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Topic: Electrochemical cell / electrolytic cell  (Read 5989 times)

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Offline Capital

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Electrochemical cell / electrolytic cell
« on: June 06, 2011, 02:03:21 AM »
Please click on the following link for the two questions

http://i1008.photobucket.com/albums/af204/John132456/Untitled18.jpg

The problem I am having is that because there is a porous barrier, I can't tell whether it is an electrochemical cell or an electrolytic cell because the two beaker (sign it is an electrochemical cell) and the one beaker (sign it is an electrolytic cell) rule does not work. Is there a way to tell just by looking? (not figuring whether the reaction will be spontaneous)

There is a light bulb so can I assume it is an electrochemical cell instead of an electrolytic cell. (electrolytic cell has a dc power source instead right?)

I got question 46 (answer C) but I am unsure of question 47.

The anode reaction is Ag --> Ag+ + e-
The cathode reaction is Cl2 + 2e- --> 2Cl-

I'm unsure what precipitates would form because I don't know how the ions would migrate.

Offline Borek

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Re: Electrochemical cell / electrolytic cell
« Reply #1 on: June 06, 2011, 03:48:51 AM »
Take a look at your reactions - what ions will be present in both solutions after the reaction is over?
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Offline Capital

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Re: Electrochemical cell / electrolytic cell
« Reply #2 on: June 06, 2011, 06:43:56 PM »
Take a look at your reactions - what ions will be present in both solutions after the reaction is over?

Ag+ will be present and so will Cl-.

I'm unsure of this but I think the Ag+ travels through the porous barrier toward the cathode. And the Cl- travels toward the anode. (because cations to cathode and anion to anode)

So in the anode do we have Ag+ and Na+ ? And in the cathode do we have Cl- and NO3-

But a precipitate cannot be formed with only cations.

Offline DevaDevil

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Re: Electrochemical cell / electrolytic cell
« Reply #3 on: June 06, 2011, 07:35:47 PM »
of the ions present, which compound (cation + anion) has limited solubility?

Offline Capital

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Re: Electrochemical cell / electrolytic cell
« Reply #4 on: June 06, 2011, 08:01:41 PM »
AgCl. But the answer to the original question is B. I don't see why a precipitate would form in both cells.

Offline Borek

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Re: Electrochemical cell / electrolytic cell
« Reply #5 on: June 07, 2011, 05:31:18 AM »
Porous barrier just slows down mixing, so ions could diffuse both ways.

But I don't like the question. Too much depends on the details. I am not convinced B is the only possible correct answer. Only D is wrong for sure.
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Offline Capital

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Re: Electrochemical cell / electrolytic cell
« Reply #6 on: June 07, 2011, 06:32:37 PM »
Why is D wrong for sure?

And does the light bulb mean it is an electrochemical cell because electrochemical cells produce energy?

Offline Borek

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Re: Electrochemical cell / electrolytic cell
« Reply #7 on: June 07, 2011, 06:43:32 PM »
Why is D wrong for sure?

Sorry, I got it reversed. Or at 0:40 a.m. it is too late and I no longer know what I am doing.

First of all - think what is the overall reaction taking place, that's the starting point for further analysis. Write it. Think, what ions are present, think what they can do. Remember that the barrier just slows mixing, but ions will sooner or later cross it.
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Offline Capital

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Re: Electrochemical cell / electrolytic cell
« Reply #8 on: June 07, 2011, 07:05:30 PM »
The overall reaction is Cl2 + 2Ag --> 2Ag+ + 2Cl-

The Ag+ travels through the porous barrier to the anode
The Cl- travels through the porous barrier to the cathode.

In the anode, there are Ag+, Na+ (is there Cl- too? ... because if there is AgCl is a precipitate)
In the cathode, there are Cl-, NO3- (is there Ag+ too? (how far to completion has the reaction gone? ... because if there is AgCl is a precipitate)


Offline Borek

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Re: Electrochemical cell / electrolytic cell
« Reply #9 on: June 08, 2011, 05:17:37 AM »
is there Cl- too?

Quote
is there Ag+ too?

Any reason why they should be NOT there?
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Offline Capital

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Re: Electrochemical cell / electrolytic cell
« Reply #10 on: June 09, 2011, 12:25:51 AM »
Good point.

Thanks for the help. I understand it now.

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