Thermodynamics of cell reactions
We have just learnt that in a galvanic cell, the chemical energy is
converted into electrical energy. The electrical energy produced by the cell is
equal to the product of the total charge of electrons and the emf of the cell
which drives these electrons between the electrodes.
If ‘n’ is the number of moles of electrons exchanged between the
oxidising and reducing agent in the overall cell reaction, then the electrical
energy produced by the cell is given as below.
Electrical energy = Charge of ’n’ mole of electrons × Ecell ......(9.20)
Charge of 1 mole of electrons = one Faraday (1F)
∴ Charge of ’n’ mole of electrons = nF
Equation (9.20) ⇒ Electrical energy = nFEcell .......
(9.21)
Charge of one elctron = 1.602 × 10-19 C
∴ Charge one mole of elctron = 6.023 × 1023 ×1.602 ×10−19 C
= 96488 C
i.e., 1F 96500 C
This energy is used to do the electric work. Therefore the maximum work
that can be obtained from a galvanic cell is
(Wmax)cell = - nFEcell .....(9.22)
Here the (-) sign is introduced to indicate that the work is done by the
system on the surroundings.
We know from the Second Law of thermodynamics that the maximum work done
by the system is equal to the change in the Gibbs free energy of the system.
i.e., Wmax = ∆G .....(9.23)
From (9.22) and (9.23),
∆G = - nFEcell
.....(9.24)
For a spontaneous cell reaction, the ∆G should be negative. The above expression (9.24)
indicates that Ecell should be positive to get a negative ∆G value.
When all the cell components are in their standard state, the equation
(9.24) becomes
∆Go = - nFEcello
.....(9.25)
We know that the standard free energy change is related to the
equilibrium constant as per the following expression.
∆Go = - RT
lnKeq .....(9.26)
Comparing (9.25) and (9.26),
nFEcell = RT lnKeq
⇒ Ecell = [2.303 RT /nF] log Keq .....(9.27)
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