Question

If $E{{{}^\circ }_{F{{e}^{+2}},Fe}}$ is ${{x}_{1}}$ , \[E{{{}^\circ }_{F{{e}^{+3}},Fe}}\] is ${{x}_{2}}$ , then \[E{{{}^\circ }_{F{{e}^{+3}},F{{e}^{+2}}}}\] will be
A. $3{{x}_{2}}-2{{x}_{1}}$
B. ${{x}_{2}}-{{x}_{1}}$
C. ${{x}_{2}}+{{x}_{1}}$
D. $2{{x}_{1}}-3{{x}_{2}}$

Answer 1

Hint: Standard cell potential is defined as the potential of the cell under standard conditions, that are, concentrations of one mole per litre, pressure of $1\,atm$ at $25{}^\circ C$ . To get the overall standard cell potential, the potentials of half – cells are added.

Complete step by step answer:
Here, it is given that $E{{{}^\circ }_{F{{e}^{+2}},Fe}}$ is ${{x}_{1}}$ and \[E{{{}^\circ }_{F{{e}^{+3}},Fe}}\] is ${{x}_{2}}$
So, firstly we will write the chemical reaction as follows:
$F{{e}^{2+}}+2{{e}^{-}}\to Fe$ $E{{{}^\circ }_{F{{e}^{+2}},Fe}}={{x}_{1}}$ $-(1)$
$F{{e}^{3+}}+3{{e}^{-}}\to Fe$ \[E{{{}^\circ }_{F{{e}^{+3}},Fe}}={{x}_{2}}\] $-(2)$
Now, we will reverse the $(1)$ equation, we will get
$Fe\to F{{e}^{2+}}+2e$ $E{{{}^\circ }_{Fe,F{{e}^{+2}}}}=-{{x}_{1}}$
And now we will write both the equation $(1)$ and $(2)$
\[Fe\to F{{e}^{2+}}+2{{e}^{-}}\] $E{{{}^\circ }_{Fe,F{{e}^{+2}}}}=-{{x}_{1}}$
\[F{{e}^{3+}}+3{{e}^{-}}\to Fe\] $E{{{}^\circ }_{F{{e}^{+3}},Fe}}={{x}_{2}}$
Solving both the equations, we get,
\[F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}}\]
As we know $E{}^\circ $ is an intensive property, therefore,
\[\Delta G{}^\circ =\Delta G{{{}^\circ }_{1}}+\Delta G{{{}^\circ }_{2}}\]
Also , $\Delta G{}^\circ =-nFE{}^\circ $
\[-nFE{}^\circ =-{{n}_{1}}FE{{{}^\circ }_{1}}-nFE{{{}^\circ }_{2}}\]
\[\Rightarrow nFE{}^\circ ={{n}_{1}}FE{{{}^\circ }_{1}}+nFE{{{}^\circ }_{2}}\]
on solving , we get
\[\Rightarrow E{}^\circ =\dfrac{{{n}_{1}}E{{{}^\circ }_{1}}+nE{{{}^\circ }_{2}}}{n}\]
where, \[G{}^\circ \] is the standard Gibbs free energy,
$E{}^\circ $ is the standard reduction potential,
$F$ is the Faraday’s constant, and,
$n$ is the number of electrons transferred.
Now, we will substitute the above values in the above formula.
 $E{}^\circ =\dfrac{-2\times {{x}_{1}}+3\times {{x}_{2}}}{1}$
 $\Rightarrow E{}^\circ =-2{{x}_{1}}+3{{x}_{2}}\Rightarrow 3{{x}_{2}}-2{{x}_{1}}$

So, the correct answer is Option A.

Note: 1.\[\Delta G{}^\circ \] is defined as the energy change under standard conditions like pressure at $1atm$ .
2.\[\Delta G\] is the Gibbs free energy which is used for the measurement of energy content. If \[\Delta G\] is less than zero then, it gives an exothermic reaction. If \[\Delta G\] is more than zero, then it gives an endothermic reaction.
3.Standard cell potential $(E{}^\circ )$ is defined as the potential of a cell under concentration of $1$mole per litre and pressure of $1atm$at $25{}^\circ C$ .
4.Faraday’s constant is denoted with the symbol $F$ and is defined as change in coulombs. One Faraday constant is equal to $96500$ $Cmo{{l}^{-1}}$.
$1F=96500$ $Cmo{{l}^{-1}}$
Here, $n$ is the number of electrons transferred in a reaction.
5.The relationship between $\Delta G{}^\circ ,\Delta E{}^\circ ,n,F$ is:
$\Delta G{}^\circ =-nFE{}^\circ $
When we reverse a chemical reaction, its $E{}^\circ $ value gets opposite of the previous $E{}^\circ $ value given.