Standard Potentials, super confused

[TinaBina22]

I'm super confused about standard electrode potentials. I always thought I understood them, but apparently I don't.

So in a chart of standard electrode potentials, how do you know which species is oxidized or reduced? I always thought that the most negative potential meant that species was the most likely to get oxidized, and then visa versa for reduced. And then the strongest oxidizing agent would be that species that's most positive, whereas the strongest reducing agent is most positive. But that line of reasoning doesn't work when I'm answering questions in EK and TBR...So could some please explain it to me?

Thanks!
Tina

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[TinaBina22]

For instance, there's an EK chemistry question, #21 on page 99.

Essentially it gives us that:

2F- --> F2 + 2e-, -2.87V
2Cl- --> Cl2 + 2e-, -1.36V

And I thought that the strongest reducing agent would be F-, because it has the lowest electrode potential. However, the answer is Cl-. Why is that? It's really confusing me...

 

[superman3]

You understand electrode potentials. However, in this particular case you are applying the concept incorrectly. It is important to distinguish between a reduction potential and an oxidation potential. The question you are asking about provides you with oxidation potentials, since the equations show oxidation reactions occuring.This means that the strongest reducing agent (and the substance that wants to be oxidized the most), will be the substance with the highest oxidation potential, which in this case is Cl-.

 

[piojita63]

For instance, there's an EK chemistry question, #21 on page 99.

Essentially it gives us that:

2F- --> F2 + 2e-, -2.87V
2Cl- --> Cl2 + 2e-, -1.36V

And I thought that the strongest reducing agent would be F-, because it has the lowest electrode potential. However, the answer is Cl-. Why is that? It's really confusing me...

Remember that F- is more electronegative so it doesn't "want" to lose it's electrons (be oxidized) when compared to Cl-

 

[rjosh33]

I'm super confused about standard electrode potentials. I always thought I understood them, but apparently I don't.

So in a chart of standard electrode potentials, how do you know which species is oxidized or reduced? I always thought that the most negative potential meant that species was the most likely to get oxidized, and then visa versa for reduced. And then the strongest oxidizing agent would be that species that's most positive, whereas the strongest reducing agent is most positive. But that line of reasoning doesn't work when I'm answering questions in EK and TBR...So could some please explain it to me?

Thanks!
Tina

I always found it helpful to first note whether the half reaction as written was a reduction (e- on the left of the rxn arrow, oxidation state getting more negative) or an oxidation (e- on the right, oxidation state getting more positive). Then, it was very important to note the sign of the standard potential (Eo).

We can relate standard potentials to free energy through the equation delta G = -nFE. Because of the negative sign, a negative standard potential will give a positive (nonspontaneous) delta G, and a positive E will give a negative (spontaneous) delta G. From this information you can tell if the reaction as written "wants" to happen or not.

For example, in your problem from EK, they have:

2F- --> F2 + 2e-, -2.87V
2Cl- --> Cl2 + 2e-, -1.36V

Because each standard potential is negative, they both result in positive delta G's -- they both don't "want" to be oxidized. That is, F2 and Cl2 much prefer to be reduced to their respective anions. However, the question asks which is the best reducing agent, which is basically the same as asking which most wants to be oxidized (reducing agents get oxidized, oxidizing agents get reduced). Well, they both don't want to happen, but because fluorine has the most negative standard potential as an oxidation, it really doesn't want to be oxidized. So, between F- and Cl- (the two species that are getting oxidized), Cl- would prefer to be oxidized, meaning it would be the better reducing agent.

 

[Temperature101]

For instance, there's an EK chemistry question, #21 on page 99.

Essentially it gives us that:

2F- --> F2 + 2e-, -2.87V
2Cl- --> Cl2 + 2e-, -1.36V

And I thought that the strongest reducing agent would be F-, because it has the lowest electrode potential. However, the answer is Cl-. Why is that? It's really confusing me...

You got to be careful here because MCAT will do that...superman3 and rjosh33 give you good expanation why Cl- is the answer. Try to make the distinction as to when they give you an oxidation or reduction potential half reaction.

 

[Dasypus]

And sometimes, on these sorts of questions, 'stronger' means 'less weak'....

 

[MedPR]

For instance, there's an EK chemistry question, #21 on page 99.

Essentially it gives us that:

2F- --> F2 + 2e-, -2.87V
2Cl- --> Cl2 + 2e-, -1.36V

And I thought that the strongest reducing agent would be F-, because it has the lowest electrode potential. However, the answer is Cl-. Why is that? It's really confusing me...

For X ---> Y + e-: X is being oxidized, Y is being reduced.
For X + e- ----> Y: X is being reduced; Y is being oxidized

In the potentials you listed, F- is being oxidized, and F is being reduced. The larger number means more likely to happen and -1.36 is larger than -2.87. So Cl- is more likely to be oxidized than F- and therefore Cl- is the stronger reducing agent.