# Chemical Forums

## 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
OK I'll try it 0.059V for Hg+ and half that for Hg22+

For Sophia and others - it is actually very simple.

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

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

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

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

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

$$E_1=E_0+\frac{59mV}{n}logC_1$$

$$E_2=E_0+\frac{59mV}{n}logC_2$$

and the cell potential is

$$E=E_1-E_2=E_0 + \frac{59mV}{n}logC_1 - E_0 - \frac{59mV}{n}logC_2$$

or

$$E=\frac{59mV}{n}(logC_1 - logC_2)=\frac{59mV}{n}log\frac{C_1}{C_2}$$

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

$$E=\frac{59mV}{n}$$

and it depends only on n.

Ogg measured the potential to be 28.9-29.0 mV.