Question

The charge required for reducing 1 mole of $Mn{O}_4^{-}$ to $M{n}^{2+}$ is:
A. $1.93\times {10}^{5}C$
B. $2.895\times {10}^{5}C$
C. $4.28\times {10}^{5}C$
D. $4.825\times {10}^{5}C$

Answer 1

Hint: Charge has SI unit coulomb. We can define Coulomb as the quantity of electricity transported in one second by a current of one ampere. We can calculate the total charge on n mole of the electron by the formula: Q= nF, where Q denotes the charge, n is the number of moles and F is the value of Faraday.

Complete step by step solution:
- First of all, we can write the reaction as:
     $$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$$
We can see that here, $Mn{O}_4^{-}$ is 1 mole and there is 5 mole of the electron.
- Here, we can see that in $Mn{O}_4^{-}$ the oxidation state is +7 and in $M{n}^{2+}$ oxidation state is +2, so there is a change in oxidation number taking place.
- As we know that on one mole of the electron the charge is about 1 F, where F is Faraday. So, we can say that the total charge on n mole of electron Q= nF
where Q denotes the charge, n is the number of moles and F is the value of Faraday.
 - So, we will get the value of Q from here as
$$\begin{array}{rl}& Q=5mol\times 96500Cmol\\ & Q=482500C\\ & Q=4.825\times {10}^{5}C\end{array}$$

Hence, we can say that the correct option is (D), that is the charge required for reducing 1 mole of $Mn{O}_4^{-}$ to $M{n}^{2+}$ is $4.825\times {10}^{5}C$.

Note: It is found that one Faraday of electric charge corresponds to the charge on a mole of electrons, that is $6.022\times {10}^{23}$electrons, Hence, we can say that one Faraday of charge is equal to 96485.33 coulombs.