# The purpose of C/2 in aqueous solution equation

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.

• 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. – Molx Apr 18 '15 at 0:22
• 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 – Hexacoordinate-C 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:
$\left[\ce{Hg^{2+}}\right]=\left[\ce{Hg_2^{2+}}\right]=\frac{C}{2}$