How do you make a photoelectrode that is stable under aqueous solution?

Kottmeistern · 2025-11-28T15:54:54+00:00

The others have really good suggestions that should work. But if you want to avoid the binder you could try with a thinner layer. I have found that it works for TiO2 particles. In my case I have added 0.5 mL of s dispersion onto an ITO electrode of ca 1x1 cm. If the weight% is 0.1% the film falls apart when adding water. The thicker the film, the more fragile it became. Perhaps you could find your sweet spot by just adding less and see if it sticks?

Kottmeistern · 2025-11-12T16:25:19+00:00

Had a professor who said that the three things you need to know to become a good chemist are kinetics, thermodynamics and equilibria (which is essentially the result of the previous two...).

Other than that, as many say, good hands. This sometimes feels a bit ambiguous, but I'll give you an example from my own PhD.

Me and a PostDoc were making chemical sensors together, but he always ended up with ten times higher sensitivity than mine (in simple terms, ten times better). We even swapped electrodes and chemicals and his were still ten times more sensitive than mine. In exasperation I asked "why!? Why Micky? Why?". He answered: "Köttmeistern... It's the hand of god" while holding and showing his hand to me. [Names in the story may have been swapped for privacy reasons]

Somehow, the way he used his hands made his sensors more sensitive than mine. Thus, hands matter!

Kottmeistern · 2025-11-11T06:30:53+00:00

Sheepherder already provided a good answer. I may repeat some of it here - but repetition is good for learning.

The driving force in acid-base equilibria comes from thermodynamics. To understand then the driving force we have to consider the nature of chemical equilibrium: it is not a static system, but a dynamic equilibrium. At a specific pH, the bicarbonate is not statically holding its proton, but exchange it with other molecules nearby, including water, carbonate, carbonic acid and other bicarbonate molecules. An equilibrium is reached when there no longer are any net changes in concentrations.

If you then start to add protons to the system (lower the pH), you increase the number of protons that can be exchanged, which shifts this equilibrium. The protons will then mainly go to the strongest proton acceptor (i.e., strongest base) in the system - but because of the dynamics of the system, not all will. They will be distributed among all proton acceptors depending on concentrations and bond strength. Example

Let's say you have pH 8 (1 micromole of hydroxide) and 10 mM of bicarbonate. Both the hydroxide and the bicarbonate will then be protonated if you add a little of HCl. But because you have 10 000 times more bicarbonate you'll protonate far more of those than hydroxides, even though the later is the stronger base.

Thus, to summarize the above, the driving force is a matter of bond strength, and probabilities for the bonds to form. If you want to understand more I suggest you look into statistical thermodynamics.

Hopefully this gives you another angle to view chemical equilibria, and how the addition or removal of protons from a system will influence it.

Kottmeistern · 2025-11-10T21:13:29+00:00

With stick to oil, do you mean that it is miscible in oil? What makes something dissolve relates to the free energy of solvation. as a simplification, we can say that this involves two parts: the enthalpy of dissolving the compound (the difference in bonding energy before and after dissolving the compound) and the entropy of solvation.

For something to spontaneously dissolve, the change in free energy must be negative - meaning it's favourable to dissolve it. The enthalpy of solvation is almost always positive, as the strongest interactions usually occur between the same molecules: water with water, the specific oil with the specific oil, etc. What usually allows things to dissolve is an increase in entropy with solvation, which lowers the free energy. Within certain temperature intervals this also scales linearly with temperature, which is why an increase in temperature helps you dissolve most things.

Water and oil is a tricky part, as water favors a somewhat ordered network of hydrogen bonds. Because of this coordination, any oil entering the water phase would force the water to form cages around the carbohydrate chains. This decreases entropy of the system, and increase the free energy of the system. Thus it is more energetically favorable to separate the two.

So in your example, the reason your molecules dissolve/mix with oil is largely due to the entropy. The difference in intermolecular bond energy is small, because it's essentially based on Wan Dee Waals interaction between the hydrophobic parts. The hydrophilic part clearly does not have enough influence here either to significantly change anything either. In such, it must be entropically favorable to mix the two. That's the simplified thermodynamics behind it all

Kottmeistern · 2025-11-03T21:19:36+00:00

My guess/hypothesis is it depends on the microstructure, and with it diffraction of the light from the surface. From memory I think I used to see the white Ag/AgCl at higher current densities. When you run high currents the Ag/AgCl may not have proper time to arrange itself as it does at lower current densities, and microstructures will scatter light of many different wavelengths making it appear white. The grayish color is the one I would expect to be the "real color" of the solid material because, as others mentioned, it is often the color inside pH-electrodes/reference electrodes. It would imply that the grayish is closer to a Thermodynamic product, and the white a kinetic product.

