Question

If ${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 2}}/{\text{Fe]}}}$ is ${{\text{x}}_1}$ , ${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{Fe]}}}$ is ${{\text{x}}_2}$ then ${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{F}}{{\text{e}}^{ + 2}}]}$ will be
A. $3{{\text{x}}_2} - 2{{\text{x}}_1}$
B. ${{\text{x}}_2} - {{\text{x}}_1}$
C. ${{\text{x}}_2}{\text{ + }}{{\text{x}}_1}$
D. $2{{\text{x}}_1} + 3{{\text{x}}_2}$

Answer 1

Considering the Gibbs free energy, to get the total free energy, all the free energy is added together. The potential and Gibbs free energy is related. The Gibbs free energy is equal to the product of electrode potential and nF, where n is the number of electrons transferred and F is the Faraday constant.

Complete answer:
We are given the following details.
${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 2}}/{\text{Fe]}}} = {{\text{x}}_1}$
${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{Fe]}}} = {{\text{x}}_2}$
Now we can calculate the Gibbs free energy,
$\vartriangle {{\text{G}}^0}{\text{ = - nF}}{{\text{E}}^0}$
Where $\vartriangle {{\text{G}}^0}$ is the standard Gibbs free energy
${\text{n}}$ is the number of electrons transferred
${\text{F}}$ is the Faradays constant
${{\text{E}}^0}$ is standard electrode potential or the emf under standard conditions
Calculating the Gibbs free energy of first half reaction, the half reaction is given as
${\text{F}}{{\text{e}}^{2 + }} + 2{\text{e - }} \to {\text{Fe}}$
Therefore $\vartriangle {{\text{G}}^0} = - 2{\text{F}} \times {{\text{X}}_1}$ is equation 1
Now the Gibbs free energy of second half reaction is calculated
${\text{F}}{{\text{e}}^{3 + }} + 3{\text{e - }} \to {\text{Fe}}$
Therefore $\vartriangle {{\text{G}}^0} = - 3{\text{F}} \times {{\text{X}}_2}$ is equation 2
Now the last half reaction is taken
${\text{F}}{{\text{e}}^{3 + }} + {\text{e - }} \to {\text{F}}{{\text{e}}^{2 + }}$
$\vartriangle {{\text{G}}^0} = - 1{\text{F}} \times {{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{F}}{{\text{e}}^{ + 2}}]}$ is equation 3
The total Gibbs free energy is given by,
$\vartriangle {{\text{G}}^0} = \vartriangle {{\text{G}}^0}_1 + \vartriangle {{\text{G}}^0}_2$
Now the equation 1 is reversed and added with equation 2 to get the equation 3
$- 1{\text{F}} \times {{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{F}}{{\text{e}}^{ + 2}}]} = 2{\text{F}} \times {{\text{X}}_1} + ( - 3{\text{F}} \times {{\text{X}}_2})$
The constant F gets cancelled from both the sides
Solving the above equation, we get,
${{\text{E}}^0}_{[{\text{F}}{{\text{e}}^{ + 3}}/{\text{F}}{{\text{e}}^{ + 2}}]} = 3{{\text{X}}_2} - 2{{\text{X}}_1}$
And hence the answer is option A.

Additional information: The exponent 0 denotes that the certain value is at standard conditions. If instead of the unknown values, a number was given then the value of faraday is used or can be eliminated as it can be cancelled from both the sides. The other two values are not constant and change with the half reaction. The value of faraday constant is 96485 denoted by the symbol F.

Note:
Write the half reactions. Then calculate the Gibbs free energy of the half reaction. The total Gibbs free energy gives the electrode unknown potential.