Determine Oxidizing and Reducing Behavior of Hydrogen Peroxide

To determine the behavior of hydrogen peroxide ($\text{H}_2\text{O}_2$) in its reactions with potassium periodate ($\text{KIO}_4$) and hydroxylamine ($\text{NH}_2\text{OH}$), we analyze the changes in oxidation states for each reaction.

1. Reaction with Potassium Periodate ($\text{KIO}_4$): The reaction between $\text{KIO}_4$ and $\text{H}_2\text{O}_2$ proceeds as follows: $\text{KIO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{KIO}_3 + \text{H}_2\text{O} + \text{O}_2$

   • Oxidation state of iodine ($\text{I}$) in $\text{KIO}_4$ is $+7$.
   • Oxidation state of iodine ($\text{I}$) in $\text{KIO}_3$ is $+5$.
   • Since the oxidation state of iodine decreases from $+7$ to $+5$, $\text{KIO}_4$ is reduced.
   • The oxidation state of oxygen in $\text{H}_2\text{O}_2$ increases from $-1$ to $0$ in $\text{O}_2$.
   • Since $\text{H}_2\text{O}_2$ reduces $\text{KIO}_4$, it acts as a reducing agent.

2. Reaction with Hydroxylamine ($\text{NH}_2\text{OH}$): The reaction between $\text{NH}_2\text{OH}$ and $\text{H}_2\text{O}_2$ can be represented as: $\text{NH}_2\text{OH} + \text{H}_2\text{O}_2 \rightarrow \text{HNO}_2 + \text{H}_2\text{O} \quad \text{or} \quad 2\text{NH}_2\text{OH} + \text{H}_2\text{O}_2 \rightarrow \text{N}_2 + 4\text{H}_2\text{O}$

   • Oxidation state of nitrogen ($\text{N}$) in $\text{NH}_2\text{OH}$ is $-1$.
   • In the products (such as $\text{N}_2$ or $\text{HNO}_2$), the oxidation state of nitrogen increases to $0$ or $+3$, respectively.
   • Nitrogen is oxidized, while the oxygen in $\text{H}_2\text{O}_2$ is reduced from an oxidation state of $-1$ to $-2$ in $\text{H}_2\text{O}$.
   • Since $\text{H}_2\text{O}_2$ oxidizes $\text{NH}_2\text{OH}$, it acts as an oxidising agent.

Therefore, hydrogen peroxide acts as a reducing agent with $\text{KIO}_4$ and an oxidising agent with $\text{NH}_2\text{OH}$.

Correct Option: A