Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kaylaalicia on March 23, 2009, 02:30:15 AM
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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.
Eo= 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 :)
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What is Q? How does it look for the half reaction you have wrote?
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OH! I made such a stupid mistake. Thanks for that hint =]