Detailed derivation of the Nernst equation

Consider the following reaction at equilibrium: \({{\rm{A}}^{z + }} + {\raise0.5ex\hbox{$\scriptstyle z$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{{\rm{H}}_2} \Leftrightarrow {\rm{A}} + z{{\rm{H}}^ + }\)

This can be expressed as two half equations:

 A^z+ + ze^- = A         and    \({\raise0.5ex\hbox{$\scriptstyle z$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{{\rm{H}}_2} = z{{\rm{H}}^ + } + z{e^ - }\)

The left hand reaction represents the equilibrium between atoms of A on a metal surface and A^z+ ions in solution. The term ‘equilibrium’ refers to the fact that the rate of reaction in one direction equals the rate of the reverse reaction.

For the above reaction, the free energy change, ΔG, is given by

\[\Delta G = \Delta {G^0} + RT\ln \frac{{{{{X_A}} \mathord{\left/ {\vphantom {{{X_A}} {X_A^0}}} \right. \kern-0em} {X_A^0}}}}{{{{{C_{{A^{z + }}}}} \mathord{\left/ {\vphantom {{{C_{{A^{z + }}}}} {C_{{A^{z + }}}^0}}} \right. \kern-0em} {C_{{A^{z + }}}^0}}}}\]

where X_A is the mole fraction of A in the metal and C_A is the concentration of A^z+ in solution. When the metal, A, is pure, X_A = 1. Also, the standard states X⁰_A and C⁰_Az+ can be omitted, as the standard state for the metal phase is unit mole fraction, and for the dissolved ions is 1 mol dm^-3. Thus, the free energy change can be expressed as

\[\Delta G = \Delta {G^0} + RT\ln \frac{1}{{{C_{{A^{z + }}}}}}\;\;\;\;\;\ (1)\]

At equilibrium, the chemical driving force, ΔG is always equal to the electrical driving force, E_e. As discussed previously, this can be expressed as

ΔG⁰ = -z F E_e

where z is the number of moles of electrons exchanged in the reaction and F is Faraday’s constant, 96 485 coulombs per mole of electrons. Under standard conditions,

ΔG⁰ = -z F E

From the fundamental thermodynamic equation

ΔG⁰ = -R T ln K

E⁰ can therefore be expressed as

\[{E^0} = \frac{{RT}}{{zF}}\ln K\]

where K is the equilibrium constant for the reaction.

However, only the standard equilibrium potential, E⁰, is related directly to K. The non-standard potential, E_e is not.

Equation (1) can now be expressed in terms of electrode potential by substituting for ΔG and ΔG⁰

\[ - zF{E_e} = - zF{E^0} + RT\ln \frac{1}{{C{}_{{A^{z + }}}}}\]

The equilibrium potential is therefore given by

\[{E_e} = {E^0} - \frac{{RT}}{{zF}}\ln \frac{1}{{C{}_{{A^{z + }}}}}\]

It is conventional to work in decadic logs rather than natural logs, since this is arithmetically more convenient and the pH scale is expressed in the decadic form: ln X = 2.303 log X so the equation may be written as:

\[{E_e} = {E^0} - \frac{{2.303RT}}{{zF}}\log \frac{1}{{C{}_{{A^{z + }}}}}\]

Note that for this simple reaction, the Nernst equation shows that the equilibrium potential, E_e is independent of the pH of the solution. Many half-cell reactions contain H⁺ ions and their Nernst equations therefore depend on the pH.

For the half-cell reaction

MnO₄^- + 4H⁺ + 3e^- = MnO₂ + 2H₂O , E = 0.588 V(SHE)

the Nernst equation appears as

\[{E_e} = {E^0} - \frac{{2.303RT}}{{3F}}\log \frac{{[Mn{O_2}]{{[{H_2}O]}^2}}}{{[MnO_4^ - ]{{[{H^ + }]}^4}}}\]

\[ = {E^0} - 0.0197\log \frac{1}{{[MnO_4^ - ]{{[{H^ + }]}^4}}}\]

= 0.588 + 0.197 log [MnO₄^-] - 0.789 pH

The Nernst equation can therefore be generalised as

\[{E_e} = {E^0} - \frac{{2.303RT}}{{zF}}\log \frac{{[reduced]}}{{[oxidised]}}\]

The notation [reduced] represents the product of the concentrations (or pressures where gases are involved) of all of the species that appear on the reduced side of the electrode reaction, raised to the power of their stoichiometric coefficients. The notation [oxidised] represents the same for the oxidised side of the electrode reaction.

Solutions in which components, such as Cl^- ions, can complex with the metal ions require a different treatment.