TBR ox/red problem

[greenseeking]

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Given that the conversion of silver chloride into silver metal and chlorine gas requires 0.56 volts, what is true about silver?

Given that Cl2 + 2e- --> 2 Cl- emf=1.36 Volts
(Equation and emf for silver not given in the problem)

A. oxidiation of silver metal has an emf of -.80 V
B. oxidation of silver cation has an emf of -.80 V
C. Reduction of silver metal has an emf of -.80 V
D. reduction of silver cation has an emf of -.80 V

Correct Answer is A
I got D... I was able to eliminate B and C because it's hard to oxidate a cation and hard to reduce a metal. Then I thought that it would be D because Cl- gets OXIDIZED to Cl2 (1.36) and Ag+ will get REDUCED to Ag. Thus, 0.56= 1.36 + (emf for Ag+); emf for Ag+=-.80.

Can someone please explain what is wrong with my reasoning? Thanks in advance.

 

[ashtonjam]

2AgCl -> Ag(s) + Cl2(g) -0.56V overall reaction.

The way this reaction is written, silver is reduced and chlorine is oxidized. The voltage is negative because it is unfavorable and requires at least 0.56V of external energy to get going.

In the half cell reaction of chlorine, Cl2 + 2e- --> 2 Cl- 1.36V, chlorine is being reduced. This is the reverse of what is shown above. Flipping it gives 2Cl- -> 2e- + Cl2 -1.36V (chlorine oxidation).

Now we want silver to be reduced. Ag+ + e- -> Ag(s) ??V (not known yet)

Adding Ered (unknown EMF of silver half reaction) and Eox (chlorine half reaction EMF) together will give the right answer.

Etot = Ered + Eox
-0.56 = Ered + -1.36V
0.8V = Ered for silver reduction

The half reaction of silver ion reduction Ag+ + e- -> Ag(s) is therefore 0.8V from the little math problem above.

The reverse of this is Ag(s) -> e- + Ag+. Silver metal oxidation is -0.8V.

 

[greenseeking]

Brilliant! Thanks.
Just one question- how do you know that it's -.56 V and not .56 V?? Does it have to do with G=-nfE?

 

[ashtonjam]

Brilliant! Thanks.
Just one question- how do you know that it's -.56 V and not .56 V?? Does it have to do with G=-nfE?

It's mostly because you said that the reaction requires 0.56V, so I just went ahead and made that assumption. If overall EMF is positive, then no outside push is needed; it is spontaneous and just gets started on its own. dG = -nFE; E is positive. dG is negative, meaning that it's spontaneous.

If the reaction requires a minimum external voltage to get started, then the E of the system must be negative. If the external voltage is big enough to get the total voltage to a positive number, then the cell will finally run in the unfavorable direction. dG = -nFE. E is negative. dG is a positive number, and the reaction will run in the opposite direction unless an outside emf is applied to make 1) E positive again and 2) dG negative again.

 

[greenseeking]

Thanks so much. I did not know that the word requires means negative voltage. Your explanation was great!

 

[BerkReviewTeach]

Fantastic explanation AshtonJam. Well thought, clearly reasoned, and systematic. Absolutely nothing to add besides "thank you."