KMnO_4 reacts in alkaline medium as follows:2 | vedclass

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(D) In alkaline medium,the reduction reaction is:$MnO_4^- + e^- 	o MnO_4^{2-}$The oxidation state of $Mn$ changes from $+7$ to $+6$.The change in oxi

Vedclass · Accepted Answer

$158$

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(D) In alkaline medium,the reduction reaction is:$MnO_4^- + e^- 	o MnO_4^{2-}$The oxidation state of $Mn$ changes from $+7$ to $+6$.The change in oxidation number ($n$-factor) is $|7 - 6| = 1$.The equivalent weight of $KMnO_4$ is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$Molar mass of $KMnO_4 = 39 + 55 + (4 	imes 16) = 158 \ g/mol$.$	ext{Equivalent weight} = \frac{158}{1} = 158$.

$KMnO_4$ reacts in alkaline medium as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
What is the equivalent weight of $KMnO_4$?
(Atomic weights: $K = 39, Mn = 55, O = 16$)

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$KMnO_4$ reacts in alkaline medium as follows:$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$What is the equivalent weight of $KMnO_4$?(Atomic weights: $K = 39, Mn = 55, O = 16$)