The Concentration Dependence of Cell Emf - The Nernst Equation

The Concentration Dependence of Cell Emf - The Nernst Equation

Consider an electrochemical equilibrium whose overall cell reaction is represented by the scheme -
aA+bB cC+dD                                                                                                           (3)
where a, b, c and d are the number of moles of the reagents taking part in the reaction The reagent or reagents could be gases, ions or molecules in solution. Solids may also be involved but they are treated as constants.
 For 1mole of species A, the free energy is given by
GA=GA°+RT in aA                                                                                           (4)
where GA is the electrochemical free energy, and GA° is the standard free energy and aA is the activity of reagent A.
For a moles of reagent A,
aGA = aGA + aRT ln aA
aGA + RT in (aA)a
Similar-expressions can be written for the other three reagents.
The overall free energy change is the free energy of the products minus free energy of the reactants:
G =Gproducts - Greactants
G = G0 + RT In
Recall  G =-nEF and G° = -nE°F
Hence
 E=E0 -                                                                                     (5)
Equation (5) is known as the Nernst equation and gives the dependence of the electrode or cell potential in terms of activities or concentrations of the reactants and products. E° is known as the standard e.m.f of the cell and it is the cell potential when the activities are all unity.
At 25°C equation (5) reduces to
E=E0

E=E0 –                                                                                                                                                                                                                                                 (6)
Also by definition,
G° = -RT In K = -nE0F, therefore
E0 =                                                                                         (7)
where K is the equilibrium constant for the electrode or the cell reaction and hence equation (6) can also be written as:
E=                                                                     (8)
Exercise 1
For the following cell at 298K,

Zn(s)/Zn(aq)2+ (a = I )//Pb(aq)2+ (a = 1 )/Pb(s)

(i) Write individual electrode reactions and the overall cell reaction.
(ii) Calculate E°cell.
(iii) Calculate G° of the cell reaction.
(iv) In which direction is the cell reaction spontaneous?
Given E0Zn2+/Zn = 0.762V and E0Pb2+/Pb =-0.126V


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