An oxidation-reduction reaction in which 3 electrons are Transferred has ∆G° = +18.55 kJ at 25°C? What is the value of E°?

The correct answer is: 0.064 V

Explanation:

In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:

where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:

+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell

Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)

Eºcell= (18550 J)/ (289500 C)

Eºcell= 0.064 J/C

Since 1 Volt= 1 Joule/1 Coulomb, thus:

Eºcell= 0.064 J/C = 0.064 V

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