Relation between EMF and free energy

Relation between EMF and free energy

When a cell produces a current, the current can be used to do work - to run a motor, for instance. Thermodynamic principles can be employed to derive a relation between electrical energy and the maximum amount of work, W_max, obtainable from the cell. The maximum amount of work obtainable from the cell is the product of charge flowing per mole and maximum potential difference, E, through which the charge is transferred.

W_max    = -  n FE

where n is the number of moles of electrons transferred and is equal to the valence of the ion participating in the cell reaction. F stands for Faraday and is equal to 96,495 coulombs and E is the emf of the cell.

According to thermodynamics, the maximum work that can be derived from a chemical reaction is equal to the free energy (DG) for the reaction,

W_max = DG

Therefore, from (1) and (2), we can write

DG = -  n FE

Thus only when E has a positive value, DG value will be negative and the cell reaction will be spontaneous and the e.m.f. of the cell can be measured.

Here E refers to the E_cell.

Thus, the electrical energy supplied by the cell is (nFE) equal to the free energy decrease (- DG) of the cell reaction occurring in the cell.

Example : Determine the standard emf of the cell and standard free energy change of the cell reaction.

Zn, Zn²⁺ || Ni²⁺, Ni. The standard reduction potentials of Zn²⁺, Zn and Ni²⁺, Ni half cells are - 0.76 V and - 0.25 V respectively.

E^o_cell  =   E^oR - E ^oL = - 0.25 - (- 0.76)

= + 0.51 V  E^o_cell is + ve. \ DG^o = - ve.

DG^o     =   - n FE^o_cell

              =   2 electrons

 DG^o   =   -2  ´ 96495 ´ 0.51 = -97460 Joules = - 97.46