Question

# Find the n-factor in the following chemical changes :$$KMnO_4\overset{H^+}{\rightarrow} Mn^{2+}$$

A
3
B
2
C
4
D
5

Solution

## The correct option is D 5In 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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