I have to build the stability diagram of mercury and I have a problem with this couple:

$\ce{Hg^2+}/\ce{Hg2^2+}$ $E^\circ=0.91\ \mathrm{V}$

The exercise says that a the border the concentration is $C=0.10\ \mathrm{mol \cdot L^{-1}}$ for all ions.

So I have : $\ce{2Hg^2+ +2e^- <=> Hg2^2+}$

Then by Nernst relation I have : $E=E^\circ+0.03\ \mathrm{V} \times \log\left(\frac{\left[\ce{Hg^2+}\right]^2}{\left[\ce{Hg2^2+}\right]}\right)$

And in the solution of the exercise they write at the border $\left[\ce{Hg^2+}\right]=\left[\ce{Hg2^2+}\right]=\frac{C}{2}$

I don’t understand why it is $C/2$.

I think that’s a “stupid” question but I really don’t understand.

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    $\begingroup$ I'm not sure about what's going on in this example, but apparently it was considered that the state is at the equilibrium at standard conditions, which makes $E = Eº$. Considering that, the log term has to equals zero, and that happens for log(1). This then means that the ions concentrations are equal, and therefore they are C/2. I won't post this as an answer because I'm not sure about how to justify this chemically correctly. $\endgroup$ – Molx Apr 18 '15 at 0:22
  • $\begingroup$ I don't undersand can you explain a bit more ^^ write the egality with the log please like you think may be i will understand it too $\endgroup$ – ParaH2 Apr 18 '15 at 0:34

The total concentration of all ions is $C=0.10\, \mathrm{mol.L^{-1}}$. The conservation equation of element mercury requires that:


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    $\begingroup$ Generally, it's a convention to make the concentration of ions at the border equal. On the other hand, matter conservation implies that the concentration of each ion is C/2 $\endgroup$ – Yomen Atassi Apr 18 '15 at 12:48
  • $\begingroup$ I know this convention but I don t understand C/2 for me its like if I don t understand why 1+1=2 and you answer 1+1=2 because 1+1=2 ... :/ $\endgroup$ – ParaH2 Apr 18 '15 at 12:50
  • $\begingroup$ It is a simplifying assumption. So, the fraction within the "log" in the Nernst equation will become 1 and we get rid of the log (log1=0). $\endgroup$ – Yomen Atassi Apr 18 '15 at 17:44
  • $\begingroup$ I don t understand what you mean ... $\endgroup$ – ParaH2 Apr 18 '15 at 17:48
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    $\begingroup$ I am sorry thta you are confused and I really want to help you. If I am not clear, I hope someone else could explain this to you.There aren't any steps. You have [Hg2+]+[Hg2+2]=C at the border we assume having only these two forms. In order to have a simplifying formula. They make the assumption that:[Hg2+]=[Hg2+2]=C/2. I hope it's clear now. $\endgroup$ – Yomen Atassi Apr 19 '15 at 7:56

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