OK I'll try it 0.059V for Hg^{+} and half that for Hg_{2}^{2+}

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.