The equivalent weight of $K M n O_{4}$ (formula weight $=$ M), when it is used as an oxidant in neutral medium, is:

M

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\frac{M}{2}

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\frac{M}{3}

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\frac{M}{5}

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Solution

The correct option is C $\frac{M}{3}$
In neutral medium, $K M n O_{4}$ gets reduced to the brown +4 oxidation state of $M n O_{2}$ .

So, here the change in oxidation state is 3. Therefore, the equivalent weight of $K M n O_{4} = \frac{M}{3}$ .

Equivalent weight $= \frac{M o l e c u l a r w e i g h t}{C h a n g e i n o x i d a t i o n s t a t e}$

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Similar questions

K M n O_{4}

K M n O_{4}

M

List - I	List - II
A) Equivalent weight of $K M n O_{4}$ in acidic medium	1) M/6
B) $K M n O_{4}$ in neutral medium	2) M/5
C) $K M n O_{4}$ in basic medium	3) M/3
D) Phosphorous acid	4) M/2
	5) M/1

K_{2} C r_{2} O_{7}

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

K_{2} C r_{2} O_{7}

= M

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