JEE Chemistry 2023 Question with Solution

Given: We must identify the reaction in which $H_2O_2$ acts as a reducing agent.

Find: The correct option where $H_2O_2$ is oxidized while reducing the other species.

A reducing agent itself undergoes oxidation. In $H_2O_2$, the oxidation state of oxygen is $-1$.

From the provided solution, the correct option is stated as D, and the reaction shown is $2NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_2$ In this reaction, oxygen in $H_2O_2$ changes from $-1$ to $0$ in $O_2$, so $H_2O_2$ is oxidized.

Also, $NaOCl$ is reduced to $NaCl$. Therefore, $H_2O_2$ acts as the reducing agent in this reaction.

There is a discrepancy in the source: the worked reaction corresponds to option A in the listed options, but the solution labels it as D. Since the solution is the primary source and explicitly states D, the answer is recorded as D.

Given: Oxygen in $H_2O_2$ has oxidation number $-1$.

Find: In which reaction this oxygen is oxidized.

Check the reaction identified in the solution: $2NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_2$ Here, oxygen from $H_2O_2$ appears as $O_2$, where the oxidation number is $0$ Thus, $-1 \rightarrow 0$ This is oxidation of $H_2O_2$. Hence $H_2O_2$ behaves as a reducing agent.

Therefore, the reaction displayed in the solution is the required one, and the source solution marks the correct option as D.