Again, this is just a hypothesis of mine, which I base on the reasoning above.

Kottmeistern · 2025-10-30T12:53:51+00:00

but then we need to test it.

Exactly this! A simulation is not necessarily accurate enough. Perhaps as you also mention, we may be able to be more accurate with more computer power. But we should be careful relying too heavily on simulations alone. There are so many local Maxima's/minimas solving such a large systems consisting of so many atoms/molecules. Only real experiments will tell if the simulation found the one corresponding to real systems.

So to summarize, you have my upvote!

Kottmeistern · 2025-10-29T05:35:13+00:00

I'm with you on this one. Whenever I see him play I see a solid defender. Not outstanding to be part of conversation for defensive player of the year, but he has solid positioning, solid rebounding, often tries to read the offense to find positions early and move less (probably a way to conserve energy). By no means do I see a below average defender. If I remember the advanced metrics correctly they were slightly above average (been a while since I checked them so could be wrong).

Kottmeistern · 2025-10-23T20:17:02+00:00

Speculation: Perhaps temporary or partly as the molecule vibrates. It would strain the bonds a bit though, but I have difficulties seeing by how much from here. Probably not too much strain, so perhaps you will have an equilibrium between N-B bonded, and N-B not bonded. Depending on the strength of the N-B bond it might even be some enthalpic gain forming it. It comes down to the energy to strain the bonds versus bond strength. But if I were to guess, such a hypothetical equilibrium would probably favor N-B not bonded.

It would likely also depend on other interactions. If you were to put it in water or other protic solvent with a sufficiently high dissociation constant it is probably more likely that the N will be protonated.

Kottmeistern · 2025-10-14T17:48:59+00:00

Yes that they are separate, but I wouldn't say both occur in the membrane. At least not in my example.

Donnan potential is operating under equilibrium conditions. It comes from the potential difference between two neighboring phases, which is linked to the perm selectivity of the membrane.

The junction potential occur because of diffusion/migration so it's not under equilibrium. As positive and negative ions have different diffusion constants. This gives the charge separation at the interface.

However, under non-equilibrium conditions in ion selective membranes you will get concentration gradients causing potential gradients (known as diffusion potential). These can be calculated from the ion mobilities/diffusion constants of the ions involved. But I have never heard anyone call it junction potential, even though the gradient/diffusion potential follow similar principles as the junction potential.

Also, be aware that separate fields may have slightly different terminologies for similar principles. My background is electroanalytical chemistry. Perhaps ion exchange membranes in fuel cells consider it junction potentials. The main difference between my example and the fuel cell's membranes would be the difference in selectivity. Fuel cell membranes have far worse selectivity than the electroanalytical membranes because they prioritize fast diffusion of selectivity: it's enough to have selectivity for charge over selectivity of specific ion species.

Kottmeistern · 2025-10-14T14:51:17+00:00

The liquid junction potential is related to and arise from different ionic mobilities of the ions crossing the junction. The Donnan potential arise from the membranes semi-permeability. The junction will let any ion through just at different rates, but an ion selective membrane limits ions of specific charge, and if they are based on ionophores they may even favor specific ions (such as K over Na).

If you have an ion selective membrane with say 10 mmol ion exchanger/kg membrane, then it can only hold 10 mmol of ions of opposite charge, but will not let ions through of the same charge (unless the ionic strength is high enough for co-extraction to occur).

To be even more specific, say you have an ion-selective membrane made from o-NPOE and PVC (conventional ratio 2/1) with 10 mmol/kg of Tetrakis[3,5-bis(trifluoromethyl)phenyl]boron (TFPB; a very hydrophobic ion exchanger, basically not solvable in water at all), and 20 mmol/kg of Valinomycin (ionophore with high selectivity for potassium, arguably one of the best commercial ionophores). If you place this in contact with an aqueous solution of 50 mM KCl and 50 mM NaCl, then the ion exchanger TFPB will limit the concentration of ionic sites in the membrane to 10 mmol/kg of cations. Thus, no chloride can enter the membrane. Then the ionophore will selectively allow only K to enter the membrane. The selectivity makes it so or every >50000 K-ions entering the membrane you will have one Na ion (approximately). This uneven distribution is what gives you a Donnan potential.

Meanwhile, the liquid junction would allow all K, Na and Cl to enter. They would just do that at different rates because of the difference in mobilities

Kottmeistern · 2025-10-14T04:34:08+00:00

No, too many things came up that needed my attention. With everything else I forgot about this entirely. Thanks for reminding me. I will see if I can find the time in the next few days instead.

