Predicting Standard Reduction Potential - NAD+

[WhiskeyTangoFoxtrot]

Khan Academy, Bioenergetics Questions Set 1

Both Khan and Kaplan say that a larger standard reduction potential means a greater tendency for a species to be reduced. The question is attached with the 3 hints shown beneath the answer choices. I feel like their first two hints indicate an SRP greater than 1 ?

Is it that NAD+ is just more "comfortable" unreduced? Or is there something more to the "half-reaction" that I'm missing?

---

Members don't see this ad.

 

[chemist16]

It says that NADH (the reduced form of NAD), is the key electron-donating molecule, i.e. it is reducing. Since it is reducing, NADH must be readily oxidized. That is, the oxidation reaction NADH to NAD is facile and thus the E of that reaction is greater than zero. Therefore, the E of the opposite reaction - reduction of NAD - must be equal but negative.

 

[WhiskeyTangoFoxtrot]

I was with you until

thus the E of that reaction is greater than zero

Not sure where you got that? Assuming this is true, of course the rest makes sense.

And I still don't understand how the answer fits with the definition of the SRP: a more positive standard reduction potential means a greater tendency for a species to be reduced. I would think NAD+ would have a greater tendency to be reduced and therefore more positive. So how do you predict/assign positive or negative E?

NAD+ + 2H+ + 2 e– --> NADH + H+

 

[chemist16]

Not sure where you got that? Assuming this is true, of course the rest makes sense.

Can you see why the oxidation reaction of NADH to NAD+ is facile? Because NADH is a good electron-donating molecule, i.e. reducing agent, the oxidation of NADH to NAD+ is facile. This is the reverse of this reaction:

NAD+ + 2H+ + 2 e– --> NADH + H+

Saying that the oxidation of NADH is facile is the same thing as saying the reduction of NAD+ to NADH is not facile. If something really wants to be oxidized, then the oxidized product isn't going to want to be reduced. This is a simple thermodynamic argument.

In other words, when you're moving from right to left, E of that half reaction (which is an oxidation) is positive. But when you're going the other direction - from left to right - the E is equal in magnitude but opposite in sign. If the forward E is somehow not equal in magnitude to the reverse E, then you could theoretically set up a cell that produces infinite energy.

 

[chemist16]

And I still don't understand how the answer fits with the definition of the SRP: a more positive standard reduction potential means a greater tendency for a species to be reduced. I would think NAD+ would have a greater tendency to be reduced and therefore more positive. So how do you predict/assign positive or negative E?

Here is the error in your thinking. NAD+ has a greater tendency to be reduced compared to what? You can't compare NAD+ and NADH because they are part of the same half cell - this won't tell you anything meaningful about the sign of E. When you're calculating a standard reduction potential, you're comparing only to the standard hydrogen electrode and whether NAD+ is more likely to be reduced relative to that. Furthermore, you are implicitly assuming that NAD+ is more likely to be reduced than NADH which is not true (read above post).

 

[WhiskeyTangoFoxtrot]

Point taken; relative tendency doesn't work for this problem.

I do agree/understand that NADH is a good reducer/e- donating molecule, but I'm not sure why it would rather be oxidized (ie why the backward reaction is stronger(?more favored) than the forward reaction) and going back to my first point of confusion, how this relates to predicting the SRP. Or in this case predicting SOP for NADH and flipping the sign for SRP of NAD+.

Forgot to say before - thank you for the reply. I really appreciate it!

 

[chemist16]

I do agree/understand that NADH is a good reducer/e- donating molecule, but I'm not sure why it would rather be oxidized (ie why the backward reaction is stronger(?more favored) than the forward reaction) and going back to my first point of confusion, how this relates to predicting the SRP. Or in this case predicting SOP for NADH and flipping the sign for SRP of NAD+.

The passage states clearly, "NAD, in its reduced form as NADH, is a key electron-donating molecule..." If you see that NADH is a good reducing agent, then you must agree that it would rather be oxidized than reduced - that's the very definition of a reducing agent. If it liked being reduced and not oxidized, then it would be an oxidizing agent and not a reducing agent. Then if you can see that it would rather be oxidized, then you would also agree that the reverse reaction is facile because the reverse reaction is nothing other than the oxidation of NADH. In other words, the E of the reverse reaction (some would call it the standard oxidation potential) is positive. So then from knowing a little electrochemistry, you know that if you flip the reaction around as written, the E must change signs. Therefore, the E for the forward reaction must be negative. Let me illustrate using a concrete example:

Say you have an electrochemical cell where the anode is aluminum. Since it's the anode, oxidation is occurring there and let's say aluminum is oxidized to Al+++. So the half reaction you look up in a standard reduction potential table is like this: Al+++ + 3e- ----> Al, E = -1.66 V. What's the E of the oxidation reaction? Well, you know from gen chem that E of the reaction Al ----> 3e- + Al+++ is just the same E as before but with the opposite sign: +1.66 V. How do you know this? Conservation of energy.

Now let's go the other way around. I give you this: NADH + H+ -----> 2H+ + 2e- + NAD+, E = +X. What's the E of the reduction reaction?

