Electrode potential depends upon the concentration of the species in solution, so also emf of a galvanic cell is controlled by the concentration of the ions. Nernst equation gives a relation between emf of a galvanic cell and the concentration of ions in solution. The equation is deduced as follows:

Let us consider a reversible reaction,

α A + βB <–> γC + δD

Equilibrium constant $\large K = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}$

We know from thermodynamics that decrease of free energy (ΔG) measures the tendency of a chemical reaction to reach the equilibrium state.

If G_{1} = total free energy of the reactants and

G_{2} = total free energy of the products, the

ΔG = G_{2} – G_{1} , If G_{1} > G_{2} , then ΔG = negative

A negative ΔGof a reaction implies its spontaneity. ΔG is a measure of the ability of a system to do useful work. Since we can draw electricity from a galvanic cell, it implies that a spontaneous reaction is going on inside the cell and hence ΔG of such reaction must be negative.

Electrical work is the result of the decrease of free energy of the cell reaction.

We know that, the electrical work obtained from any spontaneous reaction supplying nF coulomb of electricity at a potential E is given by,

Net electrical work = nFE

Again net electrical work = −ΔG (decrease of free energy)

ΔG = − nFE

Applying standard state,

ΔG° = − n F E°

Again from thermodynamics, ΔG is related to K by the following equation

ΔG = ΔG° + RT ln K

Substituting the value of ΔG ,

− n F E = − nFE° + RT ln K

$\large E = E^o – \frac{RT}{nF}lnK $

$\large E = E^o – \frac{RT}{nF} ln\frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}$ … (i)

Putting R = 8.314J/k–/mole^{−1} T = 298K and F = 96500 Coulomb.

We get, $\large E = E^o – \frac{0.059}{n} log\frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}$ …(ii)

This is the fundamental form of Nernst equation. This can be applied in both oxidation and reduction process as well as in galvanic cell.

(i) If we represent an oxidation reaction as follows.

Reductant <—> oxidant + ne

Here product is OX and reactant is Red.

∴ E = E° − (0.059/n) log[OX]/[RED] ….(iii)

Equation (iii) is one from of Nernst equation. When [OX] = [Red], then E = E°. E° is called standard oxidation potential.

(ii) if we represent a reduction reaction in the following way

oxidant + ne^{–} <—> reductant

then by applying Nernst equation,

E = E° − (0.059/n) log[RED]/[OX]

(or) E = E° + (0.059/n) log[OX]/[RED] ———- (iv)

Equation (iv) is another form of Nernst equation.

Here also, when [OX] = [Red], E = E°. In this case, E° is called standard reduction potential.

(iii) Equation (iii) and (iv) are both infact , two forms of Nernst equation,

but must be remembered that E_{OX} = − E_{red} , like E°_{OX} = − E°_{red}