# Reduction potential as a function of pH

I have read that Latimer diagrams can be represented both in an acidic medium (pH = 0) and in a basic medium (pH = 14).

I would like to know how to calculate the reduction potentials as a function of pH. That is, if I know the reduction potential for the pair $$X^{+}/X$$ at pH = 0, calculate it for another pH.

I thought the Nerst equation could be applied, but I'm not sure about that.

Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is: $$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}$$

The potential can be written as:

$$E = E^\circ + \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$

If you want to see explicitly the $$\mathrm{pH}$$, you can rewrite the preceding expression, which gives:

$$E = E^\circ + 0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - \left(\frac{8\times 0.059}{5} \right) \mathrm{pH}$$

or : $$E = E^\circ + 0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - 0.094 \: \mathrm{pH}$$

• So, for example, what would be the reduction potential at pH = 0? And, another question, if I want to determine the potential for calcium reduction is $\mathrm{Eº(Ca^{2+}/Ca)} = -2.760 V$ under normal conditions. What would it be at pH = 10? Thanks – aprendiendo-a-programar Oct 22 at 17:39