you characterize " the commercial catalyst?" Ask this question to the manufacture of this material.

RuO2 is used in almost every article, and I cannot achieve the 10 mA/cm2 which is one of the main parameters used for OER. I assume that the 99,999% RuO2 should achieve the literature results

Dear Rafael,

Did you correct your results for IR drop? On some potentiostats it can be corrected automatically, or use R determined by impedance, E= E(measured)-jR

Hello,

I used the iR drop correction in the potentiostat program I use in some experiences to see if it would solve the problem, but it did not work.

Check the reference electrode potential, AgCl is not stable in alkaline solutions. Small amount of KOH could penetrated in ref electrode forming AgOH with different potential?

You should increase your potential range to 2.0 V or larger vs. RHE, perhaps even 2.5 V. The evolution of oxygen would be very small for your present voltage range vs. RHE.

I'll try to do that, but normally articles in this area only go to 1.7 vs RHE

Branimir N. Grgur I used two different AgCl electrodes and they gave both the same results, I have used AgCl for almost 2 years in the oxygen reduction reaction and the results are normal.

I have already mentioned why are you sure that your "commercial" material should be active? You might waste your time working with a wrong material. Be sure that the surface areaof your Pt counter electrode is large enough. What is your membrane? It might have too high resistance. Yours " I have used AgCl for almost 2 years in the oxygen reduction reaction." At what pH? I would not trust to Ag/AgCl reference electrode in 0.1M KOH

I use a glass cell and the electrodes are submersed in about 100 mL of electrolyte, there is no membrane. The pH is 13. I think the material should be active because it is the reference of almost all literature, dont know if there is an activation process, never read about it. It is the benchmark catalyst for OER. The surface area of the counter electrode is not something I had thought about, the area is not really that big (I'll measure it as soon as possible and put here), the working electrode has a glassy carbon surface area of 0.07 cm2.

For ORR, the system works just fine in pH 13 with a carbon rod as a counter electrode. If it helps, I can attach one Nova file so everyone can see how OER is performed.

You need to use two compartment cell with stirring in each compartment. In one compartment cell without stirring the reaction is controlled by diffusion. With a stirring you have a cross reaction.

However, the main problem is the most likely to be too small surface area of your cathode and H2 formation is a limiting factor

I don't have a two compartment cell at the moment. This is the cell I use. Left side is the Ag/AgCl electrode, in the middle the glassy carbon working electrode and on the right the Pt counter electrode. The area of the Pt counter electrode is 0.8x0.7 cm. I'll try to obtain an electrode with higher surface area to see if that is the cause of the low intensities.

The Nova file is an example of a test I performed with iR drop correction (I just wrote the value and it automatically corrects it), with a 0.4 mg/cm2 loading of RuO2. To calculate the mA/cm2 you have to multiply by 1000 and divide by 0.07067, which is the area.

Melvin Henry Miles I increased the potential range from 1.85 to 2.25 and, although literature usually goes only to 1.7V, the intensity pratically tripled (screenshot). However, I was told that above the 1.85V the glassy carbon electrode is damaged and at around 2.2 the electrolyte decomposes (maybe that is why the intensity increases so much)

your cell seems to be OK.

How do you convert the potential measured versus Ag/AgCl to the one versus RHE?

E =E(measured) + 0.21 (Ag/AgCl to RHE) + 0.059 pH (in this case 13)

Yurii V Geletii could it be the area I'm dividing the intensity for? The calculation I do is:

j (mA/cm2) = jmeasured (A) *1000 / 0.07067 (which is the area of the glassy carbon electrode)

Am I using the wrong area for my calculations?

Is this correct?

E =E(measured) + 0.21 (Ag/AgCl to RHE) + 0.059 pH (in this case 13), should it be E =E(measured) + 0.21 (Ag/AgCl to RHE) - 0.059 pH (in this case 13).?

Surface area. It's OK for a rough estimation

I used the initial first formula where potential shifts to more positive values with the increase of pH, just like all literature I've seen.

