Calculate the reduction potential (in V) at 25°C of each of the following half-cells.

Question

The Nernst equation can be applied to half-reactions. Consider the following half-reactions.

Cu2+ + 2 e− → Cu

MnO4− + 8 H + + 5 e− → Mn2+ + 4 H2O

Calculate the reduction potential (in V) at 25°C of each of the following half-cells.

(a)

Cu/Cu2+ (0.11 M)

(b)

MnO4− (0.10 M)/Mn2+ (0.010 M) at pH = 2.90

I honestly do not know where to start on this problem so any guidance would help at this point. In terms of using the Nernst equation, I'm getting lost when it comes to using the concentrations and plugging them into Q. thank you 

Answer 1

Cu²⁺(aq) + 2e- ==> Cu(s) Eº = 0.34V

Q = 1/0.11 = 9.09

MnO₄^-(aq) + 8H⁺(aq) + 5e- ==> Mn²⁺(aq) + 4H₂O(l)

Q = 0.01/(0.1)(1.26x10^-3)⁸ Note: the 1.26x10^-3 is derived from the pH since pH = -log [H⁺]

Q = 1.57x10²²

These are my calculated values of Q. You can use them in the Nernst equation to solve for the reduction potential of the half cell.