[Train]

I'm having trouble wrapping my mind around the membrane potential of a pH electrode.  I understand that there are two reference electrodes at constant potential and a junction potential that is (hopefully) negligable, leaving only the membrane potential to be affected by the analyte solution.  The inside and the outside surfaces of the membrane have different amounts of absorbed H+ based on the H+ activity of the solutions they're in contact with.  So there's a charge difference across the membrane.

But how does that effect the measured potential difference between the two electrodes?  I'm still thinking in terms of an electrochemical cell in which the charge is carried by ions.  So say for example that the analyte solution has lower H+ activity than the internal solution.  Then the inside of the membrane will have more positive charge than the outside.  Am I right in thinking that this would tend to drive positive ions away from the reference electrode out into the analyte solution? (Of course no net charge is moving because this is a potentiometric measurement.)  So then since cations move toward the cathode an electromotive force acting in the opposite direction would tend to lower the effective potential of that electrode with respect to the other electrode.

Sorry for the longish post.  That's the best I've been able to come up with in terms of a physical explanation.  I don't think it's complete though and I'm not sure it makes sense to me because it seems like cations would tend to be repelled from both surfaces.  And since there's not supposed to be any actual moving charges, I'm not sure how the tendancy of reduction to occur at the electrode would be affected by the membrane potential.  Am I on the right track?  Can someone please clarify?

[Borek]

I am not sure I understand correctly your idea, but - assuming I got right - my bet is that that we are talking about so small charges adsorbed that they don't have a long range coulombic effects.

[Arkcon]

Briefly scanning your post, I'd like to focus on one point: 

And since there's not supposed to be any actual moving charges, I'm not sure how the tendancy of reduction to occur at the electrode would be affected by the membrane potential.

As I understood it, one of the models for the glass membrane electrode is that it functions as a capacitor -- an accumulation of charged atoms on (um ... or within, to a certain depth of) the glass surface can affect the equilibrium at the center electrode.  As I said, that's just the model I'd heard of, the ways that the glass electrode works are not very well known.

[Train]

According to my textbook (Fundamentals of Analytical Chemistry), the potential on the inner surface of the membrane adds to the overall electrode potential while the potential on the outer surface subtracts from it.  So the effective potential is something like E=E(ref) + E(inner) - E(outer) plus the asymetric potential.

But I don't understand how what's going on at the membrane affects the reduction potential at the electrode.  I could understand it if current were flowing requiring net ionic charge to pass through the membrane to make up for what's lost/gained in the redox reaction.

Does the potential at the membrane create an electric field which could then influence the behavior of ions and electrons at the electrode?

[Arkcon]

Does the potential at the membrane create an electric field which could then influence the behavior of ions and electrons at the electrode?

That is how a capacitor works, yes.  So that is, I assume, how the glass electrode works.  Or at least, that is a good, if incomplete, explanation for it.

Don't be surprised that ions, separated by a capacitance membrane, alter chemical equilibrium at a distance by their electric fields.  This is how animal cells function.  If it were not for this phenomena, happening in neurons, muscle cells and other specialized cells, you would not be reading, typing or thinking about the topic.

[Train]

According to my textbook (Fundamentals of Analytical Chemistry), the potential on the inner surface of the membrane adds to the overall electrode potential while the potential on the outer surface subtracts from it.  So the effective potential is something like E=E(ref) + E(inner) - E(outer) plus the asymmetric potential.

OK, I think I figured out where my confusion was coming from.  What I wrote above is actually kind of the opposite of what the book says, but I think the book's wrong.

According to my book, the potential at the membrane surface E=0.0592*log a, where a is H+ activity.   Then they write the equation for the E=E(ref) + E(outer) – E(inner), where E(inner/outer) = 0.0592*log a.  The effect of writing it this way is to have the cationic flow of the cell moving from areas of high potential to areas of low potential.
 
If they instead write E=-0.0592*log a and E=E(ref) + E(inner) – E(outer) then you end up with the same equations ultimately but it makes more sense conceptually (to me anyway) because then a higher inner potential than outer would add to the potential of the electrode by encouraging the flow of cations in that direction.

[Train]

My professor says I have it all wrong.   


Moving on before my head explodes . . . I had another question about measuring the pH with pH meters.  I once worked in a lab where the standard practice when measuring the pH of very dilute samples (eg from freshwater aquariums) was to let the probe equilibrate in the sample for 15 minutes before taking a reading.  Do you agree with this practice?  What's the reasoning?  What ionic strenght qualifies as dilute enough to require this?

[Borek]

My professor says I have it all wrong.   

If it helps you - there is no consensus about what is really happening at the surface and why the measurements are possible. Here goes another approach to explain the situation:

http://www.ph-meter.info/pH-electrode-potential

I have not posted it earlier as I feel it doesn't touch things that you are trying to envision.

I once worked in a lab where the standard practice when measuring the pH of very dilute samples (eg from freshwater aquariums) was to let the probe equilibrate in the sample for 15 minutes before taking a reading.  Do you agree with this practice?  What's the reasoning?  What ionic strenght qualifies as dilute enough to require this?

It must had something to do with the high electrical resistance of the pure water. Ultra pure water has a specific resistance of 18 MOhms/cm² and that makes maesurements of the eletrode potential impossible. Fresh water (or tap water) is not that pure, but its resistivity can be still relatively high. Some procedures call for addition of an inert salt that should not change pH (much) - like NaCl or KCl. That should speed up process of reaching the equilibrium.

[Train]

http://www.ph-meter.info/pH-electrode-potential

Hey, that's a cool site!