Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Rutherford on July 28, 2012, 11:30:47 AM
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http://portal.acs.org/portal/PublicWebSite/education/students/highschool/olympiad/pastexams/CTP_005486
It is the 4th f).
I don't understand how they got the formula in the answer, so I tried a different approach, calculating the potential for every electrode separately and the to calculate the potential of the cell:
E(Cr3+)=-0.744+0.059/3*log(0.1)2=-0.783V
E(Ni2+)=-0.236+0.059/2*log(0.1)3=-0.324V
Then the cell potential is E(Ni2+)-E(Cr3+)=0.458V, but it is 0.488V in the answer. Where am I wrong?
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Check concentration that you used.
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The Nernst Equation is commonly written as [tex]E = E^o -\frac{RT}{nF}\ln{(Q)}[/tex]. The 0.0257 you see is simply RT/F, when R = 8.314, T = 298 K, and F = 96485. The 6 is n, the number of moles of electrons involved in the reaction. Since so many calculations take place at 25 °C it isn't uncommon to see the constants simplified in that way.
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The Nernst Equation is commonly written as [tex]E = E^o -\frac{RT}{nF}\ln{(Q)}[/tex]. The 0.0257 you see is simply RT/F, when R = 8.314, T = 298 K, and F = 96485. The 6 is n, the number of moles of electrons involved in the reaction. Since so many calculations take place at 25 °C it isn't uncommon to see the constants simplified in that way.
That doesn't bother me. I don't understand Q.
Check concentration that you used.
Yes, it is 0.01M, but even then I don't get the right result (got 0.41V).
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http://en.wikipedia.org/wiki/Reaction_quotient
Write overall reaction equation and you should get the answer as they listed it. Your approach - separate calculation of each half cell potential - is also correct, but your calculation of each half cell potential is not.
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I shouldn't degree the concentrations in the half reactions. I solved it correctly now, and I also derived their equation. Thanks for the help.