Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: everydaygame on February 28, 2010, 06:29:51 PM
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Hi
the standard electrode potential of water is 1.23V vs NHE
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O 1.23V
However when the oxidation of water is used in photochemical process such as a gratzel cell or in honda and fujishimas work this is also quoted as 1.23V
2 H2O ⇄ O2(g) + 4 H+ + 4 e− 1.23V
I would have expected the oxidation potential to be -1.23V relative to the reduction potential of 1.23V
Can anyone give any insight as to why this is? It's got me hugely confused
Thanks
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No, potential at which reaction occurs is the same no matter in which direction it goes.
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Thanks for the response
Yes i do understand this
my question is that why is it quoted as 1.23V by Gratzel and Honda and fujishima when in the context of being split by oxidation which to me seems totally wrong when the reduction potential is 1.23V. Its unlikely they have made a mistake being such an established author but there must be some assumption that I dont understand. I was hoping someone might know
Thanks
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Think about it this way - high reduction potential means strong oxidizer. And O2 is a strong oxidizer, no dobut about it.
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Im not doubting what your saying, im agreeing completely with you
the problem is the convention in photochemical work to state the oxidation potential as 1.23V when it should be -1.23V to my eyes
There must be a reason why this is so common place I am just looking for an explanation as to why all the major authors in this field do this
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Heres an image of what i mean:
http://img69.imageshack.us/img69/3804/wrong2.png
On the left is a diagram from a paper showing a potential for oxidation of 1.23V and on the right is data form a table showing the reduction potential to be 1.23V
I agree someone is wrong but i dont understand how this and many other papers could be wrong. i assume there sis a convention i dont understand
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The sign on the potentials are agreed upon by convention. So yeah, if you reverse the reaction the sign is supposed to change:
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O 1.23V
2 H2O ⇄ O2(g) + 4 H+ + 4 e− -1.23V
Having said that, we can just as easily use the opposite sign convention and do things this way (and some people do):
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O -1.23V
2 H2O ⇄ O2(g) + 4 H+ + 4 e− 1.23V
All that matters in terms of being correct is that the sign conventions are carried though consistently when using other formulas, such as the Gibbs, so that you can accurately describe which way the direction of the reaction is going (which is the point of the sign to begin with). So if the authors are following the latter approach for their sign conventions you might be able to trace this sign decision by studying the signs of their :delta: G to see if this is a typo or not. Or if they actually list both reactions in the same paper with the same sign that might be considered a good indicator of a typo as well. Unfortunately, as a scientist you have to learn to be flexible when it comes to things such as sign conventions because in different places of the world different sign conventions are being followed. Some places even use different sign conventions and units for certain things which really makes things a pain.
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Hi
thanks thats a clear explanation, im not sure that fully answers the issue though
in this case when the reduction and oxidation reactions are being described they are in relation to the flat band potential of semiconductors as in the image
http://img69.imageshack.us/img69/3804/wrong2.png
The flat band potential of TiO2 shows a conduction band potential of aprox -0.5V vs NHE and an oxidation potential of 2.5V that have not be reversed
as the conduction band potential of TiO2 is negative as expected and has not be inversed along with the oxidation potential of water then still dont understand why it is +1.23 for oxidation when the expected value would be -1.23V
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As Borek said, the potential doesn't change because of the direction of the reaction. Think of a regular voltammogram, for a nice reversible couple the reduction and oxidation waves are centered around the standard potential.
The change in signs arise when you want to calculate the difference between the potentials of two half reactions.
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Everydaygame, I did look at that image you attached more carefully, and I do think I see where the confusion is.
If you look at the axis of the graph you will see that they are indeed using the opposite sign convention scheme as 2H+ + 2e- :rarrow: H2 appears at 0.0V and 2 H2O :rarrow: O2(g) + 4 H+ + 4 e− appears at a value on the graph of positive ~+1.2 V. The problem here if you notice is that the *entire y-axis* on the graph has been inverted as values of -.05 and -1.0V are going up whereas positive quantities such as 2.0V and 3.0V are going down. So if we look at the image on the right and expect -1.23V because one half reaction appears "lower" than the other, we might suspect they made an error but actually what they did is flip the entire y-axis upside down from what we are used to.
They aren't incorrect to do it this way. Because the numerical value remains the same (as Borek and Bee have mentioned) either way you go, we can easily just revert the y-axis back to it's normal orientation if we want by switching the positive and negatives on the y-axis while leaving everything else in the image the same. Then the gap between the half reactions becomes -1.23V like you might want. But since the y-axis is inverted on the image, then the value of 1.23V is correct.
Why did they invert the y-axis on the graph? I don't know, but at least from the y-axis flip you can conclude that they did know what they were doing and probably had some compelling reason to do it this way (perhaps to fall in line with sign conventions? Notice that the x-axis is also backwards from the way we do things typically).
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Hi
thanks for the help everyone
I had this sent to me that explains what is happening, it agrees mainly with what you've been saying-
Confusion is caused because some stick too IUPAC convention for electrode reactions while others use the a flexible/pragmatic description.
The H2O/O2 electrode reaction as per IUPAC convention is always described as a reduction reaction whose Standard Potential is +1.23 V (with respect to standard hydrogen electrode). In water splitting water is oxidized to O2, hence the actual potential for oxidation is -1.23 V, however the Standard Potential by the convention is still quoted as +1.23 V even if the reaction direction is opposite.
The Authors have used the convention when quoting the potential but disregarded it for writing the reaction. Thus they are being practical with the reactions and can simply add the reactions and thus e + h to cancel charge, while with potential they will carry out an subtraction, since all reduction potential are used for all electrodes with only one electrode reaction being reduction, the other has to be oxidation.
It is not incorrect but schizophrenic.
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everydaygame:
I am having the same confusion when studying photocatalysts and photoelectrochemical cells. Did you ever clear this up, because it still doesn't seem very clear to me from the posts. For example, if the H2O/O2 is reported as a positive value (when the oxd potential is really negative) then does that mean that on the same scale, such as the figure you have posted, that negative values E red values would be put as positive? Such as Cr3+/Cr(s) is -0.740, where I would put this on the plot as +0.740?
In other words, the reduction half reactions are now reverse sign?