Use the standard half-cell potentials listed below to calculate the standard Gibbs energy (in kJ) for the following reaction occurring in an electrochemical cell at 25°C and 1 atm.

Question

(The equation is balanced.)

6 Cl⁻(aq) + 2 Fe³⁺(aq) → 3 Cl₂(g) + 2 Fe(s) 

Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E° = +1.36 V

Fe³⁺(aq) + 3 e⁻ → Fe(s) E° = -0.04 V

 

Answer 1

Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E° = +1.36 V ... cathode (reduction)

Fe³⁺(aq) + 3 e⁻ → Fe(s) E° = -0.04 V ... anode (oxidation)

Eºcell = 1.36 V - (-0.04 V) = 1.40 V

∆Gº = -nFEºcell

n = mols electrons transferred = 6

F = Faraday constant = 96,500 C/J

Eºcell = 1.40 V

∆Gº = -(6)(96,500 C)(1.40 J/C) x 1 kJ/1000 J = -811 kJ