Question

A voltaic cell utilizes the following reaction:
4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l).

A)

What is the emf of this cell when [Fe2+]= 2.0 M , [Fe3+]= 1.0×10⁻² M , PO2=0.60 atm and the pH of the solution in the cathode compartment is 3.7?

Express your answer using two significant figures.

Answer 1

emf of this cell = -0.037 V

Explanation

Oxidation half reaction : Fe²⁺ (aq) Fe³⁺ (aq) + e^- : E^o = -0.771 V

Reduction half reaction : O₂ (g) + 4 H⁺ (aq) + 4 e^- 2 H₂O (l) : E^o = 0.82 V

Overall reaction : 4 Fe²⁺ (aq) + O₂ (g) + 4 H⁺ (aq) 4 Fe³⁺ (aq) + 2 H₂O (l)

Standard cell potential, E^o_cell = (-0.771 V) + (0.82 V)

E^o_cell = 0.049 V

According to Nernst equation,

E_cell = E^o_cell - (0.0591 V / n) * log([Fe³⁺]⁴ / [Fe²⁺]⁴(pO₂)[H⁺]⁴)

where n = number of electrons transferred = 4

[H⁺] = 10^-pH

[H⁺] = 10^-3.7

[H⁺] = 2.0 x 10^-4 M

Substituting the values,

E_cell = 0.049 V - (0.0591 V / 4) * log[(1.0 x 10^-2 M)⁴ / (2.0 M)⁴ * (0.60 atm) * (2.0 x 10^-4 M)⁴]

E_cell = 0.049 V - 0.086 V

E_cell = -0.037 V