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

Calculate the reduction potential (at 25°C) of the half-cell MnO

_{4}^{-} (5.00×10-2 M)/ Mn

^{2+} (3.60×10-2 M) at pH = 2.00.

(The half-reaction is MnO

_{4}^{-} + 8H

^{+} + 5e

^{-} --> Mn

^{2+} + 4H

_{2}O.)

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