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### Topic: What is the correct form of Nernst equation?  (Read 6174 times)

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#### Tdha

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• Gender:  ##### What is the correct form of Nernst equation?
« on: November 20, 2014, 09:29:40 PM »
Hi!

I have a minor problem, connected to Nernst's equation. In english literature the equation is noted the following form:
E=E0-RT/(zF)*ln[a(ox)/a(red)]
But in other literatures, such as german, french, hungarian (and so on) it is noted the following form:
E=E0+RT/(zF)*ln[a(ox)/a(red)]

The difference is that some people use "+", another "-".

Which form is correct? I searched for derivations, and I have found for both of them.

For "-" version:
http://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html

For "+" version:
http://www.doitpoms.ac.uk/tlplib/pourbaix/nersnt_detailed.php (note that ln(x/y)=-ln(y/x))
http://oszkdk.oszk.hu/storage/00/00/59/45/dd/1/fizkemia_animaciok_nelkul.pdf
http://de.wikipedia.org/wiki/Nernst-Gleichung
http://daten.didaktikchemie.uni-bayreuth.de/umat/nernst_gleichung/nernst_gleichung.htm#Hintergrundinformation

Please somebody explain it for me, because I cannot see, where is the difference in the different derivations. I suppose that both of them should be correct, only the interpretation of "a(ox)" and "a(red)" differs. Thanks a lot.
My first language is Hungarian, so... Please, be tolerant.

#### Hunter2

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« Reply #1 on: November 21, 2014, 01:21:58 AM »
That is logarithm mathematics.

Definition is ln(Ox/Red) and this has to be added to E 0. In some of the papers they talking about ln(Red/Ox) and this has to be subtracted from E0. So both are correct.

#### Tdha

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« Reply #2 on: November 21, 2014, 08:45:38 AM »
Now I realised, what did I not understand.
The "+" form is written for cathodic reaction (Mn+ + ne- M), and the "-" form is written for anodic reaction (M Mn+ + ne-). Correct?

So... This is what I think:

General reaction for a cell, where the stochiometric factors are equal:

General equation for cathodes: Mcz+ + ze- M, Q=[Mc]/[Mcz+]
General equation for anodes: Ma Maz+ + ne-, Q=[Maz+]/[Ma]

Thus we can write the following half-cell equation for cathode:
Ec=Ec0+R*T/z*F*ln([Mc]/[Mcz+])
According to the convention, [Mc] must be 1, thus:
Ec=Ec0+R*T/z*F*ln(1/[Mcn+])
That is equals to:
Ec=Ec0-R*T/z*F*ln([Mcn+])

For anode:
Ea=Ea0-R*T/z*F*ln([Maz+]/[Ma])
According to the convention, [Ma] must be 1, thus:
Ea=Ea0-R*T/z*F*ln([Maz+])

The electromotive force is:
E=Ec-Ea

Mix the half-cell equations:
E=Ec0-R*T/z*F*ln([Mcz+])-[Ea0-R*T/z*F*ln([Maz+])]
E=Ec0-R*T/z*F*ln([Mcz+])-Ea0+R*T/z*F*ln([Maz+])

Ecell0=Ec0-Ea0, thus:

E=Ecell0+R*T/z*F*ln([Maz+]/[Mcz+])

In other words:
E=Ecell0+R*T/z*F*ln(a(that have been oxidised)/a(that will be reduced))

Correct?
« Last Edit: November 21, 2014, 12:46:03 PM by Tdha »
My first language is Hungarian, so... Please, be tolerant.

#### mjc123

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• Mole Snacks: +246/-11 ##### Re: What is the correct form of Nernst equation?
« Reply #3 on: November 21, 2014, 11:50:53 AM »
I don't think that's right. I think it will be easier if you try to think physically and apply Le Chatelier's principle.
Standard electrode potentials are defined for the reduction half-reaction. The electrode potential will be increased by any change that tends to push the reaction to the right, e.g. an increase in concentration of the oxidised form or a reduction in concentration of the reduced form.
So E = E° + RT/nF*ln([ox]/[rd])   {Insert exponents as required by stoichiometry, but you get the point I hope}
This applies to both cathode and anode. Then
Ecell = Ec - Ea = Ec° - Ea° + RT/nF*ln([ox]c[rd]a/[rd]c[ox]a)
= Ecell° - RT/nF*lnQ
where Q is the reaction quotient for the overall reaction oxc + rda rdc + oxa
The negative sign is because Q has products on top and reactants on the bottom, so an increase in Q would tend to push the reaction to the left, and so reduce the cell emf.

You are getting confused with your anode potentials. Your statement that Ea=Ea0-R*T/z*F*ln([Maz+]/[Ma]) is correct if Ea° is the potential for the oxidation, i.e. minus the standard reduction potential. But then Ecell = Ec + Ea (as so defined). In short, you can regard the overall reaction as the cathodic reduction plus the anodic oxidation, or as the cathodic reduction minus the anodic reduction, but not as the reduction minus the oxidation.

This all looks complicated, but I hope it helps to clarify things a bit. Briefly, I suggest:
(i) Think in terms of reduction half-reactions and potentials, subtract as appropriate (anode from cathode).
(ii) Apply Le Chatelier.

#### Tdha

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« Reply #4 on: November 21, 2014, 01:30:33 PM »
Thank you for your fantastic explanation, everything is clear. You are awesome.
« Last Edit: November 21, 2014, 03:53:40 PM by Tdha »
My first language is Hungarian, so... Please, be tolerant.

#### mjc123

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• Mole Snacks: +246/-11 ##### Re: What is the correct form of Nernst equation?
« Reply #5 on: November 21, 2014, 06:07:51 PM »
Well, one does one's best, and hopes to get it right occasionally.

#### Tdha

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« Reply #6 on: November 24, 2014, 02:12:30 PM »
Does this mean, that the first equation at the wikipedia article "Nernst equation" is incorrect?
It shows the following form for the half-cell reduction potential:
Rred=Ered0-R*T/z*F*ln(aox/ared).
That must be incorrect.
« Last Edit: November 24, 2014, 02:29:52 PM by Tdha »
My first language is Hungarian, so... Please, be tolerant.

#### Borek ##### Re: What is the correct form of Nernst equation?
« Reply #7 on: November 24, 2014, 05:19:26 PM »
Looks incorrect to me. Doesn't matter if we talk about reduction or oxidation half cell, potential is

$$E=E_0 + \frac{RT}{nF}\ln{\frac{a_{Ox}}{a_{Red}}}$$

where aOx and aRed are activities of the oxidized and reduced form of the substance involved in the half cell reaction (actually it should be Q for the half cell reaction, but when written the way I did it, it is easier to remember that it is "+" if oxidized side of the reaction is in the nominator).

Then, the cell potential is difference between two half cells potentials. You can combine them into one equation if you want, but typically it is much easier to deal with these half cell potentials separately.

Edit: note that the wikipedia page is inconsistent, in the "Derivation" section it derives the equation in two different ways, yielding two contradicting forms.
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#### Tdha

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« Reply #8 on: November 24, 2014, 05:45:26 PM »