Question

If $\text{E}{{+}_{\text{F}{{\text{e}}^{2}}/\text{Fe}}}$ is ${{\text{x}}_{1}}$ $\text{E}{{+}_{\text{F}{{\text{e}}^{3}}/\text{Fe}}}$ is ${{\text{x}}^{2}}$ then what will be ${{\text{E}}^{+}}_{\text{Fe}/\text{F}{{\text{e}}^{2+}}}$ ?

Answer 1

${{\text{E}}^{+}}$ denotes the electrode potential of a cell. And energy produced in a cell is called the Gibbs free energy. Gibbs free energy is given by expression
$-\Delta \text{G}=$ nEF
Electrical energy produced in a cell. When $\Delta \text{G}$ denotes Gibbs free energy
Where $\Delta \text{G}$ denotes Gibbs free energy
nf= Number of n faraday's electricity generated by cell.
E=6MF of cell.

Complete step by step solution
The EMF of ${{\text{E}}^{+}}_{\text{F}{{\text{e}}^{2+}}/\text{Fe}}={{\text{x}}_{1}}$
The EMF of ${{\text{E}}^{+}}_{\text{F}{{\text{e}}^{3+}}/\text{Fe}}={{\text{x}}_{2}}$
We have to find ${{\text{E}}^{+}}_{\text{F}+/\text{F}{{\text{e}}^{3+}}}$ =?
Let ${{\text{E}}^{+}}_{\text{F}+/\text{F}{{\text{e}}^{3+}}}=\text{z}$
The half reaction of ${{\text{E}}^{+}}_{\text{F}{{\text{e}}^{2+}}/\text{Fe}}$ is given by
$\text{F}{{\text{e}}^{2+}}+\text{2}{{\text{e}}^{-}}\to \text{Fe}$
Gibbs free energy of cell $=-\text{nEF}$
$\Delta {{\text{G}}^{+}}=-2\text{F}{{\text{x}}_{1}}$ …… (1)
The half reaction of ${{\text{E}}^{+}}_{\text{F}{{\text{e}}^{3+}}/\text{Fe}}$ is given by
$\text{F}{{\text{e}}^{3+}}+\text{3}{{\text{e}}^{-}}\to \text{F}e$
$\Delta {{\text{G}}^{+}}=-3\text{F}{{\text{x}}_{2}}$ ……. (2)
The half reaction for ${{\text{E}}_{\text{Fe}+/\text{F}{{\text{e}}^{3+}}}}$ A
$\text{F}{{\text{e}}^{+}}-{{\text{e}}^{-}}\to \text{F}{{\text{e}}^{2+}}$
$\Delta {{\text{G}}^{+}}=-\left( \text{nFz} \right)$
$=-\text{Fz}$ ….. (3)
Adding equation (1) & (2) and comparing with (3)
$=-\text{Fz}=-2\text{F}{{\text{x}}_{1}}+\left( -\text{3}{{\text{x}}_{1}} \right)$
$=-\text{Fz}=-\text{F}\left[ 2{{\text{x}}_{1}}+3{{\text{x}}_{2}} \right]$
$\text{z}=2{{\text{x}}_{4}}+3{{\text{x}}_{2}}$
$\therefore {{\text{E}}^{+}}_{\text{F}{{\text{e}}^{+}}/\text{F}{{\text{e}}^{2}}^{+}}=2{{\text{x}}_{1}}+3{{\text{x}}_{2}}$ .

Note
To solve this equation the expression for the Gibbs free energy must be remember, i.e. $\Delta \text{G}=-\text{nEF}$ where (F) Faraday's constant and the value of faraday’s is equal to $96485\text{ C mo}{{\text{l}}^{-1}}$ .
This constant represents the magnitude of electric charge per mole of electrons. Gibbs free energy is associated with a chemical reaction that can be used to do work. It is a Thermodynamic quantity. Negative sign in Gibbs free energy tells us that reactant has more free energy than product.