Nernst's Equation
Category : JEE Main & Advanced
(1) Nernst’s equation for electrode potential
The potential of the electrode at which the reaction,
\[{{M}^{n+}}(aq)+n{{e}^{-}}\to M(s)\]
takes place is described by the equation, \[{{E}_{{{M}^{n+}}/M}}=E_{{{M}^{n+}}/M}^{0}-\frac{RT}{nF}\ln \frac{[M(s)]}{[{{M}^{n+}}(aq.)]}\]
or\[{{E}_{{{M}^{n+}}/M}}=E_{{{M}^{n+}}/M}^{0}-\frac{2.303\,\,RT}{nF}\log \frac{[M(s)]}{[{{M}^{n+}}(aq)]}\]
above eq. is called the Nernst equation.
Where,
\[{{E}_{{{M}^{n+}}/M}}\]= the potential of the electrode at a given concentration,
\[E_{{{M}^{n+}}/M}^{0}\] = the standard electrode potential
R = the universal gas constant, \[8.31\ J\,{{K}^{-1}}\,mo{{l}^{-1}}\]
T= the temperature on the absolute scale,
n = the number of electrons involved in the electrode reaction,
F = the Faraday constant : (96500 C),
\[[M(s)]\]= the concentration of the deposited metal,
\[[{{M}^{n+}}(aq)]\]= the molar concentration of the metal ion in the solution,
The concentration of pure metal M(s) is taken as unity. So, the Nernst equation for the \[{{M}^{n+}}/M\] electrode is written as,
\[{{E}_{{{M}^{n+}}/M}}=E_{{{M}^{n+}}/M}^{0}-\frac{2.303\,\,RT}{nF}\log \frac{1}{[{{M}^{n+}}(aq)]}\]
At 298 K, the Nernst equation for the \[{{M}^{n+}}/M\] electrode can be written as,
\[{{E}_{{{M}^{n+}}/M}}=E_{{{M}^{n+}}/M}^{0}-\frac{0.0591}{n}\log \frac{1}{[{{M}^{n+}}(aq)]}\]
For an electrode (half - cell) corresponding to the electrode reaction,
Oxidised form \[+n{{e}^{-}}\to \]Reduced form
The Nernst equation for the electrode is written as,
\[{{E}_{half-cell}}=E_{half-cell}^{0}-\frac{2.303\,RT}{nF}\log \frac{[\text{Reduced form }]}{\text{ }\!\![\!\!\text{ Oxidised form }\!\!]\!\!\text{ }}\]
At 298 K, the Nernst equation can be written as,
\[{{E}_{half-cell}}=E_{half-cell}^{0}-\frac{0.0591}{n}\log \frac{[\text{Reduced form }]}{\text{ }\!\![\!\!\text{ Oxidised form }\!\!]\!\!\text{ }}\]
(2) Nernst’s equation for cell EMF
For a cell in which the net cell reaction involving n electrons is, \[aA+bB\to cC+dD\]
The Nernst equation is written as,
\[{{E}_{cell}}=E_{cell}^{0}-\frac{RT}{nF}\text{ln}\frac{{{\text{ }\!\![\!\!\text{ C }\!\!]\!\!\text{ }}^{\text{c}}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
Where, \[E_{cell}^{0}=E_{cathode}^{0}-E_{anode}^{0}\].
The \[E_{cell}^{o}\] is called the standard cell potential.
or \[{{E}_{\text{cell}}}=E_{cell}^{o}-\frac{2.303\,RT}{nF}\log \frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
At 298 K, above eq. can be written as,
or \[{{E}_{\text{cell}}}=E_{cell}^{o}-\frac{0.0592}{n}\log \frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
It may be noted here, that the concentrations of A, B, C and D referred in the eqs. are the concentrations at the time the cell emf is measured.
(3) Nernst’s equation for Daniells cell : Daniell’s cell consists of zinc and copper electrodes. The electrode reactions in Daniell’s cell are,
At anode : \[Zn(s)\to Z{{n}^{2+}}(aq)+2{{e}^{-}}\]
At cathode : \[C{{u}^{2+}}(aq)+2{{e}^{-}}\to Cu(s)\]
Net cell reaction : \[Zn(s)+C{{u}^{2+}}(aq)\to Cu(s)+Z{{n}^{2+}}(aq)\]
Therefore, the Nernst equation for the Daniell’s cell is,
\[{{E}_{cdll}}=E_{cell}^{0}-\frac{2.303\,RT}{2F}\log \frac{[Cu(s)][Z{{n}^{2+}}(aq)]}{[Zn(s)][C{{u}^{2+}}(aq)]}\]
Since, the activities of pure copper and zinc metals are taken as unity, hence the Nernst equation for the Daniell’s cell is,
\[{{E}_{cdll}}=E_{cell}^{0}-\frac{2.303\,RT}{2F}\log \frac{[Z{{n}^{2+}}(aq]}{[C{{u}^{2+}}(aq)]}\]
The above eq. at 298 K is,
\[{{E}_{cdll}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}(aq]}{[C{{u}^{2+}}(aq)]}V\]
For Daniells cell, \[E_{cell}^{0}=1.1\,V\]
(4) Nernst's equation and equilibrium constant
For a cell, in which the net cell reaction involving n electrons is, \[aA+bB\to cC+dD\]
The Nernst equation is
\[{{E}_{Cell}}=E_{cell}^{0}-\frac{RT}{nF}\ln \,\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\] .....(i)
At equilibrium, the cell cannot perform any useful work. So at equilibrium, \[{{E}_{Cell}}\]is zero. Also at equilibrium, the ratio
\[\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}={{\left[ \frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}} \right]}_{equilibrium}}={{K}_{c}}\]
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