JEE Main & Advanced Chemistry Electrochemistry Nernst's Equation

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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