[Borek]

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⁺?

[Calicum]

B-But I just an undergrad

[Borek]

Should be enough.

[Sophia7X]

B-But I just a high schooler

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

[UG]

-2.3 V for Hg⁺?

[Borek]

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.

[DrCMS]

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

[Sophia7X]

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

[Borek]

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.