calculate the equivalent weights of kmno4 when it acts as an oxiding agent in acidic , basic and neutral media plzz give proper answer of this question
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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
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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