Question

When 0.1mol $Mn{O_4}^{2 - }$ is oxidized the quantity of electricity required to completely $Mn{O_4}^{2 - }$ to $Mn{O_4}^ -$ is:
(A) 96500 C
(B) $2 \times 96500{\text{ C}}$
(C) 9650 C
(D) 96.50 C

Answer 1

First find the number of electrons involved in the reaction with the help of the oxidation states of the atoms. We can say that 1 mole of electrons are equivalent to a charge of 96500 C.

Complete answer:
Here, we need to give the quantity of electricity required for the given concentration of species to get converted to another compound. So, we first need to give the complete reaction first.
- We will first find the oxidation states of Mn atom in the species in order to find the number of electrons involved in the reaction. Here, the oxidation state of O-atom is taken as (-2).
So, we can write that
Overall charge on $Mn{O_4}^{2 - }$ = Oxidation state of Mn + 4(Oxidation state of O)
-2 = Oxidation state of Mn + 4(-2)
Oxidation state of Mn = -2 + 8 = +6
Now, for $Mn{O_4}^ -$, we can write that
Overall charge on $Mn{O_4}^ -$ = Oxidation state of Mn + 4(Oxidation state of O)
-1 = Oxidation state of Mn + 4(-2)
Oxidation state of Mn = -1 +8 = +7
- So, we can write the overall reaction as below.
$Mn{O_4}^{2 - } + {e^ - } \to Mn{O_4}^ -$
So, we can write that one mole ${e^ - }$ are required to convert one mole of $Mn{O_4}^{2 - }{\text{ to Mn}}{{\text{O}}_4}^ -$.
- Faraday given that 1 mole of electrons has a charge of 96500 C.
- Thus, we can say that 1 mole of $Mn{O_4}^{2 - }$ requires 96500 C to get reduced. So, 0.1mol of $Mn{O_4}^{2 - }$ will require $\dfrac{{0.1 \times 96500}}{1} = 9650{\text{ C}}$
Thus, we obtained that the required electricity will be 9650 C.

So, the correct answer is (C).

Note:
Note that here, the oxygen atoms are in their normal form. So, their oxidation states are taken as (-2). When oxygen atoms are in the form of superoxide, their oxidation number becomes (-0.5) and when in peroxide form their oxidation number becomes (-1).