 

[tturchi51]

Add this to your notes
- LEO GER- Loss of e- is oxidation and Gain of e- is reduction.

- oxidizing agent-is the molecule, element, letters, etc. that gains e- (GER)

- reducing agent-is the molecule that loses e- (LEO)

-NAD+ MUST be available in order for glycolytic pathway to proceed normally (if not then nothing is there to carry the e-)

-The first molecule of NAD+ is reduced to NADH during the conversion of

isocitrate to alpha keto-glutarate (enzyme is isocitrate dehydrogenase)








1. Would anything with a (+) after it want to gain or lose an e-?

Gain e- because it wants to be neutral (aka have no + sign)

2. Did the NAD+ gain or lose an e- in order to become neutral NADH?

IT GAINED an e-!!!! According to GER the NAD+ was reduced to NADH. Also it reduced to NADH because the H+ followed the e- to NAD+


3. Since NAD+ is reduced to NADH, is the NAD+ an ox agent or red agent?

-oxidizing agent-because it helps whatever molecule wants to be oxidized by gaining e- (reducing itself)


Also here are my gen chem notes about ox red potential

Strongest oxidizing agent-F2

Strongest Reducing agent-Li(s)

The element that will most readily undergo oxidation is the one with the most negative reduction potential.

The more positive the reduction potential is, the more readily they will undergo reduction.

IDK what happened to the table but Al is the most - reduction potential and Ag has the most + reduction potential for the 2 examples


According to the table of reduction potentials, which of the following is the strongest reducing agent?


Al

Ag

Ag+

Al3+

The strongest reducing agent is the species that is most spontaneously oxidized (most positiveoxidationpotential). The oxidation reactions are the reverse of those listed in the table of reduction potentials and the oxidation potentials are the negative of those listed. The most positive oxidation potential would be +1.66Vand Al (not Al3+) is the reactant that is getting oxidized and is the strongest reducing agent (choice B).
Al Al3++ 3e- +1.66V








According to the table of reduction potentials, which of the following is the strongest oxidizing agent?


Al

Ag

Al3+

Ag+

The strongest oxidizing agent is the species that is most spontaneously reduced (most positive reduction potential). The most positive reduction potential listed is 0.80V and Ag+ (not Ag) is the reactant that is getting reduced and is the strongest oxidizing agent.








Also I highly recommend you look up LiAlH4 and NaBH4 for organic chemistry.

 

[amy_k]

To summarize all of the above responses for OP:

A more positive reduction potential indicates the species readily gains electrons (is reduced), and the reduction half equation equilibrium lies to the right. These species can act as oxidizing agents as they readily take electrons from other species, oxidizing the other species in the process.

A more negative (less than zero) reduction potential indicates the species more readily releases electrons, (is oxidised), and the reduction half equation equilibrium lies to the left. These species can act as reducing agents as they release electrons which can be used to reduce other species. This is the category that NADH falls into.

 

[amy_k]

Reading the question explanation - "electron accepting molecules such as NAD+ have a high tendency to be reduced under physiological conditions" could be misinterpreted... i.e. that they are saying it must have a positive reduction potential and be more readily reduced than NADH is oxidized.

NAD+ and NADH are used as both oxidizing and reducing agents respectively during respiration, but that doesn't mean that they are both equally spontaneous processes.

NAD+ acts as an oxidizing agent by accepting electrons and H+ to form NADH in the TCA. BUT it is important to realize that NAD+ is the low energy form and NADH is the high energy form so NAD+ must be raised to the higher energy form NADH in the energy investment phase of metabolism. I.e. this only readily occurs under physiological conditions as the explanation states. Then, NADH acts as an electron donating molecule by being oxidized later in the ETC, this process releases energy and occurs spontaneously. Thus NADH is more readily oxidized into NAD+ than NAD+ is reduced into NADH, but physiological conditions do promote NAD+ reduction to NADH at earlier stages of the metabolic pathway.

 

[WhiskeyTangoFoxtrot]

So here is the crux of my problem:

A more negative (less than zero) reduction potential indicates the species more readily releases electrons, (is oxidised), and the reduction half equation equilibrium lies to the left. These species can act as reducing agents as they release electrons which can be used to reduce other species. This is the category that NADH falls into.

If NADH falls into this category of negative SRPs, how did @chemist16 assign it a positive E? What is the line of reasoning between NADH readily releases electrons and NADH has positive E? Is there a maxim/rule/relationship that I'm missing like, all good reducers have positive E?

Or again, this line

That is, the oxidation reaction NADH to NAD is facile and thus the E of that reaction is greater than zero.

makes it sound like positive E is conditioned on facility of oxidation. Is that the relationship?

 

[chemist16]

If NADH falls into this category of negative SRPs, how did @chemist16 assign it a positive E? What is the line of reasoning between NADH readily releases electrons and NADH has positive E? Is there a maxim/rule/relationship that I'm missing like, all good reducers have positive E?

