Question

Equivalent mass of $ Fe{S_2} $ in the following reaction (M: Mole Mass of $ Fe{S_2} $ ) is
 $ Fe{S_2} + {O_2} \to FeS{O_4} + S{O_2} $
A. $ \dfrac{M}{{12}} $
B. $ \dfrac{M}{{11}} $
C. $ \dfrac{M}{6} $
D. $ \dfrac{M}{4} $

Answer 1

Hint: To calculate the equivalent mass of $ Fe{S_2} $ we have to calculate the total number of electrons involved in the redox reaction. Redox reaction means both the oxidation and reduction takes place in the same reaction.
 $ Equivalent\,weight = \dfrac{{Molecular\,weight}}{{n\,factor}} $
 $ {\text{n }}factor $ of a redox reaction is the number of electrons involved by one mole of that compound.

Complete answer:
Now, we have to calculate the oxidation number $ \left( {O.N} \right) $ in the reaction. Oxidation number of an atom is the total number of electrons involved to form the bond. So, the oxidation number of an atom depends on the atom or molecule with which it was attached. Oxidation number of free atoms is equal to $ 0 $ .
O.N of $ Fe $ in $ Fe{S_2} = + 2 $
O.N of $ S $ in $ Fe{S_2} = - 1 $
O.N of $ Fe $ in $ FeS{O_4} = + 3 $
O.N of $ S $ in $ S{O_2} = + 4 $
 $ F{e^{ + 2}}{S_2}^{ - 1} + {O_2} \to F{e^{ + 3}}S{O_4} + {S^{ + 4}}{O_2} $
 $ F{e^{2 + }} \to F{e^{3 + }} $ 1 electron lost in this reaction (means 1 electron involved in this part of the reaction)
 $ {S^{ - 1}} \to {S^{ + 4}} $ 5 electrons gained in this reaction (means 5 electrons involved in this part of the reaction)
But in the question it was asked that molecular weight of $ Fe{S_2} $ where no of $ S $ atoms is two
So number of electrons involved by $ S $ atom is 10 ( $ {S^{ - 1}} \to {S^{ + 4}} $ 5 electrons gained in this reaction $ \times 2 $ )
Now total electron involved in $ Fe{S_2} $ is electron involved in $ Fe $ part $ + $ electron involved in $ S $ part
 $ = 1 + 10 = 11 $
 $ equivalent\,weight = \dfrac{{Molecular\,weight}}{{11}} $
So the correct answer is (B) $ \dfrac{M}{{11}} $ .

Note:
In gravimetric analysis, the phrase "equivalent weight" meant the quantity of precipitate that corresponded to one gram of analyte (the species of interest). The various definitions arose from the habit of expressing gravimetric results as mass fractions of the analyte, which were frequently reported as percentages.