[scientizscht (OP)]

Hello!

In fuel cells, you have precious metal catalysts that gets the oxygen from atmosphere and does what with it?
I do not understand its role, can you explain please?
Also, what is its efficiency, eg for a square centimetre of precious metal catalyst electrode?

Thanks

[chiralSPO]

Here are some accessible articles that discuss the role of Pt in the oxygen reduction reaction in fuel cells, they might answer some of your questions:

http://orbit.dtu.dk/files/12640977/rsc.pdf
https://www.nature.com/articles/ncomms15938

[scientizscht (OP)]

Here are some accessible articles that discuss the role of Pt in the oxygen reduction reaction in fuel cells, they might answer some of your questions:

http://orbit.dtu.dk/files/12640977/rsc.pdf
https://www.nature.com/articles/ncomms15938

Thanks but my dyslexia is killing me, would you skim through and give me any relevant info?

[chiralSPO]

The half reaction required for oxygen reduction is:

In acid:
O₂ + 4 e^– + 4H⁺ --> 2 H₂O

or in base:
O₂ + 4 e^– + 2H₂O --> 4 OH^–

In either case this is a complex reaction for electrochemistry, where elementary steps usually involve only one electron transfer at a time. Most of the intermediates along the way for an un-catalyzed system are very high in energy, like superoxide (O₂^–), and can actually require more energy to happen at any reasonable rate than could be supplied by oxidation of the fuel (If one put hydrogen and oxygen in a bottle and stuck a copper wire in there, nothing would ever happen).

Using a catalyst, these high-energy intermediates can either be stabilized, or avoided all together (put a platinum wire in that bottle of hydrogen and oxygen, and it will immediately react, consuming both the hydrogen and oxygen to make water).

[scientizscht (OP)]

The half reaction required for oxygen reduction is:

In acid:
O₂ + 4 e^– + 4H⁺ --> 2 H₂O

or in base:
O₂ + 4 e^– + 2H₂O --> 4 OH^–

In either case this is a complex reaction for electrochemistry, where elementary steps usually involve only one electron transfer at a time. Most of the intermediates along the way for an un-catalyzed system are very high in energy, like superoxide (O₂^–), and can actually require more energy to happen at any reasonable rate than could be supplied by oxidation of the fuel (If one put hydrogen and oxygen in a bottle and stuck a copper wire in there, nothing would ever happen).

Using a catalyst, these high-energy intermediates can either be stabilized, or avoided all together (put a platinum wire in that bottle of hydrogen and oxygen, and it will immediately react, consuming both the hydrogen and oxygen to make water).

What base and acid? I thought the catalyst grabs the oxygen from the atmosphere? And converts it to O2- ?

What is the rate per cm2 of catalyst that does that?

[chiralSPO]

The acid or base is the environment in which the electrode operates--yes it's open to the atmosphere, which is where the oxygen comes from, but there is also the side where the oxygen goes, and that's what matters.

Rates will depend on many, many variables: like what is the temperature, pH, pressure of oxygen, conductivity and geometry of the cell, load on the fuel cell, etc. etc. etc.

[scientizscht (OP)]

The acid or base is the environment in which the electrode operates--yes it's open to the atmosphere, which is where the oxygen comes from, but there is also the side where the oxygen goes, and that's what matters.

Rates will depend on many, many variables: like what is the temperature, pH, pressure of oxygen, conductivity and geometry of the cell, load on the fuel cell, etc. etc. etc.

Can you find me more info on these rates please? I am very struggling to find anything relevant.

[scientizscht (OP)]

The acid or base is the environment in which the electrode operates--yes it's open to the atmosphere, which is where the oxygen comes from, but there is also the side where the oxygen goes, and that's what matters.

Rates will depend on many, many variables: like what is the temperature, pH, pressure of oxygen, conductivity and geometry of the cell, load on the fuel cell, etc. etc. etc.

Can you find me more info on these rates please? I am very struggling to find anything relevant.

Anyone please?
Isn't this forum supposedly run by Cambridge university people?

[Colin2B]

Anyone please?
Isn't this forum supposedly run by Cambridge university people?

Everyone here gives their time free of charge, so it’s up go them whether they reply or not and it will depend on how busy they are with the day job.
In @chiralSPO you have a recognised expert in chemistry, but a very busy one.

[chiralSPO]

Honestly, I have tried... Apparently my own explanations lack the necessary detail, while academic literature is too detailed...

Here are some accessible articles that discuss the role of Pt in the oxygen reduction reaction in fuel cells, they might answer some of your questions:

http://orbit.dtu.dk/files/12640977/rsc.pdf
https://www.nature.com/articles/ncomms15938

Thanks but my dyslexia is killing me, would you skim through and give me any relevant info?