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

Hint: $ {{\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.