Trends in Cell (Redox) Potentials
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Trends in Cell (Redox) Potentials
On page 531, the text makes the statement that "the most negative [standard reduction potentials]...are usually found toward the left of the periodic table, and the most positive...are found toward the upper right." But the elements found on the left of the periodic table often have positive oxidation states meaning that they would more likely be reduced correct? Wouldn't this indicate a more positive standard reduction potential instead of a negative one? And the elements found on the right of the periodic table often have negative oxidation states meaning that they would more likely be oxidized correct? Wouldn't this indicate a more negative standard reduction potential instead of a positive one? Is there an error in my reasoning or are these facts just a contradiction in of themselves?
Re: Trends in Cell (Redox) Potentials
I think it may help you if you think about the natural properties of the elements in each group (e.g. ionization energy, electron affinity, etc) and how they are usually found. Think about the energy levels of each of the possible states. The alkali metals on the far left of the periodic table are most commonly found as cations (e.g. Na+, Li+) as they can easily donate their electrons, while the halogens are generally seen in a diatomic state (e.g. F2, Cl2) until they get more electrons, in which case a reduction would result in lower energy (but the key point is that the extra negative charge may not exist in the solution/gas!).
If you think about it this way, then there is no way that an Na+ or Li+ would go back to Na or Li, the resulting state would not be stable and the process would require a lot of energy (deltaGº>0). Because the Gibbs free energy is positive, then Eº<0 (by deltaGº=-nFEº). As for the halogens, although they are most commonly seen in their diatomic states, as I mentioned above, that is most likely because they are not given the opportunity to undergo reduction because there is no negative charge present. However, when in contact with an electron donor, they still have very strong electron affinity. For example, consider diatomic fluorine being reduced through the equation F2 + 2e- --> 2F-. This has one of the most positive Eº values, because the fluorine can release so much extra energy. Thus strong electron affinity-->high standard reduction potential.
Hope that helps,
Iris
If you think about it this way, then there is no way that an Na+ or Li+ would go back to Na or Li, the resulting state would not be stable and the process would require a lot of energy (deltaGº>0). Because the Gibbs free energy is positive, then Eº<0 (by deltaGº=-nFEº). As for the halogens, although they are most commonly seen in their diatomic states, as I mentioned above, that is most likely because they are not given the opportunity to undergo reduction because there is no negative charge present. However, when in contact with an electron donor, they still have very strong electron affinity. For example, consider diatomic fluorine being reduced through the equation F2 + 2e- --> 2F-. This has one of the most positive Eº values, because the fluorine can release so much extra energy. Thus strong electron affinity-->high standard reduction potential.
Hope that helps,
Iris
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Re: Trends in Cell (Redox) Potentials
Essentially the way you want to look at it is like this:
Oxidation is loss of electrons (OIL) and reduction is gain (RIG).
F is very electronegative so it will want to gain electrons. Thus, it will very much want to be reduced and likewise have a very positive and large E0red.
Now consider something like Li. It is electronegativity is pretty weak compared to that of F so that means it will have its electrons taken away pretty easily. In more chemical terms, it gets oxidized really easily so it will have a very negative (large magnitude negative number) reduction potential. Thus, Li is a lousy electrode for reduction but conversely a great one for oxidation!
Oxidation is loss of electrons (OIL) and reduction is gain (RIG).
F is very electronegative so it will want to gain electrons. Thus, it will very much want to be reduced and likewise have a very positive and large E0red.
Now consider something like Li. It is electronegativity is pretty weak compared to that of F so that means it will have its electrons taken away pretty easily. In more chemical terms, it gets oxidized really easily so it will have a very negative (large magnitude negative number) reduction potential. Thus, Li is a lousy electrode for reduction but conversely a great one for oxidation!
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