Question

Write Nernst equation for Daniel cell at equilibrium condition and derive relationship between ${\text{E}}_{{\text{cell}}}^{\text{o}}$ and equilibrium constant ${{\text{K}}_{\text{C}}}$.

Answer 1

The Daniel cell is an electrochemical cell consisting of zinc anode and copper cathode. Zinc oxidised at anode and copper reduce at cathode. Electrochemistry gives the relation between Gibbs free energy and standard reduction potential of the cell. Thermodynamics give the relation between Gibbs free energy and equilibrium constant, by comparing both equation we can determine the relation in ${\text{E}}_{{\text{cell}}}^{\text{o}}$ and equilibrium constant${{\text{K}}_{\text{C}}}$.

Complete step by step solution: The Nernst equation is used to determine the electromotive force (emf) of a cell. The Nernst is represented as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{{\text{n}}}\log \,\dfrac{{\left[ {{\text{reduced}}} \right]}}{{\left[ {{\text{oxidized}}} \right]}}$
${{\text{E}}_{{\text{cell}}\,}}$ is the electromotive force of the cell.
${\text{E}}_{{\text{cell}}}^ \circ \,$ is the standard reduction potential of the cell.
${\text{n}}$ is the number of electrons involved in a redox reaction.
The zinc-copper electrochemical cell is known as Daniel cell.
The Nernst equation for Daniel cell at equilibrium condition is as follows:
The Denial cell is represented as follows:
${\text{Zn/Z}}{{\text{n}}^{{\text{2 + }}}}{\text{//C}}{{\text{u}}^{{\text{2 + }}}}{\text{/Cu}}$
In this cell, zinc electrode works as anode and copper electrode works as cathode.
We can write the half-cell reaction to determine the oxidised and reduced species.
The half-cell reaction are as follows:
Oxidation reaction:${\text{Zn}}\left( {\text{s}} \right)\, \to {\text{Z}}{{\text{n}}^{2 + }}{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\,$
Zinc is releasing two electrons to form zinc ions.
Reduction reaction: ${\text{C}}{{\text{u}}^{\text{ + }}}\,{\text{ + }}\,\,{\text{2}}{{\text{e}}^ - }\, \to {\text{Cu}}\,\left( {\text{s}} \right)$
Copper ions are accepting two electrons to form copper metal.
So, the number of electrons exchanged during the reaction is$2$.
So, the Nernst equation for Daniel cell is as follows:
${{\text{E}}_{{\text{cell}}\,}}{\text{ = }}\,{\text{E}}_{{\text{cell}}}^ \circ \, - \dfrac{{0.0591}}{2}\log \,\dfrac{{\left[ {{\text{Z}}{{\text{n}}^{2 + }}} \right]}}{{\left[ {{\text{C}}{{\text{u}}^{2 + }}} \right]}}$

The relationship between ${\text{E}}_{{\text{cell}}}^{\text{o}}$ and equilibrium constant${{\text{K}}_{\text{C}}}$ is as follows:
The relation between Gibbs free energy and ${\text{E}}_{{\text{cell}}}^{\text{o}}$is as follows:
$\Delta {{\text{G}}_{\text{o}}} = \, - {\text{nF}}\,{\text{E}}_{{\text{cell}}}^{\text{o}}$ …..$(1)$
According to thermodynamics, the relation between Gibbs free energy and equilibrium constant is as follows:
$\Delta {{\text{G}}_{\text{o}}} = \, - {\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}}$….$(2)$
So, according to equation$(1)$and $(2)$,
$\, - {\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}} = \, - {\text{nF}}\,{\text{E}}_{{\text{cell}}}^{\text{o}}$
${\text{E}}_{{\text{cell}}}^{\text{o}} = \,\dfrac{{{\text{RT}}}}{{{\text{nF}}}}{\text{ln}}\,{{\text{K}}_{\text{C}}}$
In term of log,
${\text{E}}_{{\text{cell}}}^{\text{o}} = \,\dfrac{{{\text{2}}{\text{.303RT}}}}{{{\text{nF}}}}{\text{log}}\,{{\text{K}}_{\text{C}}}$

Note: We can calculate the emf of the cell by using Nernst equation. Nernst equation tells the relation in electromotive force (emf) and concentration of oxidised and reduced species. At ${\text{25}}{\,^{\text{0}}}{\text{C}}$, the value of $\,\dfrac{{{\text{2}}{\text{.303RT}}}}{{\text{F}}} = \,0.0591$so, the value ${\text{E}}_{{\text{cell}}}^{\text{o}}$is,
${\text{E}}_{{\text{cell}}}^{\text{o}} = \,\dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log}}\,{{\text{K}}_{\text{C}}}$