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Question

Find the n-factor in the following chemical changes :

$$KMnO_4\overset{H^+}{\rightarrow} Mn^{2+}$$


A
3
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B
2
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C
4
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D
5
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Solution

The correct option is D 5

In acidic medium, $$Mn^{7+}$$ goes to $$Mn^{2+}$$ state and hence there is a net gain of 5 electrons.

In $$KMnO_4 $$ the oxidation state of Mn is $$+7$$. In an acidic medium, the reaction is:


$$Mn{O}_4^¯ +8H^+ +5e^- \rightarrow Mn^{2+} +4H_2O$$


As you can see, $$Mn^{7+}$$ gains 5 electrons. (Making it an oxidizing agent.)

So the 'n factor' is 5 for $$KMnO_4$$


Chemistry

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