The cell EMF and the free energy of the overall cell reaction

The great utility of the equilibrium electrochemical cell of Fig. 10.2c is that its EMF is proportional to the difference in the free energies of the mixtures in the two electrodes - that is, s is a direct measure of AG of the overall cell reaction. This relationship is derived in this section.

If the z moles of electrons pass through a voltage drop s in the external circuit of the galvanic cell in Fig. 10.2a, the motor produces a quantity Wei of work1:

Wel = SZeNAv where e = 1.6x10-19 Coulombs is the electronic charge, NAv = 6 x 10 is Avogadro's number and the product eNAv is the charge of one mole of electrons. This product is called Faraday's constant, F = 96500 Coulombs/mole. Since a Coulomb is one Joule per Volt, and there are 4.184 Joules in a calorie, a more convenient value of Faraday's constant is 96.5 kJ/mole-Volt.

If the cell operates reversibly, Eq (1.20) shows that the external work is accompanied by a decrease AG in the free energy of the entire cell:

Combining the above two equations yields:

If all species in both half-cells are in their standard states (i.e., pure liquids or solids, as in reaction (10.1)), the superscript o is appended to the quantities in Eq (10.5a) to indicate this special case:

Equation (10.5b) applies to the equilibrium cell of Fig. 10.2c because all four species in reaction (10.1) are in their standard states.

1 This equation is the time-integrated form of the familiar electrical formula: power = voltage x current

Example What is the EMF of the equilibrium cell X|XO||electrolyte||Y|YO at 1000oC with X = Ni and Y = Fe?

Converting kcal to kJ in the Ellingham diagram of Fig. 9.8, the free energies of formation of the oxides are:

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