Question

Total charge required for the oxidation of two moles of $M{n_3}{O_4}$ into $MnO_4^{2 - }$ in presence of alkaline medium is:
A) 5F
B) 10F
C) 20F
D) None of these

Answer 1

Hint: To solve this question we’ll first have to write the balanced chemical formula for the Oxidation of $M{n_3}{O_4}$ to $MnO_4^{2 - }$ . The charge required can be given by the formula: Charge = $n \times F$, where n is the no. of electrons transferred and F is the faraday unit which is equal to 96500 C.

Complete answer:
First, we’ll find the Oxidation numbers of Mn in $M{n_3}{O_4}$ and $MnO_4^{2 - }$ individually. Consider the Oxidation no. of Mn to be ‘x’ in both. For $M{n_3}{O_4}$ we can say that: $3x + 4( - 2) = 0 \to x = \dfrac{8}{3}$
For $MnO_4^{2 - }$ the Oxidation Number for Mn (x) will be equal to: $x + 4( - 2) = - 2 \to x = + 6$
The basic equation hence will be: $\mathop {M{n_3}{O_4}}\limits_{ + 8/3} \to \mathop {MnO_4^{2 - }}\limits_{ + 6} $
Given that the reaction occurs in the basic medium we will add $O{H^ - }$ on the LHS. Also, balancing the number of Mn we will get the equation as:
$M{n_3}{O_4} + O{H^ - } \to 3MnO_4^{2 - }$
Balancing the number of Oxygen on both sides and adding water to compensate for the additional Oxygen and Hydrogen. The equation becomes: $M{n_3}{O_4} + 16O{H^ - } \to 3MnO_4^{2 - } + 8{H_2}O$
Now to balance the charges on both sides we’ll first find out the change in the charge on both sides. The difference in the charges $\Delta $ can be given as the difference between the charges of Mn on the product side and the reactant side.
$\Delta = |3 \times \left( { + \dfrac{8}{3}} \right) - 3 \times ( + 6)| = 10$
Therefore, the number of electrons transferred will be 10. Adding the electrons in the above reaction we get the final reaction as: $M{n_3}{O_4} + 16O{H^ - } \to 3MnO_4^{2 - } + 8{H_2}O + 10{e^ - }$
From this reaction we conclude that 1 mole of $M{n_3}{O_4}$ will require 10 electrons. Hence two moles of $M{n_3}{O_4}$ will require: $2 \times 10 = 20{e^ - }$
For one mole of $M{n_3}{O_4}$ the charge needed is: $n \times F = 10 \times F$
For two moles of $M{n_3}{O_4}$ the charge required is: $2 \times 10 \times F = 20F$
The correct answer is Option (C).

Note:
To balance equations in the basic medium, remember to balance them in the acidic medium first. Then adding the equivalent amount of $O{H^ - }$ on the opposite side will give us the reaction in Basic medium.