Why is oxygen the final electron acceptor in cellular respiration

[laczlacylaci]

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In cellular respiration, electrons are transferred from glucose to oxygen through a series of chemical species. At the end of the ETC, we produce H2O (at complex IV) and ATP (at ATPase). For O2 to form H2O, isn't there a transfer of electrons (coming from NADH&FADH2)? Why do we say O2 is the final electron acceptor rather than H2O?

 

[premedkid1994]

Oxygen accepts the electrons to become H20... Water isn't the one getting reduced
The reaction is C6H6O6 + Oxygen ----> CO2 and H2O.
Key combustion reaction to know btw

 

[wizzed101]

Because the final reaction of the ETC doesn't produce water!

The reaction is ONE electron (this is important- just ONE electron) + O2 -> [O2].

That radical can oxidize a lot of things. Consequently, we are evolved to be protected from its deleterious effects. Forming water is an adaptation.

I don't think it is a good idea to think of glycolysis + Krebs + ETC as combustion. That is too much of an oversimplification imo. Like, I bet only 0.1% (or nobody?) here can account for all the water that are produced in the process. (I tried and failed) If you can, please enlighten me!

 

[DoctorCakes]

If H20 were the final electron acceptor then hypothetically it would form H2O2. This is not the case. Remember that the transfer of electrons is usually accompanied by a H+ so O2 accepts an electron which also brings a H+ ion forming H20. Another example: Reduction of NAD+ forms NADH.

 

[betterfuture]

Is this a philosophical 'why'?

Edit: Nvm, I can't read😳

 

[aldol16]

If H20 were the final electron acceptor then hypothetically it would form H2O2. This is not the case. Remember that the transfer of electrons is usually accompanied by a H+ so O2 accepts an electron which also brings a H+ ion forming H20. Another example: Reduction of NAD+ forms NADH.

Reducing water doesn't form H2O2. You would need to oxidize water. H2O2 is an intermediate oxidized form of oxygen - the oxidation state here is -1, which is between that of oxygen in molecular O2 and water. So to get H2O2, you need to oxidize water by 1 unit. That's how two H2O2 can disproportionate to form O2 and water - one molecule is reduced while the other one is oxidized.

 

[laczlacylaci]

Oxygen accepts the electrons to become H20... Water isn't the one getting reduced
The reaction is C6H6O6 + Oxygen ----> CO2 and H2O.
Key combustion reaction to know btw

Yes, I agree. So what I'm thinking is since O2 accepts the electrons, those electrons eventually are inside H2O bonds, correct? So wouldn't H2O be the FINAL electron acceptor? Or maybe I'm confusing acceptor with "keeper"
I understand that 1/2 O2 was reduced to form water. But in reduction reactions, we usually say that there is a gain of electrons for the oxidation agent (in this case, H2O)

Reducing water doesn't form H2O2. You would need to oxidize water. H2O2 is an intermediate oxidized form of oxygen - the oxidation state here is -1, which is between that of oxygen in molecular O2 and water. So to get H2O2, you need to oxidize water by 1 unit. That's how two H2O2 can disproportionate to form O2 and water - one molecule is reduced while the other one is oxidized.

Your thoughts? Is it that I'm confusing electron acceptor with electron keeper?

Thanks all!

 

[aldol16]

Yes, I agree. So what I'm thinking is since O2 accepts the electrons, those electrons eventually are inside H2O bonds, correct? So wouldn't H2O be the FINAL electron acceptor? Or maybe I'm confusing acceptor with "keeper"

I understand that 1/2 O2 was reduced to form water. But in reduction reactions, we usually say that there is a gain of electrons for the oxidation agent (in this case, H2O)

Your thoughts? Is it that I'm confusing electron acceptor with electron keeper?

In reduction reactions, you say that there is a gain of electrons for the oxidizing agent. The oxidizing agent here is O2. Maybe you're confusing the definition of an oxidizing agent. The oxidizing agent oxidizes something else and therefore gets reduced itself.

Okay, let's think of a simpler scenario. You have a dollar. Your friend has a dollar. You give that dollar to your friend. Who has accepted the dollar? Your friend. You wouldn't say that the system of your friend and the dollar (i.e. your friend with the dollar) has accepted the dollar. You would say that your friend is the one doing the accepting. This is a matter of using the correct direct object.

 

[laczlacylaci]

In reduction reactions, you say that there is a gain of electrons for the oxidizing agent. The oxidizing agent here is O2. Maybe you're confusing the definition of an oxidizing agent. The oxidizing agent oxidizes something else and therefore gets reduced itself.

Okay, let's think of a simpler scenario. You have a dollar. Your friend has a dollar. You give that dollar to your friend. Who has accepted the dollar? Your friend. You wouldn't say that the system of your friend and the dollar (i.e. your friend with the dollar) has accepted the dollar. You would say that your friend is the one doing the accepting. This is a matter of using the correct direct object.

Ah oops, I meant to say reducing agent. Got it with the dollar analogy, that makes more sense. Thanks @aldol16 !

 

[DoctorCakes]

Reducing water doesn't form H2O2. You would need to oxidize water. H2O2 is an intermediate oxidized form of oxygen - the oxidation state here is -1, which is between that of oxygen in molecular O2 and water. So to get H2O2, you need to oxidize water by 1 unit. That's how two H2O2 can disproportionate to form O2 and water - one molecule is reduced while the other one is oxidized.

oops your right, sorry bout the half-ass analogy.