What @amy_k means is that the half reaction of NAD+ being reduced to NADH has a negative E - consistent with the answer to the question. In other words, the reduction half reaction: 2H+ + 2e- + NAD+ <---> NADH + H+ has an equilibrium that lies to the left. This means that NAD+ doesn't like getting reduced. Therefore, it must have a negative standard reduction potential. It wouldn't make sense to say that NADH has some E because E is associated with some electron transfer process - either an oxidation or a reduction.

Or again, this line
makes it sound like positive E is conditioned on facility of oxidation. Is that the relationship?

Whenever you encounter a question like this, you want to ask: does the species in question like being oxidized or reduced? In this case, the species is NADH. It clearly likes getting oxidized. From basic electrochemistry, you know that something that likes to be oxidized has a positive E for the oxidation half reaction. But we're not talking about the oxidation half reaction here. We're talking about the reduction half reaction - the reverse of the oxidation reaction. So if the oxidation reaction has a positive E, then the reduction reaction (the same reaction but in reverse) has to have a negative E of the same magnitude - that's the concept you need to know.

 

[WhiskeyTangoFoxtrot]

From basic electrochemistry, you know that something that likes to be oxidized has a positive E for the oxidation half reaction.

Thank you! This is exactly what I was looking for. However I'm having trouble finding support for this in my Kaplan Gen Chem chapter on Redox reactions. If it's not too much trouble would you be able to link me to something more definitive? In any case, thanks!

 

[chemist16]

Thank you! This is exactly what I was looking for. However I'm having trouble finding support for this in my Kaplan Gen Chem chapter on Redox reactions. If it's not too much trouble would you be able to link me to something more definitive? In any case, thanks!

Now I know what you're confused about. It's a matter of convention and signs. I can write out any electrochemical half reaction and have an E for that half-reaction. But according to convention, we assign everything a standard reduction potential with the understanding that the reverse of that reaction, the oxidation, has the same magnitude potential but with opposite sign. Now when you calculate the E(cell), you're usually taught that it's E(cathode) - E(anode). This is misleading and is just a terrible way to teach it because many people forget that in this particular equation, E(cathode) and E(anode) specifically refer to standard reduction potentials. It emphasizes memorization too much and not the conceptual understanding of it. When you calculate E(cell), you're basically just calculating the net potential of the cell. Therefore, an easier way to think of it is E(oxidation) + E(reduction) where E(oxidation) is the E of the oxidation reaction and E(reduction) is the E of the reduction reaction.

For example:

Say you have a Daniel cell. At the anode is the following reaction: Zn ---> Zn2+ + 2e-. At the cathode is the following reaction: Cu2+ + 2e- ---> Cu. Now calculate E(cell). If you do this the way you memorized it, you just have to look up the reduction potentials of Cu2+ and Zn2+ and subtract Zn2+ from Cu2+. But that obscures the concept. The concept is that you want the net potential of this cell. If I asked you to tell me how many steps I've taken forward when I took 2000 steps forward and 1000 steps backward, you would say 1000. You got that by adding 2000 and -1000. We're going to do the same thing.

You can look the following up in a table of standard reduction potentials:

Zn2+ + 2e- ---> Zn, E = -0.76 V
Cu2+ + 2e- ---> Cu, E = +0.34 V

How do you solve? Well, first you look at the half cell reactions I've written above. The half reaction occurring at the cathode is the exact same as the one you looked up, so you don't have to do anything to that. But the half reaction occurring at the anode is the reverse of the reaction you're given. So, as I noted above, all you have to do is flip the equation you're given, acknowledging that the E of that reaction simply reverses sign. This is because of conservation of energy. If this were not true, you could theoretically create infinite energy by reducing something and then oxidizing it again later. So now you have something like this:

Zn ---> Zn2+ + 2e-, E = +0.76 V
Cu2+ + 2e- ---> Cu, E = +0.34 V

You calculate the net potential and get 1.10 V. So let's go back to the reduction half-reaction of Zn2+ - you notice that it's negative. Specifically, it's negative relative to the standard hydrogen electrode. So what does reduction potential measure? It measures something's affinity for electrons relative to hydrogen. Negative means that the thing simply doesn't like electrons relative to hydrogen, i.e. it doesn't want to be reduced. This is the concept you must know. Again, a positive reduction potential means something wants to be reduced. This is the convention.

Now, apply the same concept here but with oxidation potentials. Oxidation potential is simply the formal reverse of reduction potentials - therefore, something that has a high affinity for electrons won't want to be oxidized (it wants to keep its electrons) and the potential for the oxidation reaction is negative and vice versa. You have 2H+ + 2e- + NAD+ <---> NADH + H+. You're told explicitly that NADH is a key electron-donating molecule. Does that mean it has a high affinity for electrons or a low affinity for electrons? It has a low affinity for electrons because otherwise, it wouldn't want to donate those electrons. Therefore, it wants to be oxidized. Why? Because you're taking electrons from something that doesn't want electrons to begin with. So NADH + H+ ---> 2H+ + 2e- + NAD+ must have a positive E. Now, just as you flipped Zn2+ + 2e- ---> Zn above, you flip the previous equation and flip the sign of the E as you do so.

 

[WhiskeyTangoFoxtrot]

Perfect, thank you!