E(RHE) = E(Ag/AgCl) + 0.059pH + E°(Ag/AgCl)

For ORR, the results I obtain for Pt/C are equal to literature, so I don't think it will be from the potential adjustments

Then, probably you have a wrong RuO2 sample?

I've tried with a new (bought a few days before) IrO2 sample, which is the other benchmark catalyst, and also didn't get the 10 mA/cm2.

Could it be because I should calculate the area of the catalyst and apply it to the equation instead of using the glassy carbon circle area?

I did a theoretical study and if I consider that RuO2 has a density of 6.97 mg/cm3 and that the material is spheric, I could calculate the volume with the mass I loaded to the electrode, and then the area of the material.

With this area I would achieve 10 mA/cm2 at around 1.55 V vs RHE. I would need to calculate the electrochemical active surface area for pratical results. Would this be correct?

The surface area of the modified electrode can't be smaller than the geometric surface are of the GC electrode. Thus, (the current )/(for the surface are of the GC electrode) will give the upper level of the current density, proving that you do not deposit something else on the electrode. As far as I understand you are using RDE technique. Are you sure that the catalyst remains on the surface at your rpm? Try to use CV or LSV.

Yurii V Geletii I'm performing LSVs at 1600 rpm and the intensity is almost the same when performing multiple LSVs consecutively. I deposit 0.0191 mg of RuO2 on the 0.07067 cm2 of GC electrode. Those 0.0191 mg correspond to a volume of 2.7376 *10^-6 cm3 which, considering the material spherical leads to a r= 0.00868 cm and an area of 9.47*10^-4. Of course this value would have to be calculate by the electrochemical surface area measurment, but it seems like a good value. The geometrical area of the GC electrode is higher, but the electroactivity observed comes almost all from the RuO2 catalyst. Is it still not valid?

I've just recalled that in one of our work we used a modified CG electrode. When O2 was released, the bubbles of O2 were formed inside of the catalytic film. These bubbles mechanically destroyed the film on the electrode. This may also happen in your case.

Dear Rafael

Formula

E =E(measured) + 0.21 (Ag/AgCl to RHE) + 0.059 pH is correct.

Erhe = -0.059pH = -0.769 V vs SHE, or

Erhe (vs AgCl) = -0.059pH - E(AgCl~0.2V) = -0.97 V. So for example 0.5 V vs AgCl is 1.47 V RHE at pH=13

Also, how you electrodeposided RuO2, I am concerns about contact resistance between GC and RuO2 so its possible that mostly GC works. If you have NAFION solution you can prepare RuO2 ink with excellent adhesions on GC

I noticed an error in my previous answer, i assumed 1 particle for the calculations when there are many particles in the surface, so the ECSA calculation would be necessary.

Yurii V Geletii it is an option that I'll have to consider, thank you for all the help.

Branimir N. Grgur Thank you, the value I use in my calculations is 0.9646 which is the 0.97V you mention.

Branimir N. Grgur I used mainly 2 solutions:

1 mg of RuO2, 96 uL nafion, 142 uL ethanol and 220 uL water

4 mg of RuO2, 80 uL nafion, 80 uL ethanol and 470 uL water

No idea for now, sorry

I have a problem too

The thermoneutral potential for H2O electrolysis based on the enthalpy change is 1.48 V. However higher cell potentials even using excellent catalytic metals, such as platinum, palladium, or ruthenium, do not give visible evolution of oxygen gas at the anode until the cell potential is near 2 Volts in 0.1 KOH. Perhaps you should try a simple H2O electrolysis cell reaction with two platinum electrodes and note the cell potential when gases are observed at the anode.

If using a 3- electrode cell, always make sure that the area of the platinum counter electrode is considerably larger than the working electrode. Otherwise, the counter electrode reaction may become the limiting reaction and affect your reference electrode voltage measurements.

