[kaylaalicia]

I'm not entirely sure how to incorporate the pH into the question.

Calculate the reduction potential (at 25°C) of the half-cell MnO₄^- (5.00×10-2 M)/ Mn²⁺ (3.60×10-2 M) at pH = 2.00.
(The half-reaction is MnO₄^- + 8H⁺ + 5e^- --> Mn²⁺ + 4H₂O.)

Using the equation Nernst equation, this is what I did. Although, since I didn't incorporate the pH into this. I'm positive that it's incorrect.
E^o= E-(0.0591/n) log Q

= 1.51 - (0.0591/5) log [(5.00E-2)/(3.60E-2)]
=1.51


For the pH, i know that 10^-2 = [H⁺] = 0.01M

So help would be much appreciated 

[Borek]

What is Q? How does it look for the half reaction you have wrote?

[kaylaalicia]

OH! I made such a stupid mistake. Thanks for that hint =]