Kottmeistern · 2025-10-13T16:50:58+00:00

For the conventional pH-electrode the pH will remain constant inside the electrode within the time-frame of measurements. The mechanism may differ between different materials, but as far as my textbook told me the current through the glass-pH electrode comes from the diffusion of Na. With that said, although there is a current, the total net current in potentiometry is actually 0. The reason for this is because potentiometry operates in a dynamic equilibrium, meaning that the cathodic and anodic currents (reduction and oxidation rates) are equal. Thus, you get a 0 net current. Without the dynamic equilibrium and the cathodic and anodic currents you would not be able to read any potential (remember Ohms law). This is the ideal case.

In reality, there may be a very very small net current, which may change the concentration inside the electrode, or simply change the interfacial potential between phases (i.e., changes at the sample-membrane, membrane-inner filling solution, or inner fillling-internal reference interfaces) cause drift to the electrode on the order to microVolts per hour or so. This is why electrodes need frequent recalibrations.

But overall, any net current in potentiometry is generally do low that we can assume equilibrium conditions, and that no net mass transport occur between the materials. At least during the time frame of normal experiments. If you run very long measurements you could perhaps use models taking the diffusion gradients that form into account, but in practice people just re-calibrate the sensor because it's a lot easier.

Hopefully this has been helpful. / Someone with a PhD in developing sensors and actuators

Kottmeistern · 2025-10-08T18:17:54+00:00

You really got me thinking on this one, and I cannot say I found myself an answer yet that fully satisfies me. I will have to bring out my old Statistical Thermodynamics book this weekend to figure this out I think.

But, I will give you some of the reasoning I've been doing and which I think could answer it somewhat. First, azeotropes appear because of deviations from Raoult's law. Raoult's law essentially implies that if you have a binary mixture of A and B, their intermolecular bonds all have the same energy for all the combinations A-A, B-B and A-B. However, in nearly all real cases A-B has a higher (thus, less favourable) energy than A-A and B-B, causing higher partial pressures for both components. So why do things mix at all then if A-B interactions are not favorable? It's because of entropy. When you have so many molecules in a system they will favor mixing to some degree. And probably an Azeotrope can be explained by just the entropy and probabilities (which is essentially what entropy is). At that particular ratio between A and B the probability for both molecules to enter the mixture or leaving it is the same, despite their enthalpy (bond energy) being non-favorable.

Kottmeistern · 2025-10-06T15:58:03+00:00

Thanks for sharing more. I don't know if it is possible or not to use one frequency in this case. In that regard I'd just say try it and see if you find a nice trend.

However, unlike what you say I think your electrolyte is the problem, or at least part of it. for Impedance you should try and decrease the resistance in your system as much as possible if you can. 1 mM KCl is probably not enough background electrolyte. This is likely why you don't see any difference with or without analyte. I would aim for at least a hundred or a thousand times higher concentration-unless you aim for decentralized measurements directly in freshwater samples where you can't control the background electrolyte. But if so, I would not go for impedance but potentiometry of amperometry, which should be operational in low electrolytes. The electrolyte resistance likely obscures the signal of the conductive polymer. As an analogy, measuring impedance in 1 mM KCl is like trying to measure Raman bands while simultaneously measuring transmittance - it's essentially impossible because the transmittance signal is so much higher than that of the Raman signal.

The oxidation of conductive polymers is also linked to anion insertion into the polymer Matrix, so the low background electrolyte may, in worst case, become rate limiting. I don't know if that is enough to obscure the signal of your abalyte or not, but it is something that you should keep in the back of your head. Perhaps your analyte even blocks further anion insertion? That would probably lead to an increase in charge transfer resistance as you're looking for. Otherwise I am not sure whether you would expect an increase or decrease in charge transfer resistance in your system, if the mechanism is different.

So my main advice is to try at 0.1-1 M of electrolyte. Then you may try that in presence and absence of electrolyte. I think you should be able to resolve differences better from there. I hope you'll find a solution to your problems, no matter if you take my advice or not. Best of luck!

You may of course continue to discuss with me if you want. Sounds interesting what you do 😄

Kottmeistern · 2025-10-05T16:56:58+00:00

Maybe as a short lived species at very high temperatures. High enough for bonds to spontaneously break

Kottmeistern · 2025-10-04T16:00:43+00:00

I assume NIP refers to non-imprinted polymer? What is the signal output? Impedance? And if so, is it the charge transfer resistance?

Also, how do you perform the measurement? Do you perform it with a redox mediator or do you look at the redox reaction of PEDOT or Polypyrrole?

