Question

For the following cell, ${\text{Zn}}\left( {\text{s}} \right)\left| {{\text{ZnS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right)} \right.\left\| {{\text{CuS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.$, when the concentration of ${\text{Z}}{{\text{n}}^{2 + }}$ is $\text{10}$ times the concentration of ${\text{C}}{{\text{u}}^{2 + }}$, the expression for $\Delta G$ (in \[{\text{J mo}}{{\text{l}}^{ - 1}}\]) is:
[$F$ is Faraday constant; $R$ is gas constant; $T$ is temperature; ${E^ \circ }\left( {{\text{cell}}} \right) = 1.1{\text{ V}}$]
A. $2.303RT - 2.2F$
B. $2.303RT + 1.1F$
C. $1.1F$
D. $ - 2.2F$

Answer 1

Hint:The Gibbs free energy change $\left( {\Delta G} \right)$ of the system is the amount of energy released when the reactants are converted to products under standard conditions. We have to derive an expression for $\Delta G$ using the Nernst equation.

Complete answer:
We know the Nernst equation is as follows:
${E_{cell}} = E_{cell}^ \circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[{\text{Reduction}}]}}{{[{\text{Oxidation}}]}}$
where
${E_{cell}}$ is the potential of cell,
$E_{cell}^ \circ $ is the standard potential of cell,
$R$ is the universal gas constant,
$T$ is the temperature,
$n$ is the number of moles of electron,
$F$ is Faraday's constant.
Calculate the potential of the cell using the Nernst equation as follows:
In the given cell, zinc is getting reduced and copper is getting oxidised. Thus,
${E_{cell}} = E_{cell}^ \circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[{\text{Z}}{{\text{n}}^{2 + }}]}}{{[{\text{C}}{{\text{u}}^{2 + }}]}}$
We are given that the concentration of ${\text{Z}}{{\text{n}}^{2 + }}$ is 10 times the concentration of ${\text{C}}{{\text{u}}^{2 + }}$. Thus, $\dfrac{{[{\text{Z}}{{\text{n}}^{2 + }}]}}{{[{\text{C}}{{\text{u}}^{2 + }}]}} = \dfrac{{10}}{1}$.
Thus, the equation becomes as follows:
${E_{cell}} = E_{cell}^ \circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{10}}{1}$
Substitute $1.1{\text{ V}}$ for the standard potential of the cell. Thus,
$\Rightarrow {E_{cell}} = 1.1{\text{ V}} - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{10}}{1}\\$
$\Rightarrow {E_{cell}} = \dfrac{{1.1nF - 2.303RT}}{{nF}} \times 1 \\$
$\Rightarrow nF\,{E_{cell}} = 1.1nF - 2.303RT \\ $
$\Rightarrow - nF\,{E_{cell}} = 2.303RT - 1.1nF \\ $
The cell potential $\left( {{E_{cell}}} \right)$ and the free energy change $\left( {\Delta G} \right)$ accompanying an electrochemical reaction are related by:
$\Delta G = - nF{E_{cell}}$
where
$\Delta G$ is the free energy change,
$n$ is the number of moles of electrons involved,
$F$ is the Faraday’s constant
${E_{cell}}$ is the cell potential.
Thus, $\Delta G = 2.303RT - 1.1nF$
Substitute $2{\text{ mol}}$ for the number of moles of electrons.
$\Rightarrow \Delta G = 2.303RT - 2.2F \\ $
Thus, the expression for $\Delta G$ (in \[{\text{J mo}}{{\text{l}}^{ - 1}}\]) is $2.303RT - 2.2F$.

Thus, the correct option is (A) $2.303RT - 2.2F$.

Note:
The Nernst equation is applicable to single electrode reduction or oxidation potentials at any condition, standard electrode potentials, electromotive force of an electrochemical cell, unknown ionic concentrations, determining feasibility of electrochemical cells, etc. The equation is not applicable to the solutions having high concentrations and the equation cannot be used to measure the cell potential when current is flowing through the electrode.