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Offline 21385

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Electrochem question
« on: February 07, 2008, 08:09:37 PM »
Problem::
An excess of finely divided iron is stirred up with a solution that contains Cu(2+) ions. The system is allowed to come to equilibrium. The solid materials are filtered off and electrodes of solid copper and solid iron are inserted into the remaining soltion. What potential develops between these two electrodes at 25C?
a) 0 V b) -0.78 V c) 0.596 V d) 0.298 V e) cannot be determined

My attempt::
Cu(2+) + 2e- => Cu   0.34 V
Fe(2+) + 2e- => Fe    -0.45 V

I am thinking that because these two top half-reactions occur, and because the cell will always tries to make its voltage the most positive, the voltage will be 0.34 V + 0.45 V = 0.79 V, for the reaction Fe + Cu(2+) => Fe(2+) + Cu. However, no answer in the answer choices even come close to 0.79 V. choice B is -0.78 V but I don't know why it is negative. Can someone explain this to me? Thanks a lot

Offline Arkcon

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Re: Electrochem question
« Reply #1 on: February 07, 2008, 09:49:34 PM »
You are missing something very fundamental.  You're lucky this is a multiple guess question -- look at each answer, is there any way for any of them to be true?

Oh, and the usual tip, when in doubt, start with a balanced equation.  What did they do to the solution before they even inserted the electrodes?
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Offline 21385

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Re: Electrochem question
« Reply #2 on: February 08, 2008, 12:00:53 AM »
Right, Right, I missed the equilibrium part. Thus, the answer is a) 0 V.

Offline Arkcon

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Re: Electrochem question
« Reply #3 on: February 08, 2008, 07:55:54 AM »
Yes, um no, I mean, right answer, wrong reason.  First off, they tell you they have a Cu solution, then they dump an excess of iron into it.  The word equilibrium is a trick here, equilibrium doesn't matter when you are in excess.  Also, the reaction:

Cu2+ + Fe0 --> Cu0 + Fe3+(I think) *EDIT* and I'm wrong, see below

is not an equilibrium, this was driven completely to the right, because it is a precipitation.  Then they filter, so, all this work, just to say, "A solution of Fe3+." *EDIT* Wrong again

Then they stick two electrodes, into the same tube, no salt bridge, and ask for potential.  That simply won't work.

Presumably, they gave you electrode potentials on a header to this entire exam, or you looked them up, but they might have provided them for you just for this question.  In which case ...

They've tricked you.

Look at all the work you've done, and I've done.  The answer was zero, for very fundamental (that's my polite way of saying: basic, first day of class) reasons. 

If this is a standardized test, or if your teacher gives you a lot of these, put some work into finding these trick questions, so you can maximize your grade.
« Last Edit: February 08, 2008, 09:04:50 AM by Arkcon »
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Offline Borek

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Re: Electrochem question
« Reply #4 on: February 08, 2008, 08:21:04 AM »
Then they filter, so, all this work, just to say, "A solution of Fe3+."

Yes - and no. It MUST contain also Cu2+ although in much lesser amount. There are two redox systems present, both require solid and ions to work. It is also more likely to contain just Fe2+ (remember Fe3+ is oxidizer strong enough to etch copper).

Still, you are right about the equilibrium being the key word here. You are removing solid mixture of copper/iron and you are adding copper and iron electrodes. What have changed? Nothing. System was at equilibrium and it is at equilbrium.
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Offline Arkcon

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Re: Electrochem question
« Reply #5 on: February 08, 2008, 09:02:38 AM »
Then they filter, so, all this work, just to say, "A solution of Fe3+."
It is also more likely to contain just Fe2+ (remember Fe3+ is oxidizer strong enough to etch copper).

Ahhh...I knew I'd screw that up. OK.  So, copper ions, you get the redox with free iron, to a solution of iron ions and copper metal.  I used to play with that, back in the day, dumping iron filings into a copper salt solution, what you got was copper, a clear liquid, and a bunch of rusty iron -- maybe the Fe2+ reacts with free iron to FeO3?  I dunno.

But equilibrium?  Is that what happens?  Some copper redissolves?  Or some Cu2+ remains unreacted?  I don't see that happening.

Anyway, if there was residual copper(which there wasn't because there was an excess of iron,) and you put an bar of iron in, the copper plates out -- on the bar, just as efficiently as on the iron filings.  So there's no Cu2+ in this solution to be reduced to Cu0 in the electrochemical redox, which, won't even happen anyway because they're trying to do it mixed in one tube instead of two with a salt-bridge so they can have separate oxidation-reduction reactions with each metal and it's ions.

Shessh, what a lot of work, all for a trick question. 21385: are you looking at this mess of an explanation?
« Last Edit: February 08, 2008, 09:25:52 AM by Arkcon »
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Offline 21385

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Re: Electrochem question
« Reply #6 on: February 08, 2008, 09:22:36 PM »
Thank you very much guys for all this explanation.
I think I get it. Since it is at equilibrium before, it has 0 V. Because the addition and removal of solids does not change the equilibrium, the rxn is still at 0 V. Is this right?

Thanks a lot

Offline Arkcon

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Re: Electrochem question
« Reply #7 on: February 08, 2008, 09:35:23 PM »
There are 0 v because the cell is built wrong.  Draw a picture of the cell described -- a beaker with Fe2+ ions, (and maybe a trace of Cu2+,) a copper electrode and an iron electrode.  Compare it with the picture of an electrochemical cell in your book.  At least 1, or maybe 2 pieces are missing.
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Offline 21385

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Re: Electrochem question
« Reply #8 on: February 08, 2008, 10:32:09 PM »
Sorry for being a bit slow

Are you saying that because there is no separation between the two electrodes, there is no reaction. In addition, the copper ions will all be reduced to copper and there is no copper ions left to continue this reaction.

Offline Arkcon

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Re: Electrochem question
« Reply #9 on: February 09, 2008, 07:27:59 AM »
Yes, and the most important thing for you to learn is:

They've tricked you.

I never change font size on these forums, but this was too important a point for you to miss.  They made you waste time doing math, when you should have just looked at the question, checked off zero, and moved on.

Unless your multiple choice exams require an explanation or a calculation.  In which case, you did have to write something.  I was just guessing this was a standardized test, where the more questions you get to, the better.
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Offline 21385

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Re: Electrochem question
« Reply #10 on: February 09, 2008, 09:56:16 PM »
Thank you very much for your help, Arkcon

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