Question

The charge required for reducing 1 mole of $MnO_{4}^{-}$ 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:
     \[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
We can see that here, $MnO_{4}^{-}$ is 1 mole and there is 5 mole of the electron.
- Here, we can see that in $MnO_{4}^{-}$ 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
\[\begin{align}
  & Q=5\text{ }mol\times 96500\text{ }Cmol \\
 & Q=482500C \\
 & Q=4.825\times {{10}^{5}}\text{ }C \\
\end{align}\]

Hence, we can say that the correct option is (D), that is the charge required for reducing 1 mole of $MnO_{4}^{-}$ 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.