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Question

Derive the Nernst equation.

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Solution

Suppose the reaction occurring in a reversible cell is represented by the equation, A+BC+D.
The decrease in free energy, ΔG accompanying the process by the well, known thermodynamic equation.

ΔG=ΔGoRTlnJ

Where ΔGo is the decreases is free energy accompanying the same process when all the reactants & products are in their standard states of unit activity & J stands for the reaction quotient of the activities of the products and reactants at any given state of the reaction.

Substituting the value of J, we have

ΔG=ΔGoRTlnaC×aDaA×aB

If E is the emf of the cell in volts and the cell reaction involves the passage of n Faradays (i.e.) nF Coulombs, the electrical work done by the cell is in nFE volt - Coulombs, (or) Joules. Hence free energy decrease of the sys, ΔG is given by the expression ΔG=nFE

nFE=ΔGoRTlnaC×aDaA×aB=nFEoRTlnac×aDaA×aB

E=EoRTnFlnaC×aDaA×aB

where Eo is the emf of the cell in which the activity, or an approximation when the concentration ion of each reactant and each product of the cell reaction is equal to unity. Eo is known as the standard emf of the cell.

E=EoRTnFlnaC×aDaA×aB

is often referred to as the Nernst equation.
Replacing activities concentration is an approximation; the Nernst equation may be written as

E=EoRTnFln[C]×[D][A]×[B]

where the quantities in parentheses represent the concentration of the species involved. Replacing.
[C]×[D][A]×[B] as equal to K, the equilibrium in the molar concentration Units.

E=EoRTnFlnK

This equation is known as Nernst equation.

E=Eo2.303RTnFlogK and

where Eo stands for electron potential.
R = gas constant
T = Kelvin temperature
n = number of electrons transformed in the half-reaction.
F = Faraday of electricity
K = equilibrium constant for the half cell reactions as in equilibrium law.

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