Question

Which of the following equation depicts the oxidizing nature of $H_2O_2$?

• A. KIO$_4$ + H$_2$O$_2$ $\rightarrow$ KIO$_3$ + H$_2$O + O$_2$
• B. I$_2$ + H$_2$O$_2$ + 2OH$^-$ $\rightarrow$ 2I$^-$ + 2H$_2$O + O$_2$
• C. 2I$^-$ + H$_2$O$_2$ + 2H$^+$ $\rightarrow$ I$_2$ + 2H$_2$O
• D. Cl$_2$ + H$_2$O$_2$ $\rightarrow$ 2HCl + O$_2$

Hint:

If $O_2$ gas is a product of an $H_2O_2$ reaction, $H_2O_2$ has acted as a {reducing agent}. If $H_2O$ is the only oxygen-containing product, it has acted as an {oxidizing agent}.

Answer 1

The Correct Option is C

Step 1: $H_2O_2$ acts as an oxidizing agent when it is reduced (Oxygen oxidation state goes from $-1$ to $-2$ in $H_2O$).
Step 2: In option (C): Iodine goes from $-1$ (in $I^-$) to $0$ (in $I_2$). This is oxidation.
Step 3: Simultaneously, Oxygen in $H_2O_2$ ($-1$) goes to $H_2O$ ($-2$). This is reduction.
Step 4: Therefore, $H_2O_2$ is oxidizing $I^-$ to $I_2$.
Step 5: In all other options, $H_2O_2$ is being oxidized to $O_2$ ($0$), meaning it is acting as a reducing agent.