Calculate the concentration ratio

AI Thread Summary
The discussion revolves around calculating the concentration ratio using electrode potentials and the Nernst equation. Participants clarify that the standard electrode potentials are not applicable, and the importance of knowing the pressure of gaseous hydrogen is emphasized, typically assumed to be 1 atm. There is a consensus that at equilibrium, both cells have the same potential, but this does not mean the individual half-cell potentials are zero. The conversation highlights a common misunderstanding in calculating cell potentials, suggesting that instead of reversing signs, one should simply subtract the half-cell potentials. The key takeaway is that the method of calculating potentials does not depend on whether the reactions are expressed as reductions or oxidations.
Kqwert
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Homework Statement
Hello,

If we have the half-reactions:

Fe3++e- -> Fe2+
and
2H+ + 2e- -> H2

Calculate the ratio [Fe3+] / [Fe2+] at pH 2.
Relevant Equations
.
Any hint? I was immediately thinking standard electrode potentials as we have two half reactions, but not sure exactly how to go on..
 
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Not standard electrode potentials. Do you know any equation that says what the real potential is?
 
I think I can use Nernst's equation, but not sure how to calculate the real potential..
 
In general question is incomplete, as it doesn't say anything about a pressure of gaseous hydrogen. I would go for a standard 1 atm.
 
How do I go on and solve it? I assume I can calculate the electrode potential of the hydrogen reaction, but I need the total electrode potential in order to find the Fe-ratio, right? Any hints?
 
At equilibrium both cells have the same potential.
 
Ok, I thought Ecell was 0 at equilibrium??
 
Yes, but it doesn't mean potentials of both half cells are zero. It means difference between them is zero, that is something entirely different.
 
Thank you. When saying "difference between them are zero". Is that when the cell potential is calculated on reduction form? I.e. both half reactions are given as reduction reactions. I know another way of looking at it is writing one reaction as an oxidation reaction and the other as a reduction rx.
 
  • #10
Doesn't matter whether the reaction is reduction of oxidation, it occurs at exactly the same potential.

Your problem stem from the fact you were probably taught to calculate the cell potential by adding half cells but reversing the sign of one of them - this is completely unnecessary. Just subtract them and math will do the trick of reversing the sign if necessary (-(-1) is 1 after all).
 
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