Textbook 6M7


Moderators: Chem_Mod, Chem_Admin

jessicasilverstein1F
Posts: 134
Joined: Wed Sep 30, 2020 9:57 pm

Textbook 6M7

Postby jessicasilverstein1F » Tue Mar 02, 2021 8:42 pm

Arrange the following metals in order of increasing strength as reducing agents for species in aqueous solution: (a) Cu, Zn, Cr, Fe; (b) Li, Na, K, Mg; (c) U, V, Ti, Al; (d) Ni, Sn, Au, Ag.

Does anyone know what the rule for this is?

JonathanSung_2G
Posts: 100
Joined: Wed Sep 30, 2020 9:49 pm

Re: Textbook 6M7

Postby JonathanSung_2G » Tue Mar 02, 2021 9:03 pm

In order to solve this problem, look at Appendix 2B where all of the reduction potentials for all of the elements are listed. The problem is asking you for increasing strength as reducing agents, which also correlates to the most likely elements to be oxidized. Since these potential values correspond to the reduction potential, the lowest (most negative) potential corresponds to the element that is most likely to be oxidized. Hope this helped!

Mary Shih 3J
Posts: 106
Joined: Wed Sep 30, 2020 9:39 pm

Re: Textbook 6M7

Postby Mary Shih 3J » Tue Mar 02, 2021 10:00 pm

since they are asking which is the best reducing agent, you need to find out which is Most likely to be oxidized or lose an electron. Someone correct me if I am wrong but I think bigger atoms on the periodic table lose atoms more easily so they could be more easily oxidized. So maybe compare atom sizes and see if that is the correct answer

Charlene D 3H
Posts: 110
Joined: Wed Sep 30, 2020 9:54 pm

Re: Textbook 6M7

Postby Charlene D 3H » Sun Mar 07, 2021 3:47 pm

Mary Shih 3J wrote:since they are asking which is the best reducing agent, you need to find out which is Most likely to be oxidized or lose an electron. Someone correct me if I am wrong but I think bigger atoms on the periodic table lose atoms more easily so they could be more easily oxidized. So maybe compare atom sizes and see if that is the correct answer

Hi Mary,
I believe the explanation above is what we're supposed to do for this problem. When I compared size, it didn't correspond to the textbook's answer. Logically I would have thought the same thing but I think we are supposed to compare the reduction potentials on the table.
The more negative the reduction potential, the more positive the oxidation potential of the metal in question. And the more positive the oxidation potential, the more likely it is to be oxidized, making it a stronger reducing agent (can give away e- more readily).

In summary: the smaller the reduction potential, the greater the oxidation potential, and the stronger the reducing power.

Also, note that there are multiple reduction potentials given for the same metal. I learned from another post that we are supposed to use the equation that has the ion with the charge closest to reactions with only one charge
(ie Zn^2+ only exists in this form so we would look at Cr^2+ equation not Cr^3+)

RitaThomas_3G
Posts: 100
Joined: Wed Sep 30, 2020 9:40 pm

Re: Textbook 6M7

Postby RitaThomas_3G » Sat Mar 13, 2021 7:36 pm

Just to add on, I do not believe we are required to memorize the standard potentials for the elements. So, if a question like this were to arise on the final, I believe we would be given the values!


Return to “Work, Gibbs Free Energy, Cell (Redox) Potentials”

Who is online

Users browsing this forum: No registered users and 3 guests