Question:
Which of the following gas readily de-colourises the acidified KMnO\(_4\) solution?
Which of the following gas readily de-colourises the acidified KMnO\(_4\) solution?
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Sulfur dioxide (SO\(_2\)) is a common reducing agent that decolourises acidified KMnO\(_4\) solution by reducing the manganese ion from +7 to +2 oxidation state.
Updated On: Apr 15, 2025
- CO\(_2\)
- SO\(_2\)
- P\(_2\)O\(_5\)
- NO\(_2\)
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The Correct Option is B
Solution and Explanation
When acidified potassium permanganate (KMnO\(_4\)) solution is treated with reducing agents, such as SO\(_2\), the purple color of KMnO\(_4\) is decolourised due to the reduction of Mn\(^7+\) (in KMnO\(_4\)) to Mn\(^2+\). - SO\(_2\) is a reducing agent and reacts with KMnO\(_4\) in acidic solution to reduce Mn\(^7+\) to Mn\(^2+\), which leads to decolorisation of the purple solution. Thus, the correct answer is SO\(_2\).
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