Question

Identify reducing agent in following reaction:
$\text{H}_2\text{O}_{2(aq)} + \text{ClO}_{4(aq)}^- \rightarrow \text{ClO}_{2(aq)}^- + \text{O}_{2(g)}$

• A. $\text{ClO}_{2(aq)}^-$
• B. $\text{H}_2\text{O}_{2(aq)}$
• C. $\text{ClO}_{4(aq)}^-$
• D. $\text{O}_{2(g)}$

Hint:

In any chemical reaction where hydrogen peroxide ($\text{H}_2\text{O}_2$) yields oxygen gas ($\text{O}_2$), it is always undergoing oxidation (from $-1$ to $0$) and thus always acts as a reducing agent.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to identify the reducing agent in the given redox reaction. A reducing agent is a substance that reduces another reactant by losing electrons, thereby causing its own oxidation state to increase.

Step 2: Key Formula or Approach:
We can determine the reducing agent by assigning oxidation numbers to the atoms involved in the reaction and locating the species whose oxidation number increases (undergoes oxidation).

Step 3: Detailed Explanation:
Let's analyze the changes in oxidation states for the elements in the reaction: In hydrogen peroxide ($\text{H}_2\text{O}_2$), oxygen is in a special peroxide state and carries an oxidation number of $-1$. In the product elemental oxygen gas ($\text{O}_2$), the oxidation number of oxygen is $0$. Since the oxidation number of oxygen increases from $-1$ to $0$, $\text{H}_2\text{O}_2$ undergoes oxidation. Conversely, the chlorine atom in the perchlorate ion ($\text{ClO}_4^-$) goes from an oxidation state of $+7$ to $+3$ in the chlorite ion ($\text{ClO}_2^-$), meaning it undergoes reduction. Because $\text{H}_2\text{O}_2$ is the reactant that gets oxidized, it acts as the reducing agent for the perchlorate ion.

Step 4: Final Answer:
The reducing agent in the reaction is $\text{H}_2\text{O}_{2(aq)}$, which corresponds to option (B).