Question

Calculate ΔG∘rxn and E∘cell at 25∘C for a redox reaction with n = 3 that has an equilibrium constant of K = 5.1×10−2.

Part A

Gxrn= ___ kJ

Express your answer using two significant figures

Part B

E*cell=____ V

Express your answer using two significant figures.

Answer 1

Gibbs free energy of reaction ∆Gr an equilibrium constant K are related as
∆Gr = - R∙T∙ln(K)
where R is the universal gas constant and T is the thermodynamic temperature

At standard conditions the temperature is
T° = 298 K
So
∆Gr° = - R∙T°∙ln(K)
= - 8.3145 J∙K⁻¹∙mol⁻¹ ∙ 298 K ∙ ln(5.0×10⁻²)
= 7422.6 J∙mol⁻¹ or  7.42 kJ


The cell potential E° is given by
E° = - ∆Gr°/(n∙F) = R∙T∙ln(K)/(n∙F)
where n is the number of electrons exchanged in the reaction and F is the Faraday constant

For this reaction
E° = - 8.3145 J∙K⁻¹∙mol⁻¹ ∙ 298 K ∙ ln(5.0×10⁻²) / ( 3 ∙ 96485 C∙mol⁻¹)
= 0.0256 V or 2.56*10^-2 V