Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: kimi85 on June 23, 2007, 10:54:14 AM
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Hi everyone.
The problem is:
The standard half-cell reduction potential for Ag+/Ag is 0.7996 V at 25 degree Celsius. Given experimental value Ksp = 1.56 x 10raised to -10 for AgCl, calculate the standard half-cell reduction potential for the Ag/AgCl electrode.
I used this equation: E = RT/nF ln Ksp, and my answer is wrong. The correct answer is 0.2198 V. I don't know how to come up with the answer.
Thank you very much.
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E = RT/nF ln Ksp
Where is E0?
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E is the En. I think that's the formula to use when you are using standard cell potentials to find the equilibrium constants
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What I am hinting at is that the formula you mention doesn't use standard potential - don't you think it should?
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What I did is 0.7996 - x = 0.0257/1 ln 1.56 x 10-10.. x is the half-cell reduction potential for the Ag/AgCl electrode. But it's wrong. Do you know what should be done?
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Write equation for E for Ag/Ag+ electrode, then put in Ag+ concentration calculated assuming [Cl-] = 1. That's all.
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Thank you very much. I'll try it
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Hi. I still don't understand how to do it.
Can you show me how?
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Show what you did.
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This is what I did:
I get the square root of the Ksp to get the concentration of Ag+ which is equal to 1.25 x 10^-5 , then
0.7996 - x = 0.0257/ n ln 1.25 x 10^-5
I still didn't get the correct answer.
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Nope.
E = E0 + RT/nF ln Q
where Q is reaction quotient.
Now, in the simplest case (metal and its ions) it translates to
E = E0 + RT/nF ln [Men+]
Now, if the silver electrode is covered with AgCl, [Ag+] on the electrode surface depends on the [Cl-]. That's where the Kso comes into play.
Kso = [Ag+][Cl-]
So
[Ag+] = Kso/[Cl-]
Insert it into Nernst equation, see what it gives.
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Your solution is correct.Thank you very much!!!! ;D
God bless you. :)
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Why is the formula a plus?
Shouldn't it be E = Eo - RT/nF In Q?
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Depends on the way you construct Q. For half reaction
Q = [Oxidised]/[Reduced]
and Nernst equation gets form
E = E0 + RT/nF ln Q
Same goes for more complicated reactions, like
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Just put oxidized form in numerator:
Q = [MnO4-][H+]8/[Mn2+]
(note that water concentration is taken from the Q, as it is assumed to be constant and not changing, thus it is moved to E0)
It is just a matter of convention used, you may as well use minus and reverse Q exchanging numerator and denominator.
Looks like I was not precise in my previous post, hopefully it is cleared now.
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Now, I got it. Thank you very much. :)