What is the value of Ecell for this reaction

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)



Answer 1

We need the standard reduction potentials for each half reaction. Looking them up on the internet, I find

Fe³⁺ + e- ==> Fe²⁺ .. Eº = 0.77 V

Cu²⁺ + e- ==> Cu⁺ .. Eº = 0.15 V

Fe³⁺ + e- ==> Fe²⁺ reduction at cathode

Cu⁺ ==> Cu²⁺ + e- oxidation at anode

---------------------------------------------------------

Fe³⁺ + Cu⁺ ==> Fe²⁺ + Cu²⁺ overall reaction .. Eºcell = 0.77 - 0.15 = 0.62 V

To find Ecell, we use the Nernst equation, and since a temperature isn't given, we can assume 298K. Then, the Nernst equation can be written as...

Ecell = Eºcell - 0.0591 / n log Q

Ecell = ?

Eºcell = 0.62 V

n = mols electrons = 1

Q = reaction quotient = [Fe²⁺][Cu²⁺] / [Fe³⁺][Cu⁺]

Solving for Ecell ...

Ecell = 0.62 - 0.591 log (2.50x10^-4)² / (1.50x10^-3)²

Ecell = 0.62 - 0.591 * -1.56

Ecell = 0.61 - (-0.919)

Ecell = 1.53 V

(be sure to check all of the math)