Question

Hydrogen peroxide in its reaction with $KI{O}_4$ and $N{H}_{4}OH$ respectively is acting as:
(A) reducing agent, oxidizing agent
(B) reducing agent, reducing agent
(C) oxidizing agent, oxidizing agent
(D) oxidizing agent, reducing agent

Answer 1

Hint: Write the reaction of hydrogen peroxide with $KI{O}_4$ and $N{H}_{4}OH$. Assign the oxidation states of iodine and nitrogen respectively on both reactant and product sides.
Recollect the concept of oxidizing agent and reducing agent and then accordingly assign the terms for hydrogen peroxide in both the reactions.

Complete step by step answer:
- Let’s start by writing the reactions.
- When hydrogen peroxide reacts with Potassium periodate, $KI{O}_4$, potassium iodate, water and oxygen are formed as products. The reaction is represented as,
$$KI{O}_4+H_2{O}_2\to KI{O}_3+H_{2}O+O_2$$
- Oxygen has -2 oxidation state and potassium has +1 oxidation state. Let’s assume the oxidation state of iodine as ‘x’. Therefore, the oxidation state of iodine in $KI{O}_4$ is calculated as follows,
$$\begin{array}{rl}& x+1+(4\times -2)=0\\ & x=+7\end{array}$$
- Therefore, the oxidation state of iodine in $KI{O}_4$ is +7.

- Similarly, oxidation state of iodine in $KI{O}_3$ is calculated as,
$$\begin{array}{rl}& x+1+(3\times -2)=0\\ & x=+5\end{array}$$
- Therefore, iodine is getting reduced from +7 to +5.
- A reducing agent is the reagent which reduces the reacting species by itself getting oxidized.
- An oxidizing agent is the reagent which oxidizes the reacting species by itself getting reduced.
- Therefore, when hydrogen peroxide reacts with $KI{O}_4$, hydrogen peroxide acts as a reducing agent.

- Now, let’s see what happens when hydrogen peroxide reacts with ammonium hydroxide.
- When hydrogen peroxide reacts with ammonium hydroxide, dinitrogen trioxide and water is formed. The reaction is given as,
$$4H_2{O}_2+2N{H}_{4}OH\to N_2{O}_3+7H_{2}O$$
- Hydrogen has +1 oxidation state and oxygen has -2 oxidation state. Let’s assume the oxidation state of nitrogen is ‘x’. Therefore, the oxidation state of nitrogen in ammonium hydroxide is given as,
$$\begin{array}{rl}& x+(4\times +1)+(-2)+1=0\\ & x+4-2+1=0\\ & x=-3\end{array}$$
- Therefore, the oxidation state of nitrogen in $N{H}_{4}OH$ is -3.

- Similarly, let’s find the oxidation state of nitrogen in $N_2{O}_3$.
$$\begin{array}{rl}& 2x+(3\times -2)=0\\ & 2x=+6\\ & x=+3\end{array}$$
- Therefore, the oxidation state of nitrogen in $N_2{O}_3$ is +3.
- So in this reaction, nitrogen is getting oxidized from -3 to +3.
- Therefore, here hydrogen peroxide will act as an oxidizing agent.
- Therefore, hydrogen peroxide in its reaction with $KI{O}_4$ and $N{H}_{4}OH$ respectively is acting as a reducing agent, oxidizing agent.
So, the correct answer is “Option A”.

Note: Remember the reagent which oxidizes other species by itself getting reduced is an oxidizing agent and the reagent which reduces other species by itself getting oxidized is a reducing agent. Oxidation is the process of losing electrons or addition of oxygen or removal of hydrogen. Reduction is the process of gaining of electrons or removal of oxygen or addition of hydrogen.