Pourbaix diagram – writing reactions
The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:

$$\begin{alignat}{3}
\ce{2H+ + 2e- \;&amp;&lt;=&gt; H2}\quad\quad &amp;&amp;E^\circ = +0.000\ \mathrm{V}\quad\quad &amp;&amp;&amp;(\mathrm{pH}=0)\\
\ce{2H2O + 2e- \;&amp;&lt;=&gt; H2 + 2OH-}\quad\quad &amp;&amp;E^\circ = -0.828\ \mathrm{V}\quad\quad &amp;&amp;&amp;(\mathrm{pH}=14)
\end{alignat}$$

The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:

$$\begin{alignat}{3}
\ce{O2 + 4H+ + 4e- \;&amp;&lt;=&gt; 2H2O}\quad\quad &amp;&amp;E^\circ = +1.229\ \mathrm{V}\quad\quad &amp;&amp;&amp;(\mathrm{pH}=0)\\
\ce{O2 + 2H2O + 4e- \;&amp;&lt;=&gt; 4OH-}\quad\quad &amp;&amp;E^\circ = +0.401\ \mathrm{V}\quad\quad &amp;&amp;&amp;(\mathrm{pH}=14)
\end{alignat}$$

Therefore, only areas between these lines are stable in aqueous solutions.

Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$

$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&amp;&lt;=&gt; H2}\quad\quad &amp;&amp;E^\circ = +0.000\ \mathrm{V}\\
\ce{Cr^3+ + e- \;&amp;&lt;=&gt; Cr^2+}\quad\quad &amp;&amp;E^\circ = -0.408\ \mathrm{V}\\
\ce{Cr^2+ + 2e- \;&amp;&lt;=&gt; Cr}\quad\quad &amp;&amp;E^\circ = -0.913\ \mathrm{V}\\\\
\ce{2Cr + 6H+ &amp;-&gt; 2Cr^3+ + 3H2}
\end{alignat}$$

and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$

$$\begin{alignat}{2}
\ce{2H2O + 2e- \;&amp;&lt;=&gt; H2 + 2OH-}\quad\quad &amp;&amp;E^\circ = -0.828\ \mathrm{V}\\
\ce{Cr(OH)3 + 3e- \;&amp;&lt;=&gt; Cr + 3OH-}\quad\quad &amp;&amp;E^\circ = -1.33\ \mathrm{V}\\\\
\ce{2Cr + 6H2O &amp;-&gt; 2Cr(OH)3 + 3H2}
\end{alignat}$$