Question

In the redox ionic equation, the equivalent weight of potassium permanganate of molar mass Mg/mol is:
 $ 2Mno_4^ - + B{r^ - } + {H_2}0 \to 2Mn{O_2} + BrO_3^ - + 2O{H^ - } $
(A) $ \dfrac{M}{5} $
(B) $ \dfrac{M}{3} $
(C) $ \dfrac{M}{6} $
(D) $ \dfrac{M}{2} $

Answer 1

Hint :As we know that the formula for Equivalent weight is equal to the molecular weight divided by the change in the number of oxidation state of the compound. We can also calculate the change in the oxidation number which is obtained by adding the charges of all the elements.

Complete Step By Step Answer:
From the reaction given in the question $ 2Mno_4^ - + B{r^ - } + {H_2}0 \to 2Mn{O_2} + BrO_3^ - + 2O{H^ - } $ ,
the oxidation number of Manganese changes from $ + 7 $ (in $ Mno_4^ - $ ) to $ + 4 $ (in $ Mn{O_2} $ ) which can be calculated as shown below respectively:
 $ x + ( - 8) = - 1 = + 7 $ and $ x + ( - 4) = 0 = + 4 $
Therefore, on subtracting we get the total change in the oxidation number which is equal to $ 3 $ .
We also know that the equivalent weight can be given by the ratio of the total molecular mass of the substance to the change in oxidation number of the compound. Thus, we will get the equivalent weight of Potassium permanganate as: $ \dfrac{M}{3} $ .
Therefore, from here we easily get the answer which is $ \dfrac{M}{3} $ .
Therefore the correct option is option (b).

Additional Information:
Equivalent weight, in chemistry, is the quantity of the substance that exactly reacts with or is equal to the combining value of any arbitrarily fixed quantity. Substances react with each other in stoichiometric, or chemically equivalent, proportions. The equivalent weight of a substance may vary with the type of reaction it undergoes. Thus, potassium permanganate reacts on the double decomposition and has an equivalent weight equal to its gram molecular weight. It acts as an oxidizing agent under different circumstances. The number of the equivalent weights of any substance dissolved in one litre of solution is referred to as the normality of the solution.
The formula for the general number of equivalents formula is:
 $ E = \dfrac{{MW}}{{oxidation\,number}} $
Where, MW is the molecular weight of the compound.

Note :
Always remember that the Equivalent weight is given by the ratio of molecular mass or molecular weight of the particular element to the change in oxidation number of that element present in the given compound.