What is the value of Ecell for this reaction when [Fe3+] = [Cu+] = 1.50 × 10–3 M and [Fe2+] = [Cu2+] = 2.50 × 10–4 M?

Question

What is the value of E_cell for this reaction when [Fe³⁺] = [Cu⁺] = 1.50 × 10^–3 M and [Fe²⁺] = [Cu²⁺] = 2.50 × 10^–4 M? 

Consider an electrochemical cell based on the following reaction:

Fe3+(aq)+Cu+(aq) Fe2+(aq)+Cu2+(aq)

1. I think I need to use the nernst equation for this problem, but the E^o cell is stumping me. Do I use the standard reduction values, Fe = .771 and Cu = .342

Answer 1

Yes, and yes. You do use the Nernst equation, and you do use the Eº_cell that you calculate from standard reduction potentials.

Fe³⁺(aq) + Cu⁺(aq) ==> Fe²⁺(aq) + Cu²⁺(aq)

Eº_cell = 0.771 - 0.342 = 0.430 V

Nernst equation: E_cell = Eº_cell - RT/nF ln Q and at 298K and using log₁₀ we can write this as...

E_cell = Eº_cell - 0.0591/n log Q

E_cell = ?

Eºcell = 0.430

n = 1

Q = [Fe²⁺][Cu²⁺] / [Fe³⁺][Cu⁺] = [2.5x10^-4][2.5x10^-4] / [1.5x10^-3][1.5x10^-3] = 0.02778

E_cell = Eº_cell - 0.0591/n log Q

E_cell = 0.430 - 0.0591/1 log 0.02778

E_cell = 0.430 - (0.0592)(-1.56)

E_cell = 0.430 + 0.092

E_cell = 0.522 V

(be sure to check all of the math)