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Topic: Problem of the week - 28/05/2012  (Read 7705 times)

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Offline Borek

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Problem of the week - 28/05/2012
« on: May 28, 2012, 05:45:49 AM »
In 1898 it wasn't yet clear whether monovalent mercury is present in the solution as Hg+ or Hg22+. To check it, Alexander Ogg (PhD student of Walther Nernst), prepared a cell Hg|L1|L2|Hg, where both solutions L1 and L2 contained 6.3 g of nitric acid per liter, and additionally L1 contained 2.63 g and L2 0.263 g of mercury (I) nitrate. What potential should he measure if the mercury was present as Hg+, and what if it was present as Hg22+?
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Offline Calicum

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Re: Problem of the week - 28/05/2012
« Reply #1 on: June 09, 2012, 02:23:28 AM »
B-But I just an undergrad :P

Offline Borek

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Re: Problem of the week - 28/05/2012
« Reply #2 on: June 09, 2012, 02:54:14 AM »
Should be enough.
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Offline Sophia7X

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Re: Problem of the week - 28/05/2012
« Reply #3 on: June 09, 2012, 09:02:31 PM »
B-But I just a high schooler :P

No idea what to do... am I supposed to use standard potentials or do something with the 6.3 g/L nitric acid?
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Offline UG

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Re: Problem of the week - 28/05/2012
« Reply #4 on: June 09, 2012, 11:46:49 PM »
-2.3 V for Hg+?

Offline Borek

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Re: Problem of the week - 28/05/2012
« Reply #5 on: June 10, 2012, 03:58:55 AM »
Nitric acid just acidifies the solution, you can safely ignore its presence. He couldn't use hydrochloric acid nor sulfuric acid to not risk precipitation.

No, it is not -2.3V, much less than that (note: sign doesn't matter).

This is a simple application of the Nernst equation for a concentration cell. But you don't even need to calculate concentrations, as it is their ratio that counts, and the ratio is obvious.
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Offline DrCMS

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Re: Problem of the week - 28/05/2012
« Reply #6 on: June 10, 2012, 06:02:44 AM »
OK I'll try it 0.059V for Hg+ and half that for Hg22+

Offline Sophia7X

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Re: Problem of the week - 28/05/2012
« Reply #7 on: June 10, 2012, 11:21:24 AM »
Oh, I didn't even realize that it was supposed to be a concentration cell, haha :O
well in that case, I got the same thing as DrCMS
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Offline Borek

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Re: Problem of the week - 28/05/2012
« Reply #8 on: June 10, 2012, 03:02:51 PM »
OK I'll try it 0.059V for Hg+ and half that for Hg22+

And that's the correct answer.

For Sophia and others - it is actually very simple.

Formal potential of the half cell composed of metal and its cation is

[tex]E=E_0+\frac{RT}{nF}ln[Me^{n+}][/tex]

This is just a Nernst' equation, often written in a simplified form

[tex]E=E_0+\frac{59mV}{n}log[Me^{n+}][/tex]

Now, when you have a concentration cell there are two potentials for each half:

[tex]E_1=E_0+\frac{59mV}{n}logC_1[/tex]

[tex]E_2=E_0+\frac{59mV}{n}logC_2[/tex]

and the cell potential is

[tex]E=E_1-E_2=E_0 + \frac{59mV}{n}logC_1 - E_0 - \frac{59mV}{n}logC_2[/tex]

or

[tex]E=\frac{59mV}{n}(logC_1 - logC_2)=\frac{59mV}{n}log\frac{C_1}{C_2}[/tex]

From the information given it is clear that [itex]\frac{C_1}{C_2}[/itex] is 10, so the log is 1, so the potential is

[tex]E=\frac{59mV}{n}[/tex]

and it depends only on n.

Ogg measured the potential to be 28.9-29.0 mV.
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