Question

Derive the relationship between Gibb’s free energy and maximum work obtained from galvanic cell and equilibrium constant.

Answer 1

1. In a galvanic cell, chemical energy is converted into electrical energy. The electrical energy produced by the cell is equal to the product of the total charge of the electrons and emf of the cell which drives these electrons between the electrodes.

2. 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 = change of ‘n’ moles of electrons x E_cell ……..(1) 

Charge of I mole of electrons = one Faraday = 1F 

Charge of n moles of electrons = nF …..(2) 

3. Electrical energy = nFE_cell …….(3)

This energy is used to do electric work. Therefore the maximum work that can be obtained from a galvanic cell is Here the – ve sign is introduced to indicate that work is done by the system on the surroundings.

4. Second law of thermodynamics states that the maximum work done by the system is equal to the change in Gibbs free energy of the system W_max = ΔG ……..(5)

5. ΔG = – nFE_cell (6) 

For a spontaneous cell reactions, the AG should be negative. The above expression indicates that E should be positive to get a negative AG value.

6. When all the cell components are in their standard state, the equation (6) becomes 

ΔG° = – nFE°_cell …………….(7)

7. The standard free energy change is related to the equilibrium constant as per the following equation. 

ΔG° = – RT In K_eg …..(8) 

Comparing equations (7) and (8)

nFE°_cell = RT In K_eq …..(9)

nFE°_cell = \(\frac{2.303RT}{nF}\) log K_eq ……(10)