Question:
What is change in oxidation number of ' Mn ' during the decolourisation of acidified potassium permanganate by moist sulfur dioxide?
What is change in oxidation number of ' Mn ' during the decolourisation of acidified potassium permanganate by moist sulfur dioxide?
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Permanganate in acidic medium is reduced from manganese in +7 state to Mn2+.
Updated On: Nov 13, 2026
increases by 2
increase by 5
decrease by 2
decreases by 5
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
In acidic solution, \(\text{MnO}_4^{-}\) is reduced by \(\text{SO}_2\) and the purple colour disappears as \(\text{Mn}^{2+}\) forms.
Step 2: Oxidation numbers
In \(\text{MnO}_4^{-}\): \(x + 4(-2) = -1\), so \(x = +7\).
In \(\text{Mn}^{2+}\) the oxidation number is \(+2\).
Step 3: Change
\(+7 \rightarrow +2\) means a decrease of 5.
Final Answer:
The oxidation number of Mn decreases by 5. \[ \boxed{\text{(D) }\text{decreases by 5}} \]
In acidic solution, \(\text{MnO}_4^{-}\) is reduced by \(\text{SO}_2\) and the purple colour disappears as \(\text{Mn}^{2+}\) forms.
Step 2: Oxidation numbers
In \(\text{MnO}_4^{-}\): \(x + 4(-2) = -1\), so \(x = +7\).
In \(\text{Mn}^{2+}\) the oxidation number is \(+2\).
Step 3: Change
\(+7 \rightarrow +2\) means a decrease of 5.
Final Answer:
The oxidation number of Mn decreases by 5. \[ \boxed{\text{(D) }\text{decreases by 5}} \]
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