Question

calculate the equivalent weights of kmnO4 when it acts as an oxidising agent in acidic , basic and neutral media

Answer 1

☆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