Question

Calculate the equivalent weight of KMnO4 and MnO2. And explain it.

Answer 1

Molecular mass of KMnO4 = 158 gm
In acidic medium, no. of electrons involved n is:
   (MnO4)^-  + 8H^+ +5e^-   ---> Mn^+2 + 4H2O
So, equivalent mass = molar mass/ electrons involved
                                 = 158/ 5 = 31.6 gm/ equivalent

 In strongly basic medium, 
(MnO4)^-  + 1e^-  ---> (MnO4)^-2
 So, n = 1 , equivalent weight = 158/1 = 158gm/ equivalent

In neutral medium (MnO4)^-  + 4H2O +3e^-  ---> MnO2 + 4OH^-
 Here, n = 3 so, equivalent weight = 158/3 = 52.66 gm/equivalent

 For, MnO2  (molar mass = 87) , here  n = 3
So, equivalent weight = molar mass/ 3 = 87/3 = 29 gm/equivalent. 


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