Melvin Henry Miles I'm using the 3-electrode system above and the platinum counter electrode has an area of 0.56 cm2 compared to the 0.07 cm2 of the working electrode, I've also tested a counter with more area and the results were the same

Melvin Henry Miles We are not discussing H2O electrolysis but a half reaction, water oxidation. Your " The thermoneutral potential for H2O electrolysis based on the enthalpy change is 1.48 V " is applicable for two compartment cell without IR-drop. The thermodynamic potential required for water oxidation is 1.24-0.059pH. Rafael uses 3-electrode configuration and controls the voltage between reference and working electrodes. He does not care about "the cell potential" ( Do you mean a voltage between anode and cathode?)

I have a feeling that you posted you answer without thorough reading all previous discussion.

If one is trying to do electrochemistry in water involving oxygen evolution, then the cell reaction: H2O = H2 + 0.5 O2 is always involved. All that the reference electrode does is allows you to focus on the current- voltage behavior of either the anode (this case involving oxygen evolution) or the cathode (if you are studying the H2 evolution reaction). The thermodynamic CELL voltage for this reaction does NOT depend on the pH. This electrolysis reaction is not even thermodynamically possible until the applied cell voltage is E = -1.229 V based on the Gibbs energy(G) change at STP. The pH does NOT affect this thermodynamic result. This cell reaction remains very slow, even using platinum electrodes, up to the thermoneutral potential of -1.48 V based on the enthalpy (H) change. Even then, the cell current densities (A/cm2) will be very small until higher cell voltages are applied. For example, look at the lead acid battery in an automobile involving water and sulfuric acid as the electrolyte. It provides a cell voltage of 2 volts. There is normally no unwanted electrolysis of H2O producing hydrogen and oxygen gases because this cell reaction is very slow - especially for lead (Pb) electrodes.

I realize that the study in question is oxygen evolution at an anode involving ruthenium particles as catalysts, but this still involves the very slow break up of water molecules to form oxygen gas at the anode. Furthermore, this oxygen evolution reaction is much slower by a factor of 10,000, based on electrochemical kinetic studies, than the hydrogen evolution reaction at the cathode for any choice of a catalyst.

Melvin Henry Miles I don't think we are talking of the same reaction setup. I'll add an example article which has the RuO2 catalyst and the results they obtained. Most articles with benchmark catalysts have similar results to this.

https://www.nature.com/articles/s41598-018-20974-1.pdf?origin=ppub&fbclid=IwAR0-rM9c7MmCw9BVQQz_F72C-u3-X6iybElBRB6z5P_K_qBjhudLwaQFKhI

At aproximately 1.6 V vs RHE the current density was 10 mA/cm2. This is what I want to obtain near that potential, but I am unable too using the same catalyst, electrolyte, scan rate and rotation speed.

Melvin Henry Miles We are not talking about water electrolysis, we are discussing water oxidation on anode. There is no sense to discuss CELL potential

You statement "Furthermore, this oxygen evolution reaction is much slower by a factor of 10,000, based on electrochemical kinetic studies, than the hydrogen evolution reaction at the cathode for any choice of a catalyst" is NOT correct. Read papers on catalytic water oxidation

Rafael Morais

I understand your experiment, but to obtain oxygen from H2O your cell is doing H2O electrolysis. If you use a RHE as your reference electrode, then your potentials are close to those for H2O electrolysis. I have worked with RuO2 as a supercapacitor material as well as a catalyst. I think your problem in reproducing someone's results is a particle size and the differences in the real effective electrochemical area. The effective area can change greatly depending on the size of the RuO2 particles. This makes it very difficult to exactly reproduce someone's electrochemical results.

As an example in my research on RuO2 as a supercapacitor material, we could obtain very small particles such that the effective electrochemical area was about one million times the geometric area. Laboratory methods can differ greatly in the RuO2 particle sizes and effective electrode areas. I am not at all surprised by your reported differences.

I've mentioned several times that water electrolysis is not the same as water oxidation on anode. The important parameter for water oxidation on anode is the overpotential for the reaction of water oxidation. Water oxidation potential is E =1.24 -0.59pH, water reduction potential is E = -0.059*pH. That's all! There is no sense to discuss the basic of water oxidation thermodynamic.

At the same time I agree with you that " This makes it very difficult to exactly reproduce someone's electrochemical results." I stated the same in almost all my posts.