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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on May 28, 2012, 05:45:49 AM

Title: Problem of the week - 28/05/2012
Post by: Borek 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⁺?
Title: Re: Problem of the week - 28/05/2012
Post by: Calicum on June 09, 2012, 02:23:28 AM: B-But I just an undergrad :P
Title: Re: Problem of the week - 28/05/2012
Post by: Borek on June 09, 2012, 02:54:14 AM: Should be enough.
Title: Re: Problem of the week - 28/05/2012
Post by: Sophia7X 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?
Title: Re: Problem of the week - 28/05/2012
Post by: UG on June 09, 2012, 11:46:49 PM: -2.3 V for Hg⁺?
Title: Re: Problem of the week - 28/05/2012
Post by: Borek 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.
Title: Re: Problem of the week - 28/05/2012
Post by: DrCMS on June 10, 2012, 06:02:44 AM: OK I'll try it 0.059V for Hg⁺ and half that for Hg₂²⁺
Title: Re: Problem of the week - 28/05/2012
Post by: Sophia7X 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
Title: Re: Problem of the week - 28/05/2012
Post by: Borek on June 10, 2012, 03:02:51 PM: Quote from: DrCMS on June 10, 2012, 06:02:44 AM
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.