Do I have to systematically show the protons in my reaction equation to find the Pourbaix diagram's line equations? I was trying to get the line equation for the Pourbaix diagram for the couple $\ce{Ni(OH)2}/\ce{Ni}.$ I can write two chemical equations for basic and acidic conditions, respectively:
$$
\begin{align}
\ce{Ni(OH)2 + 2 e- &amp;-&gt; Ni + 2 OH-}\tag{R1}\\
\ce{Ni(OH)2 + 2 H+ + 2e- &amp;-&gt; Ni + 2 H2O}\tag{R2}
\end{align}
$$
Now I can determine the line equation for both chemical equations using Nernst equation:
$$
\begin{align}
E^\circ + \frac{0.06}{2}\log\left([\ce{H+}]^2\right) &amp;= E^\circ - 0.06\,\mathrm{pH};\tag{1}\\
E^\circ - \frac{0.06}{2}\log\left([\ce{OH-}]^2\right) &amp;= E^\circ + 0.06\,\mathrm{pOH}.\tag{2}
\end{align}
$$
I need to make the $\mathrm{pH}$ appear in the previous equation. To do so, I use the relation
$$\mathrm{pH} + \mathrm{pOH} = 14 \quad\Leftrightarrow\quad \mathrm{pOH} = 14 - \mathrm{pH},\tag{3}$$
which gives
$$E^\circ + 0.06(14 - \mathrm{pH}) = E^\circ + 0.84 - 0.06\,\mathrm{pH}.\tag{4}$$
I've found two slightly different expressions depending on whether I choose basic or acidic conditions. I've spend quite a few hours trying to figured out what is wrong here, and I've found a PDF Diagrammes E-pH = Diagrammes de Pourbaix (courtesy of Académie de Clermont-Ferrand) stating on page 4 that even if the species only exists in basic conditions I need to make the $\ce{H+}$ appear in the half-reaction.
Is there a convention when writing chemical equation prior to the determination of the line's equation for Pourbaix diagram? Or did I make a mistake somewhere? If yes, could help me please?
the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $