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he equivalent weight of MnSO4 is half of its molecular weight when it converts toa. Mn¬2O3 b. MnO2 c. 〖MnO〗_4^- d. 〖MnO〗_4^(2-)

he equivalent weight of MnSO4 is half of its molecular weight when it converts toa. Mn¬2O3
b. MnO2
c. 〖MnO〗_4^-
d. 〖MnO〗_4^(2-)

Grade:upto college level

3 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Sol. Equivalent weight in redox system is defined as : E= (Molar mass)/(n-factor) Here n-factor is the net change in oxidation number per formula unit of oxidizing or reducing agent. In the present case-n-factor is 2 because equivalent weight is half of molecular weight. Also, MnSO4 → 1/2 Mn2O3 1 (+ 2 → + 3) MnSO4 → MnO2 2 (+ 2 → + 4) MnSO4 → MnO_4^- 5 (+ 2 → + 7) MnSO4 → MnO_4^(2-) 4 (+ 2 → + 6) Therefore,MnSO4 converts to MnO2.
adi
23 Points
8 years ago
firstly guys we know that net oxidation state of Mn in MnSo4  is Mn+(SO4)=0
so,Mn+(-2)=0 (net charge of SO4=-2)
Mn=2
now net charge of Mn in MnO2=
Mn+(-2)2 (net charge on O=-2
Mn=4
MnSO4-----MnO2
equivalent weight =molecular weight (let M)/change in the oxidation state
equivalent weight=M/(4-2)
=M/2 
Thus equivalent weight of MnSO4 is half the molecular weight when it is converted to MnO2
ankit singh
askIITians Faculty 614 Points
3 years ago
Also,MnSO4 → 1/2 Mn2O3 1 (+ 2 → + 3)MnSO4 → MnO2 2 (+ 2 → + 4)MnSO4 → MnO_4^- 5 (+ 2 → + 7)MnSO4 → MnO_4^(2-) 4 (+ 2 → + 6)Therefore,MnSO4 converts to MnO2.
 
 
firstly guys we know that net oxidation state of Mn in MnSo4  is Mn+(SO4)=0
so,Mn+(-2)=0 (net charge of SO4=-2)
Mn=2
now net charge of Mn in MnO2=
Mn+(-2)2 (net charge on O=-2
Mn=4
MnSO4-----MnO2
equivalent weight =molecular weight (let M)/change in the oxidation state
equivalent weight=M/(4-2)
=M/2 
Thus equivalent weight of MnSO4 is half the molecular weight when it is convert

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