (b).