Determine the equivalent masses of the following underline compounds by both oxidantion number and electrons methods:
underline`(KMnO_(4))underline+(FeSO_(4))+H_(2)SO_(4)K_(2)SO_(4)+MnSO_(4)+Fe_(2)(SO_(4))+H_(2)O`

Explanation

Oxidation number method :
`Koverset(+7)MnO_(4)+overset(+2)FeSO_(4)+H_(2)SO_(4)toK_(2)SO_(4)+overset(+2)Fe_(2)(SO_(4))_(3)+H_2O`
In this reaction ,`KMnO_(4)` undergoes reduction because the oxidation number of Mn decreases from +7 to +2 .So ,change in oxidation number of Mn =7-2=5 units .
`therefore`Equivalent mass `KNnO_(4)`
`("Molecular mass of " KMnO_(4))/("change in oxidatin per molecule of "(MnO_(4)" because of its reduction")`
=`(158)/(5)=31.6`
In the reaction ,FesO_(4)undergoes oxidation because oxdiation number of Fe increases from +2 to +3 .SO change in oxidation number of Fe =3-2=1 unit
Equivalent of mass of `FesO_(4)`
`("Molecular mass of " FeSO_(4))/("Change in oxidatin per molecule of "(FeSO_(4)" because of its reduction")`
=`(151.85)/(1)=151.85`
Electronic method :
`MnO-_(4)+8H^(+)+5eto Mn^(+)+4H_(2)O`
`therefore` Equivalnet mass of `KNnO_(4)`
`("Molecular mass of "KNno_(4))/("Number of electrons gained in reduction of one molecule of " KMNO_(4))`
=`(158.85)/(5)`
`Fe^(2+)to Fe^(3)+e`
`therefore` Equivalnet mass of `FeSO_(4)`
`("Molecular mass of "FeSO_(4))/("Number of electrons lost in oxidation of one molecule of "FeSO_(4))`
=`(151.85)/(1)=151.85`