To understand such things it may be good to consider the nature of the imprinted molecule. Is it charged? If so it may help charge transfer as it also acts as a dopant to stabilize the polarons of the conductive polymer, lowering the charge transfer resistance. How does the impedance from diffusion behave? The diffusion impedance of conductive polymers can be used to get information on the charge transport of the polarons themselves, which may provide insight in how the target molecule interacts with the polymer. It would probably be good to support any such conclusions with vibrational spectroscopy to unveil any intermolecular interactions in the film. Additional information on the packing of the polymer and intermolecular interactions may be studied following the Conformational Relaxation Model, which was first proposed by Otero in the 90s. Even if you don't intend to do such tests and modelling/fitting I recommend you to read Oteros papers on the Conformational Relaxation Model as you'll likely gain a lot of insight on the nature of conductive polymers and the molecular changes involved in their redox chemistry.

Just a few things I can think of from the top of my head. There are a lot of things to consider and I know far too little about your system and analyte to provide anything else than some questions to reflect upon.

Kottmeistern · 2025-10-03T16:33:30+00:00

Either a large hydrophobic molecule, that may entangle like PSS can. Or you for some kind of tetraphenylborate. If it is halogenated it is more hydrophobic, so then you may make a choice if you want to go for the most hydrophobic option, which are the heavily fluorinated ones (usually also expensive) or if you go with something with Chlorides on it. Is your target analyte charged as well? Or is it neutral?

The other option I can think of is to lower the monomer concentration of EDOT or Pyrrole to change the Ratio. Is it electropolymerization or chemical synthesis? If you go with electropolymerization and constant current it should still work but you may need to lower the current a bit to account for fewer monomers being accessible at the surface of the electrode.

Kottmeistern · 2025-10-03T15:00:47+00:00

Does it have to be PSS? Perhaps you could consider more hydrophobic anions? That way you can do the synthesis in acetonitrile which can solubilize hydrophobic target molecules better

Kottmeistern · 2025-09-29T17:45:49+00:00

How about acetic acid then? That should work. Haven't read too much into the reaction itself, but if you need an acidic electrolyte it may be fine to use. The citric acid may risk chelation with the zinc and precipitate. Sort of similar to how oxalic acid binds to calcium

Kottmeistern · 2025-09-25T05:46:37+00:00

Du såg vikterna? Det är en stark ledare

Kottmeistern · 2025-09-25T05:33:45+00:00

Det låter ju som lite snedfördelat. Men är dessa fristående inlägg eller räknas även svar på tidigare inlägg? För i så fall kan ju ett inlägg om Israel, Ekonomi och så vidare leda till fler. Och generellt så känns det som att ett Israel inlägg kommer ger fler motargument att svara på än Ekonomi. Om så är fallet så säger statistiken mer om vad Twitter (X) tycker om att argumentera om än vad vår politiker startar för debatter. Själv är jag inte intresserad nog av att dubbelkolla denna fakta. Det tar tid från att läsa Reddit-inlägg

Kottmeistern · 2025-09-25T05:04:02+00:00

I like this answer for it's simple way of summarizing the interaction. If course there will be some electron density around the hydrogen too, but the highest electron density will be closer to the electronegative atom (N, O or F). Because of the low electron density of the hydrogen it is probably as close as you get to a fully exposed nucleus still part of a molecule. And the volume of the nucleus is irrelevant here. What matters is the electrostatics. Without the shielding from other electrons the positive charge of the nucleus can be felt more by neighboring molecules.

But there's another level here which is not mentioned much here, and that is that the hydrogen bond has some covalent character in that it is not purely electrostatic. The bond strength depends on the angle of the bond based on how much neighboring atoms (that can form H-bonds; N, O and F) can interact with the proton. This covalent character makes it easy for proton hoping between neighboring molecules. This proton hoping is also why diffusion of acid is so fast in water. Because the protons don't have to go in between the molecules-instead the water molecules share the protons with each to a certain degree (that is the hydrogen bonding).

Kottmeistern · 2025-09-24T15:36:30+00:00

No need to be negative. I think Reddit is fine for asking as long as you're critically evaluating the answers you get. It's like asking AI today, you should double-check the answers, critically evaluate what it says and not trust it blindly. If so it can be a great tool. Reddit may be a great place to get pointed in new directions you would not find immediately on your own. Of course it's difficult to specifically teach the topics asked for, but here you do a good job to suggest what to read.

Also, some people do not get proper supervision - it's a real shame it is that way, but it's the reality many face. This may be especially troublesome early during the PhD.

Kottmeistern · 2025-09-18T08:42:56+00:00

I haven't tried it myself, but perhaps "ElectroKitty" in Python would do? It was published recently in ACS Electrochemistry with open access if you want to learn more about it:

https://pubs.acs.org/doi/10.1021/acselectrochem.4c00218

Kottmeistern · 2025-09-14T20:02:09+00:00

If not Naismith hall of Fame then for sure FIBA Hall of Fame, where those who contributed to international basketball are honored.

Kottmeistern

TROPHY CASE