[garijniv]

Hi

I'm looking at at this  link   https://core.ac.uk/download/pdf/52394289.pdf  that i'll refer to as the core.ac.uk link

it's a paper discussing water electrolysis


And they show a nernst equation



E_NERNST = 1.23 - 0.9 * 10^-3(T-298) + (RT/4F)Ln(.)

I understand that "0.9 * 10^-3(T-298)"  is a correction factor for  when temperatures  are far above 298 degrees Kelvin.  So eg at 298K that correction factor is 0. And at temperatures not far above 298K, that correction factor is negligible.

So given a temperature of around 298K I will simplify that to

E_NERNST = 1.23 + (RT/4F)Ln(.)

 But I have two issues with that

One is, here is I think the relevant half equation in the electrolysis of water, that they must be using

Anode (oxidation) 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−    E=-1.23

The Standard Electrode Potential for that oxidation half reaction,  is -1.23

i.e. http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html  has Standard Electrode Potential for the Reduction half equation as +1.23  and I think we should reverse the sign given that we want it for an oxidation half equation so we should use -1.23

So why in the core.ac.uk link,  have they written +1.23 there  in their NERNST equation (i'd expect -1.23) ?

And also, why on that core.ac.uk link have they written  + (RT/4F)Ln( .)

Shouldn't they have written 

E_NERNST = -1.23 - (RT/4F)Ln( .)  ?   

Since The NERNST equation as far as I know should be E_NERNST = SEP - (RT/nF)Ln(..)




Also, and the following isn't a sign question. But,  looking at their LnQ, (with interest in their Q), they have



([H₂]² * [O₂]) / [H₂O]

So they've done   Pressure of products / pressure of reactants.

ok

And they've got

Pressure of H2, to the power 2.
Pressure of O2, to the power of 1
Pressure of H2O, to the power of 1


From what I understand, in order to work out the Q in Ln(Q),  is  concentration of products  / concentration of reactants.  And in order to work out the concentrations of each molecule, you look at the coefficient..

And they seem to maybe be using




So H2O <--> H2 + 0.5 O2

So then shouldn't they have

Pressure of O2, to the power of 0.5, with Pressure of H2 to the power of 1?

Why do they have Pressure of O2 to the power of 1, and Pressure of H2 to the power of 2?

Thanks

[Borek]

Depending on how you write the reaction you get a different set of coefficients, but also a different number of electrons involved. These cancel out, don't they?

[garijniv]

Depending on how you write the reaction you get a different set of coefficients, but also a different number of electrons involved. These cancel out, don't they?

Are you referring to what they get for Q?  (i.e. the second half of my question)

I thought their reaction there doesn't mention electrons 'cos maybe it's an overall reaction where electrons already cancelled.

They have   H2O <--> H2 + 0.5 O2   

I guess electrons were involved while they were producing that overall reaction, but I don't see them otherwise?

[Borek]

Yes, I am talking about the second half of the question.

Electrons are still there. When you derive the overall potential of the cell (combining both half cells and rearranging top get the overall Q for the whole system under log) they don't magically disappear.

I suppose fact that it is overall reaction also answers your first question, abut the sign - we are interested in the total voltage difference between half cells. One is at -1.23, second is at zero by definition, and the "battery" is just "higher minus lower".

[garijniv]

I suppose fact that it is overall reaction also answers your first question, abut the sign - we are interested in the total voltage difference between half cells. One is at -1.23, second is at zero by definition, and the "battery" is just "higher minus lower".

ok thanks so that explains the +1.23.  It's 0 - -1.23.

so was I wrong to say  E_NERNST = SEP - (RT/nF)Ln(Q)

(SEP being "standard electrode potential)

i.e. it should be E_NERNST = OverallPotential - (RT/nF)Ln(Q)  ?

And also, why does that core.ac.uk PDF have +(RT/nF). When other sources have -(RT/nF) ?

 

[Borek]

I think you are still ignoring the point they refer to the potential difference, not the absolute potential of an electrode.

Try to derive the voltage difference from half cells voltages and see what you get. Remember: you are interested in total voltage, so your reference is - in general - the lower potential of the two.

[garijniv]

I think you are still ignoring the point they refer to the potential difference, not the absolute potential of an electrode.

Try to derive the voltage difference from half cells voltages and see what you get. Remember: you are interested in total voltage, so your reference is - in general - the lower potential of the two.

