Question

The standard half-cell potentials for three reactions are given below: Fe²+(aq) + 2e = Fe(s) Fe3+(aq) + 3e = Fe(s) Br2(aq) + 2e = 2Br (aq) E° = -0.44V E° = -0.04V E° = 1.09V Q 1(a) Determine what pair of half-cell reactions yields the most spontaneous redox process. Q 1(b) Calculate AGº for this process and comment on whether iron is plated during this process. Q 1(c) Draw a labelled energy level diagram showing the reactants, products and AGO. Q 1(d) Use the Nernst equation to determine the change in cell potential if the concentration of both reactants is decreased to 0.1 M.

Answer 1

Q1a :

Br₂|Br^- has highest reduction potential so , it would acts as cathode and F|Fe²⁺ has lowest reduction potential so it will acts as anode .

E°_cell = 1.09 - ( -0.44) = 1.53 V

Q1b:

∆G° = - n F E° = - 2 × 96485 C × 1.53 V = -295244.1 J

= - 295.2 kJ/mol

Net reaction is :

Br₂ (aq) + Fe $\rightarrow$ Fe²⁺ + 2 Br^-

iron is converted into Fe²⁺(aq) , so it is not plated .

Q1c:

free energy Reactants re+bla products 14 Reaction coordinates

Q1d:

E_cell = E° - RT/nF × ln([Br^-]²[Fe²⁺]/[Br₂])

Only Br₂ is the reactant whose concentration is included into the expression . Rest of the concentrations would remains same let 1M

E_cell = E° - 0.0591/2 × log(1/0.1M)

Change in potential of cell = 0.0591/2 = 0.02955V

= 29.55 mV