Why is the concentration halved at the intersection of Hg²⁺/Hg₂²⁺? For the couple $\ce{Hg^2+}/\ce{Hg2^2+}$: $E^\circ = \pu{0.91V}$.
  At the border the concentration is $c = \pu{0.10 mol \cdot L-1}$ for all ions. Hence build the stability diagram of mercury.


I have: 
$$\ce{2Hg^2+ +2e- &lt;=&gt; Hg2^2+}$$

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

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 $\frac{c}{2}$.
The total concentration of all ions is $c = \pu{0.10 mol L-1}$. The conservation equation of element mercury requires that
$$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}.$$

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 $\frac{c}{2}$.