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Topic: Electrochemistry Ag and Fe(II)/Fe(III)  (Read 4927 times)

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Offline THC

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Electrochemistry Ag and Fe(II)/Fe(III)
« on: March 24, 2008, 06:35:35 AM »
Okay, this question consists of multple question. I (think I've) managed to solve the first, but I'm not so sure about the last questions. Any help is appreciated.


Q

An electrochemical cell consists of:
1) A Ag electrode in 100 mL 0.100 M AgNO3 solution
2) A Pt electrode in 100 mL of Fe(II) and Fe(III), where [Fe^2+] = [Fe^3+] = 0.0500 M.
The half-cells are connected by a salt bridge, and T = 25 *C.

a) Write the cell diagram of the cell and calculate e_0 for the Ag half-cell.
b) Calculate E_cell and write the reaction that occurs (the spontaneous reaction).
c) The volume in both half cells are increased to 200 mL (water is added). What is E_cell now?
d) Calculate the equilibrium constant for the spontaneous reaction.
e) Calculate [Ag+] and [Fe^2+] at equilibrium.



A

a) I looked up the standard electrode potentials, e_0*, for the half-cells:
e_0*(Ag+,Ag) = 0.800 V
e_0*(Fe^2+,Fe^3+) = 0.771 V
So Fe^2+ is oxidized to Fe^3+ and Ag+ is reduced to Ag, the cell diagram becomes
Pt | Fe^2+, Fe^3+ || Ag+ | Ag
and e_0(Ag) = e_0* - 0.059 V/z *log 1/[Ag+] = 0.800 V - 0.059V*log 1/0.1 = 0.741 V

b) First, I calculate e_o(Fe) =0.771 V - 0.059 V * log 1 = 0.771 V.
So E_cell = 0.771 V - 0.741 V = 0.030 V.
The spontaneous reaction is therefore
Fe^3+ + Ag -> Fe^2+ + Ag+
This is a bit weird. Should I rewrite the cell diagram in a) now? Because I only took e_0* into account, not e_0? But I couldn't have known before this question?

c) Anyways, moving on. I keep getting to different results in this one.
So [Ag+] = 0.050 M and [Fe^2+] = [Fe^3+] = 0.025 M.
First "solution":
e_0(Ag) = 0.8 V - 0.059log 0.05 = 0.723 V
e_0(Fe) = 0.771 V,
so E_cell = 0.771 V - 0.723 V = 0.048 V
The accompanying reaction is Fe^2+ + Ag+ -> Fe^3+ + Ag, right?
Second "solution":
E_cell =  0.8 V - 0.771 V - 0.059log 0.05 = 0.106 V
In this "solution", I think the reaction is Fe^3+ + Ag -> Fe^2+ + Ag+

Okay, so this is all very confusing. I pretty sure I messed up somewhere, but I can't see what I did wrong. And I still haven't solved the two last, but I think it's better if I solve a, b and c first!

Offline THC

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Re: Electrochemistry Ag and Fe(II)/Fe(III)
« Reply #1 on: March 25, 2008, 02:12:19 PM »
Did I really manage to scare all of you just like that? I guess I'm not the only one who's confused about this one : ;D

Offline Arkcon

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Re: Electrochemistry Ag and Fe(II)/Fe(III)
« Reply #2 on: March 25, 2008, 02:38:52 PM »
I dunno, didn't someone else ask the exact same question a couple of days ago?  Now that I come to mention it, the Fe3+|Fe2+ half cell in this question is new, but the Ag half call, now that seems very familiar.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline THC

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Re: Electrochemistry Ag and Fe(II)/Fe(III)
« Reply #3 on: March 25, 2008, 03:44:29 PM »
I dunno, didn't someone else ask the exact same question a couple of days ago?  Now that I come to mention it, the Fe3+|Fe2+ half cell in this question is new, but the Ag half call, now that seems very familiar.

I searched the forum for "ag", " ag half cell" and "silver" and it didn't come up with anything relevant. Could you throw a link in my direction?

I understand the whole thing is a bit overwhelming... a lot of questions, and I guess my calculations can be hard to decipher :O

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