Question

Given, $E_{F{e^{3 + }}/Fe}^ \circ = - 0.036V,E_{F{e^{2 + }}/Fe}^ \circ = - 0.439V$. The value of standard electrode potential for the change, $F{e^{3 + }}_{(aq)} + {e^ - } \to F{e^{2 + }}_{(aq)}$ will be:
(A) -0.403V
(B) 0.385V
(C) 0.770V
(D) -0.270V

Answer 1

Hint: We can use the relation between the free energy change of the reaction and the standard potential of the cell which can be shown as
\[\Delta G = - nF{E^ \circ }\]

Complete step by step solution:
Here, we can see that standard potentials of two reduction reactions are given to us. Now, we need to find the standard potential of the same element for it’s another reduction reaction.
- Here, we will use the relation of free energy of the reaction and the cell potential which can be given as
\[\Delta G = - nF{E^ \circ }{\text{ }}..{\text{(1)}}\]
- For the reaction $F{e^{3 + }}_{(aq)} + 3{e^ - } \to F{e_{(s)}}$, we are given its potential which is $E_{F{e^{3 + }}/Fe}^ \circ = - 0.036V$. We can see that n (number of electrons involved in reaction) is 3 here. Putting all the values for this reaction in the equation (1), we get
\[\Delta G = - (3)(F)( - 0.036)\]
So,
\[\Delta G = 0.108F{\text{ }}.......{\text{(2)}}\]
- For the reaction $F{e^{2 + }}_{(aq)} + 2{e^ - } \to F{e_{(s)}}$, we are given that the $E_{F{e^{2 + }}/Fe}^ \circ = - 0.439V$. We can see that for this reaction, n is 2. So, for this reaction, we can write the equation (1) as
\[\Delta G = - (2)(F)( - 0.439)\]
So, we can write that
\[\Delta G = 0.878F{\text{ }}......{\text{(3)}}\]
Now, we need to find the potential for the $F{e^{3 + }}_{(aq)} + {e^ - } \to F{e^{2 + }}_{(aq)}$ reaction. As we subtract the equation (3) from equation (2), we will get
\[\Delta G = 0.108F - 0.878F = - 0.770F\]
So, we found the change in free energy for $F{e^{3 + }}_{(aq)} + {e^ - } \to F{e^{2 + }}_{(aq)}$ reaction. We can see that the number of electrons involved in this reaction is 1. So, we can write the equation (1) in terms of this reaction as
\[ - 0.770F = - (1)(F)(E_{F{e^{3 + }}/F{e^{2 + }}}^ \circ )\]
So, we can write that
\[E_{F{e^{3 + }}/F{e^{2 + }}}^ \circ = \dfrac{{0.770F}}{F} = 0.770V\]
Thus, we obtained that the potential for the reaction $F{e^{3 + }}_{(aq)} + {e^ - } \to F{e^{2 + }}_{(aq)}$ is 0.770V.

So, the correct answer is (C).

Note: Here, it is not necessary to put the value of F (Faraday’s constant) which is 96500 in the equation (1). If you put the value of F in the calculation, then also it will be correct but just to avoid more calculation, we avoided putting the absolute value of F.