Consider the reaction Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+ (aq) E° = 0.78 V. What is the value of E when [Cu2+] = 0.040 M and [Fe2+] = 0.40 M?

• A. 0.72 V
• B. 0.75 V
• C. 0.81 V
• D. 0.84 V

Answer: (A)

Using the Nernst equation to relate the cell potential to the standard potential and the reaction quotient under nonstandard conditions. For this reaction, copper(II) is reduced and iron is oxidized, so the standard potential is 0.78 V and two electrons are transferred (n = 2). The reaction quotient is determined by the activities of the species that appear in the equation, with solids having activity 1: Q = [Fe2+]/[Cu2+] = 0.40 / 0.040 = 10.

At 25°C, the Nernst equation gives E = E° − (0.05916/n) log Q. Substituting in the values, log Q = log 10 = 1 and n = 2, so E = 0.78 − (0.05916/2)(1) = 0.78 − 0.02958 ≈ 0.750 V.

The result is slightly less than the standard potential because the product concentration is higher relative to the reactant, which lowers the driving force for the reaction under these conditions.