Question:
When KI is reacted with \(O_3\) under aqueous condition the product formed is
When KI is reacted with \(O_3\) under aqueous condition the product formed is
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Ozone oxidizes iodide ions very easily:
\[
2I^- \xrightarrow{O_3} I_2
\]
This reaction is often used for the estimation of ozone due to the liberation of iodine.
Updated On: Jul 18, 2026
- \(I_2O_4\)
- \(I_2O_5\)
- \(I_4O_9\)
- \(I_2\)
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The Correct Option is D
Solution and Explanation
Step 1: Recall the oxidizing nature of ozone.
Ozone (\(O_3\)) is a very strong oxidizing agent. In aqueous solution, it readily oxidizes iodide ions to iodine.
Step 2: Write the oxidation half-reaction.
Iodide ions from KI undergo oxidation: \[ 2I^- \rightarrow I_2 + 2e^- \]
Step 3: Write the reduction half-reaction of ozone.
In aqueous medium, ozone is reduced to oxygen: \[ O_3 + H_2O + 2e^- \rightarrow O_2 + 2OH^- \]
Step 4: Obtain the overall reaction.
Adding the two half-reactions, \[ O_3 + 2I^- + H_2O \rightarrow I_2 + O_2 + 2OH^- \] Thus, iodine is liberated.
Step 5: Final conclusion.
Therefore, the product formed when KI reacts with ozone in aqueous medium is \[ \boxed{I_2} \] Hence, option (4) is correct.
Ozone (\(O_3\)) is a very strong oxidizing agent. In aqueous solution, it readily oxidizes iodide ions to iodine.
Step 2: Write the oxidation half-reaction.
Iodide ions from KI undergo oxidation: \[ 2I^- \rightarrow I_2 + 2e^- \]
Step 3: Write the reduction half-reaction of ozone.
In aqueous medium, ozone is reduced to oxygen: \[ O_3 + H_2O + 2e^- \rightarrow O_2 + 2OH^- \]
Step 4: Obtain the overall reaction.
Adding the two half-reactions, \[ O_3 + 2I^- + H_2O \rightarrow I_2 + O_2 + 2OH^- \] Thus, iodine is liberated.
Step 5: Final conclusion.
Therefore, the product formed when KI reacts with ozone in aqueous medium is \[ \boxed{I_2} \] Hence, option (4) is correct.
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