Question

What is the standard electrode potential for the electrode $Mn{{O}_{4}}^{-}/Mn{{O}_{2}}$ in solution? (Given: ${{E}^{\circ }}_{Mn{{O}_{4}}^{-}/M{{n}^{2+}}}=1.51V$ ,${{E}^{\circ }}_{Mn{{O}_{2}}/M{{n}^{2+}}}=1.23V$)

Answer 1

It will require 3 electrons to convert $Mn{{O}_{4}}^{-}$ to $Mn{{O}_{2}}$ . 5 electrons will be required to convert $Mn{{O}_{4}}^{-}$ to $M{{n}^{2+}}$ and 2 electrons will be required to convert $Mn{{O}_{2}}$ to $M{{n}^{2+}}$ . We can relate the potential of these electrodes and number of electrons involved in the reaction by an equation.

Complete step by step answer:
Here, we are given to find the potential of $Mn{{O}_{4}}^{-}/Mn{{O}_{2}}$ electrode. Let’s find the oxidation state of Mn in both of the compounds in order to give a reaction in terms of Mn atom’s oxidation state.

- We can write for $Mn{{O}_{4}}^{-}$ that
Overall charge = Oxidation state of Mn + 4(Oxidation state of O)
-1 = Oxidation state of Mn + 4(-2)
-1 = Oxidation state of Mn – 8
Oxidation state of Mn = 8 -1 = +7
And for $Mn{{O}_{2}}$, we can write that
Overall charge = Oxidation state of Mn + 2(Oxidation state of O)
0 = Oxidation state of Mn + 2(-2)
0 = Oxidation state of Mn – 4
Oxidation state of Mn = +4

Now, the reaction of electrode $Mn{{O}_{4}}^{-}/Mn{{O}_{2}}$ can be written as follows.
$M{{n}^{7+}}+3{{e}^{-}}\to M{{n}^{4+}}$
So, we need to find the electrode potential of the upper given reaction and we are given potentials of two other electrodes. So, let’s write their reactions also.
For $Mn{{O}_{4}}^{-}/M{{n}^{2+}}$ electrode whose potential is given as 1.51V, we can write its reaction as
$M{{n}^{7+}}+5{{e}^{-}}\to M{{n}^{2+}}$
For, $Mn{{O}_{2}}/M{{n}^{2+}}$ electrode whose potential is given as 1.23V, we can write its reaction as
$M{{n}^{4+}}+2{{e}^{-}}\to M{{n}^{2+}}$
From the above given three reactions, we can write that product of number of electrons involved and potential of $Mn{{O}_{4}}^{-}/M{{n}^{2+}}$ electrode will be equal to the sum of the numbers of electrons involved and the potential of $Mn{{O}_{2}}/M{{n}^{2+}}$ and $Mn{{O}_{4}}^{-}/Mn{{O}_{2}}$ electrodes. So, we can write that

 $$5\times 1.51=2\times 1.23+3\times E$$   
 $$7.55=2.46+3\times E$$   

So, we can write that $E=\dfrac{7.55-2.46}{3}=\dfrac{5.09}{3}=1.70$V
Thus, we can conclude that the potential of $Mn{{O}_{4}}^{-}/Mn{{O}_{2}}$ electrode will be 1.70V.

Note: Remember that the number of electrons required or produced in the half reaction of the electrode is necessary to be found out in order to proceed further in solving these types of problems. First finding out the oxidation state of an element in the electrode reaction makes it easy to find the number of electrons required or produced during the reaction.