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I am recently taking courses in environmental chemistry and I was introduced to Pourbaix diagram. I was taught that the lines in the Pourbaix diagrams are equilibrium lines. Lets take as an example the Pourbaix diagram of the water (diagram is taken from Ibanez):

                                      

Lets look at the equilibrium line between $\ce{H2O}$ and $\ce{H2}$. The reaction describing this equilibrium line is (with equilibrium constant $K$):

$$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}}$$

Now at equilibrium taking the $\log K$ and equilibrium constant we obtaing:

$$\log K = \log p_{_{\ce{H2}}} -2 \, \mathrm{pH} -2 \log {[\ce{e-}]} $$

and by defining $\log {[\ce{e-}]} = \text{p}E$ we can obtain (assuming $p_{_{\ce{H2}}} = 1 \, \text{atm}$):

$$\text{p}E=-\text{pH}$$

So we can plot how $\text{p}E$ varies with $\text{pH}$ at equilibrium conditions. What I can't understand is why if we leave that line and move below it in a point which doesn't satisfy the equation we will have only $\ce{H2}$. How is it possible to change $\text{p}E$ without changing $\text{pH}$ if they are both related through equilibrium constant? Why all the water turns into hydrogen?

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  • $\begingroup$ Alone, a Pourbaix daigram is of little interest because there are a lot of parameters that come into play: for example, it is not because a reaction is feasible that it will be done because if it is too slow to be done, it will not be done. For gases, like dissolved species, these are areas of predominance. So, in the example you quote, if we go below the line, the gas $\ce {H_2}$ will be able to form until saturation $\endgroup$
    – Nicolas
    Commented Mar 12, 2021 at 16:44

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