Question:
The reaction of H$_2$O$_2$ with KIO$_4$ in an alkaline medium gives:
The reaction of H$_2$O$_2$ with KIO$_4$ in an alkaline medium gives:
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H$_2$O$_2$ behaves as a reducing agent with strong oxidizers like MnO$_4^{-}$, IO$_4^{-}$, and Cl$_2$.
Updated On: Jun 10, 2026
- KIO$_3$ + H$_2$O + O$_2$
- KIO$_3$ + H$_2$O + O$_3$
- KI + H$_2$O + O$_2$
- KIO$_3$ + H$_2$O + H$_2$
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The Correct Option is A
Solution and Explanation
Concept:
Hydrogen peroxide (H$_2$O$_2$) acts as both oxidizing and reducing agent depending on the medium and the species it reacts with. With strong oxidizing agents like periodate (IO$_4^{-}$), it acts as a reducing agent.
Step 1: Redox reaction In alkaline medium: \[ IO_4^{-} + H_2O_2 \rightarrow IO_3^{-} + H_2O + O_2 \]
Step 2: Oxidation states
• Iodine: +7 in IO$_4^{-}$ → +5 in IO$_3^{-}$ (reduction)
• Oxygen in H$_2$O$_2$: -1 → 0 in O$_2$ (oxidation)
Step 3: Final products Potassium iodate (KIO$_3$), water, and oxygen gas. Thus correct option is (A).
Step 1: Redox reaction In alkaline medium: \[ IO_4^{-} + H_2O_2 \rightarrow IO_3^{-} + H_2O + O_2 \]
Step 2: Oxidation states
• Iodine: +7 in IO$_4^{-}$ → +5 in IO$_3^{-}$ (reduction)
• Oxygen in H$_2$O$_2$: -1 → 0 in O$_2$ (oxidation)
Step 3: Final products Potassium iodate (KIO$_3$), water, and oxygen gas. Thus correct option is (A).
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