Animated Solution for Chemistry - Redox Reactions: In which of the following reactions ${\text{H}}_2{\text{O}}_2$ acts as a reducing agent? A. ${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{e}}^{-}⟶2{\text{H}}_2\text{O}$ B. ${\text{H}}_2{\text{O}}_2-2{\text{e}}^{-}⟶{\text{O}}_2+2{\text{H}}^{+}$ C. ${\text{H}}_2{\text{O}}_2+2{\text{e}}^{-}⟶2{\text{OH}}^{-}$ D. ${\text{H}}_2{\text{O}}_2+2{\text{OH}}^{-}-2{\text{e}}^{-}⟶{\text{O}}_2+2{\text{H}}_2\text{O}$

Hydrogen peroxide (${\text{H}}_2{\text{O}}_2$) is a fascinating molecule. Because the oxygen atoms in peroxides are in an intermediate oxidation state of $-1$, ${\text{H}}_2{\text{O}}_2$ can act as both an oxidizing agent and a reducing agent depending on the reaction conditions.

To determine when it acts as a reducing agent, we must recall a fundamental principle of redox chemistry: a reducing agent reduces other substances by supplying electrons to them. In doing so, the reducing agent itself undergoes oxidation, which is characterized by a loss of electrons and a corresponding increase in oxidation state.

Let's systematically evaluate the given reactions by tracking the oxidation state of oxygen.

Reaction A: Here, ${\text{H}}_2{\text{O}}_2$ gains electrons. The oxidation state of oxygen changes from $-1$ in ${\text{H}}_2{\text{O}}_2$ to $-2$ in ${\text{H}}_2\text{O}$. Since the oxidation state decreases, ${\text{H}}_2{\text{O}}_2$ is undergoing reduction. Thus, it acts as an oxidizing agent here.

Reaction B: In this reaction, ${\text{H}}_2{\text{O}}_2$ loses electrons. The oxidation state of oxygen increases from $-1$ to $0$ in elemental oxygen (${\text{O}}_2$). This increase signifies oxidation, meaning ${\text{H}}_2{\text{O}}_2$ is acting as a reducing agent.

Reaction C: Similar to Reaction A, ${\text{H}}_2{\text{O}}_2$ gains electrons, and the oxidation state of oxygen drops from $-1$ to $-2$ in the hydroxide ion (${\text{OH}}^{-}$). This is a reduction process.

Reaction D: Here, in a basic medium, ${\text{H}}_2{\text{O}}_2$ again loses electrons to form ${\text{O}}_2$. The oxidation state of oxygen increases from $-1$ to $0$. This is oxidation, confirming that ${\text{H}}_2{\text{O}}_2$ acts as a reducing agent.

By carefully tracking the electron flow and oxidation states, we can confidently conclude that ${\text{H}}_2{\text{O}}_2$ undergoes oxidation in reactions B and D. Therefore, it acts as a reducing agent in these specific reactions. The correct choice is option (d).