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

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 $$\frac{c}{2}$$.

• 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 – ParaH2 Apr 18 '15 at 0:34

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 $$\frac{c}{2}$$.