Answer:
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:
[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]
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