Isn't that what I did here  "+1.23.  It's 0 - -1.23"

When I was asking you if the way to get +1.23V is to get the difference by doing 0 - -1.23 (and I was asking if perhaps that was what they did in that core.ac.uk link p2/7)

You also mention doing a sum. If we do  0 + -1.23 then we get  -1.23V

[Borek]

Utl means Ox, the line starting with "Jeżeli" says "if n₁=n₂=n 5.33 becomes 5.34". 0.059 is just RT/F and a constant to convert ln to log.

Actually with some tricks you can rearrange the sum even if n₁ doesn't equal n₂, you will just need to replace n in RT/(nF) with n₁*n₂ and change powers in Q.

Do you see now where the signs come from? It is a pretty simple algebra, nothing fancy.

[garijniv]

thanks..

The algebra with the Logs are interesting, and I see that when they take the difference of two Nernst equations then they get that + RT/nF.  So that explains that.

One puzzling part for me is for this basic part



I'm puzzled as to the values of

E_SEP_1   

and

E_SEP_2 

If it were the case that E_SEP_1 were a standard reduction potential and E_SEP_2 were a standard oxidation potential, then the formula i've seen is overall cell potential = E1 + E2.   (I've no doubt that if doing E1-E2 then that's not the case here).


If E_SEP_1  a Standard Reduction Potential for the cathode),  and E_SEP_2  a standard reduction potential for the anode),  then yeah i've seen the formula like that, of   overall cell potential = E1-E2     

The idea being that subtracting the reduction potential of the oxidation half reaction, will reverse the sign , so as to get the correct result.


Looking here http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html

Let's say the half reaction at the cathode has E(red)_SEP_1 =0V

And the half reaction at the anode has the Standard Reduction Potential of E(red)_SEP_2=1.23V

So the overall potential ,

is  0 - 1.23V = -1.23V

Or, to calculate it the other way.

E(red)_SEP_1 =0V

E(ox)_SEP_2 = -1.23V


 0 + (-1.23V) = -1.23V

So, I get -1.23V

They get +1.23V!

I can see that  0 -  (-1.23V)  would give +1.23V.  But  If they are doing E_SEP_1 - E_SEP_2  Then E_SEP_2 has to be  a reduction potential, so would have to be +1.23V.  And then they'd have 0 - (+1.23V) = -1.23V

So I can't see why they get +1.23V here

[Borek]

In reality there is no such thing as separate oxidation and reduction potentials. This is just an artefact of the way they are defined.

Half reaction always occurs at the same absolute potential (relative to the reference), no matter which way it is going, it it not like is jumps between red and ox potentials when switching the reaction direction.

[garijniv]

I'll grant that saying -1.23V or +1.23V on its own, is arbitrary and the magnitude is what matters.

But if it's in an equation of the form  A+B    or A-B

Then it makes a difference whether A is negative vs A being positive.

Even moreso if the voltage or value of A, were high like A=100V vs A=-100V

Because the result of A+B or A-B would be of a different magnitude.

[garijniv]

And actually while I see the  algebraic simplification in the image from the   book in #7,   is using LogA-LogB = Log(A/B)   hence Log (A/B / (C/D))   = Log (A/B * (D/C) ) = Log (AD/BC)

Hence explaining why when they subtract the two nernst equations they have  + K Log(Q)

It doesn't explain why they start off with  E1 = + K Log(Q).   or  E2 = + KLog(Q)

i.e. why doesn't that page from that book have  -(0.059/n)LogQ

i.e.

why doesn't the book have  E1 = E01 - 0.059/n log [utl1/red1]     and e.t.c. for E2.


Since most sources show

[Borek]

You do see Q vs Q^-1?

[garijniv]

You do see Q vs Q^-1?

I have some idea of what you might mean in that I understand that Log(A/B) = -Log(B/A)

I don't see  Q^-1 mentioned  explicitly anywhere.

I considered whether their Q was inversed in the core.ac.uk page..

First, Looking at the image you provided in post #7,  I see now you mention it, that they're doing Q as  reactants/products which is an inversed Q.  So fine that explains why in the image you provided in post #7, in that book, they have +LogQ

But I don't see why they have +Log or +Ln in the core.ac.uk article.



In that core.ac.uk article, their Q is the  normal way round..  products/reactants..  So shouldn't they have
-(RT/nF)LnQ  rather than  +(RT/nF)LnQ ?

[Borek]

Simple LeChatelier's style logic tells me - the higher the partial pressure of H₂/O₂, the more difficult the electrolysis should be. Their equation says exactly that, so I see no reason to doubt it is correctly derived.