3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\left[\ce{Hg^{2+}}\right]=\left[\ce{Hg_2^{2+}}\right]=\frac{C}{2}$

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.