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What will be the equivalent weight of KMnO4 in acidic, basic and neutral medium?

IN ACIDIC MEDIUM:

K+1Mn+7O4-2 ------ Mn+2S+6O4-2

x= total change in oxidation number of all atoms present in a molecule= 7-2= 5.

equivalent weight= (molecular weight)/x= 157/5u.

IN BASIC MEDIUM:

K+1Mn+7O4-2 -------- K2+1Mn+6O4-2

x= 7-6= 1.

Equivalent weight= (molecular weight)/x= 157u.

IN NEUTRAL MEDIUM:

K+1Mn+7O4-2 --------- Mn+4O2-2

x= 7-4= 3

equivalent weight= (molecular weight)/x= 157/3u

HOPE THIS HELPS..

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because Mn is the only atom present on both sides of the eq. but the charge changes...... so we take x as the change in the value of charge of Mn.

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IN ACIDIC MEDIUM: K+1Mn+7O4-2 ------ Mn+2S+6O4-2 x= total change in oxidation number of all atoms present in a molecule= 7-2= 5. equivalent weight= (molecular weight)/x= 157/5u. IN BASIC MEDIUM: K+1Mn+7O4-2 -------- K2+1Mn+6O4-2 x= 7-6= 1. Equivalent weight= (molecular weight)/x= 157u. IN NEUTRAL MEDIUM: K+1Mn+7O4-2 --------- Mn+4O2-2 x= 7-4= 3 equivalent weight= (molecular weight)/x= 157/3u
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