# 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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