Derivation of the Nernst Equation:

First law of the thermodynamics might be beginning as:

Δu = q + w                                                                 ... (1.10)

where Δ u is the change in the internal energy of the system and w the work done on the system consider a reversible process taking place in a system at constant temperature and pressure in which both mechanical work (- PΔV ) and electrical work ( w_elec) are done, i.e.

w = -PΔV + w_elec                                                                    ... (1.11)

We know which for a reversible process at constant temperature the entropy change is

ΔS = q/t  that is, q= TΔS                                          ... (1.12)

Substituting from Eqs. (1.11) and (1.12) into Eq. (1.10), we obtain

(Δu)_T.P = TΔS - PΔV + w_ele                                                                              ... (1.13)

Also, at constant pressure, the change in enthalpy (ΔH) of the system is given by,

(ΔH ) _P = (Δu) _{P +} PΔV                                                                  ... (1.14)

and at constant temperature, the change in Gibbs free energy ( ΔG ) is given by

(ΔG)_T.P  = (Δ H)_T - TΔS                                                                ... (1.15)

From Eqs. (1.14, and 1.15), we get

(ΔG) _T.P  = (Δu) _T,P  + PΔV - TΔ S

and now from Eqs. (1.13 and 1.16), we get

(ΔG)_T.P  = TΔ S - PΔV + w_ele + PΔV - TΔ S

That is, (ΔG )_T.P  = w_ele

Now considering a galvanic cell which has two terminals across which a potential variations is E, when a charge Q is moved through the external circuit, the work done is

w_elc  = E Q                                                                                          ... (1.18)

We know that, Q = nF                                                                     ... (1.19)

where F is the Faraday constant equal to 96490 C, and is the charge on one mole of electrons, and n is an number of moles of electrons included in the cell reaction. Thus, the work done by the system on the resistance is

w_elc = - nFE                                                                                       ... (1.20)

Combining with Eq. 1.17, we get that the free energy changes, or work completed is - nFE.

Hence,  Δ G = -  nFE                                                                        ... (1.21)

Assuming a chemical reaction such as:

aA + bB ↔    pP + rR                                                                       ... (1.22)

The modification in free energy is provided through the equation

Δ G = Δ G^o + RT ln a^P_p  a_R^r/ a^a_A  a^b_B                                               ... (1.23)

where R is the gas constant, T is the temperature on Kelvin scale and Δ G^o is the Gibbs free energy change when all the reactants and products are in their standard state. For standard conditions, Eq. 1.21 might be written as

Δ G^o = - nFE^o                                                                                ... (1.24)

where E^o is the standard redox potential. By Using Eqs. 1.21 and 1.24 to substitute for Δ G and Δ G^o gives

 - nFE = - nFE^o + RT ln a^P_p  a_R^r/ a^a_A  a^b_B                                             ...(1.25)

After dividing both the side by - nF,

E = E^o - (RT/ nF) ln a^P_p  a_R^r/ a^a_A  a^b_B                                        ... (1.26)

Instead it could be written as reactant/product which require a change of sign in before the ln terms,

E = E^o + (RT/ nF) ln a^a_A  a^b_B /a^P_p  a_R^r                                       ... (1.27)

E = E^o + (2.3RT/ nF) log a^a_A  a^b_B /a^P_p  a_R^r                                  ...(1.28)

The term 'ln' refer to natural logarithm (i.e. to the base). In practice, it is more convenient to use logarithm to base 10 (written as log). Equation (1.28) is the expression for Nernst equation for the chemical reaction 1.22.

Substituting the numerical values of R and F and making the approximation that activities and concentration are equivalent, at 25^oC we get following form of Nernst equation:

E = Eo + (0.0591/n) log ([A]^a [B]^b/[P] ^p [R ]^r)                          ... (1.29)

This is the form in which we commonl y use the Nernst equation.

To further understand Nernst equation, now we will take up few examples. Considering first a simple electrode, which is for a half-cell with metal-cation equilibriumΔ

M ⁿ⁺ + ne  ↔   M                                                                          ... (1.30)

E = Eº +(0.0591/n )log [Mⁿ⁺]/[M]

[M] =1

E= Eº +0.0591log[Mⁿ⁺]

at 25^oC, the Nernst equation can be written as

E = Eº +(0.0591/n )log [Mⁿ⁺]                                                                       ...(1.31)

as the [M] is unity. Remember that the activity/concentration of a pure substance and an electron is unity and therefore they do not appear in the equation.

To a more common form of the redox reaction

xO + mW     yR + zZ                                                                                    ... (1.32)

where O is the oxidized species which is being reduced, R is the reduction product, W and Z a few other species and x, y, m & z are the stoichiometric coefficients of the respective species and a Nernst equation for this reaction at 25^oC is

E = Eº +(0.0591/n )log [O]^x[W]^m/[R]^y[Z]^z                                                 ...(1.33)

For the Galvanic cell (Eq. 1.6), at 25^oC, Nernst equation can be written as,

E_cell = E^o_cell + (0.0591/2) log [Cu²⁺]/ [Zn²⁺]                                                  ... (1.34)

 It could be applied for evaluating the emf of the cell.

Now consider following reaction:

MnO^-₄ + 8H⁺ + 5e = Mn²⁺ + 4H₂O                                                                 ....(1.35)

A Nernst equation for this half cell is

E_MnO^-_4,/Mn²⁺ = E^o_MnO^-_4,/Mn²⁺ - (0.0591/5)log ([Mn²⁺]/[MnO^-4] [H+]⁸)      ....(1.36)