Question

Write Nernst equation for single electrode potential for the reaction ${M^{n + }}_{\left( {aq} \right)} + n{e^ - } \to \,{M_{\left( s \right)}}$

Answer 1

Hint: This question gives the knowledge about Nernst equation. Nernst equation helps in determining the cell potential under various conditions. Also helps in determining the electrode potential.

Formula used: The cell potential is determined using Nernst equation as follows:
$E = {E^0} - \dfrac{{RT}}{{zF}}\ln Q$
Where ${E^0}$ is standard potential, $E$ is cell potential, $R$ is universal gas constant, $z$ is ion charge, $F $ is Faraday constant, $Q$ is reaction constant and $T$ is temperature in kelvin.

Complete step by step answer:
Nernst equation helps in determining the cell potential under various non-standard conditions. Also helps in determining the electrode potential. Nernst equation helps in determining the relation between reduction potential or oxidation potential of an electrochemical cell reaction to the temperature, activities and standard electrode potential of the chemical species.
Single electrode potential is defined as the potential generated when the metal is dipped in the solution consisting of its own ions, at the interphase between solution and metal. Apparently, it is not possible to determine the single electrode potential because the half-cell reactions do not occur independently. Any reference should always be provided.
Consider the reaction,
${M^{n + }}_{\left( {aq} \right)} + n{e^ - } \to \,{M_{\left( s \right)}}$
According to Nernst equation, we have
$E = {E^0} - \dfrac{{RT}}{{zF}}\ln Q$
Where $Q = \dfrac{{{{\left[ a \right]}^n}}}{{{{\left[ b \right]}^n}}}$ and $a$ is the concentration of products and $b$ is the concentration of reactants and $n$ is the number of moles.
So, the Nernst equation becomes as follows:
$ \Rightarrow {E_{{M^{n + }}/M}} = {E^0}_{{M^{n + }}/M} - \dfrac{{RT}}{{zF}}\ln Q$
On simplifying, we have
$ \Rightarrow {E_{{M^{n + }}/M}} = {E^0}_{{M^{n + }}/M} - \dfrac{{2.303RT}}{{zF}}\log Q$
On further simplifying, we have
$ \Rightarrow {E_{{M^{n + }}/M}} = {E^0}_{{M^{n + }}/M} - \dfrac{{2.303RT}}{{zF}}\log \dfrac{{\left[ M \right]}}{{\left[ {{M^{n + }}} \right]}}$
Substitute $\left[ M \right]$ as $1$ in the above equation as follows:
$ \Rightarrow {E_{{M^{n + }}/M}} = {E^0}_{{M^{n + }}/M} - \dfrac{{2.303RT}}{{zF}}\log \dfrac{1}{{\left[ {{M^{n + }}} \right]}}$

Note: Always remember the values of all the constants. Nernst's equation should be remembered carefully. And also remember whenever $\Delta G$ is positive the reaction becomes non- spontaneous and the cell stops working.