Vertical lines in a Pourbaix Diagram
A Pourbaix diagram is a special kind of phase diagram. As in a phase diagram, the lines of all kinds represent phase boundaries. Each line represents the conditions in which two phases are in dynamic equilibrium. Without information on what the phases enclosed by lines look like, a phase diagram is pretty meaningless. Below is the phase diagram of water. Imagine how hard it would be to interpret without the labels. We still need to go look up what is meant by Solid IX - is it ice IX or ice 9?


Here is the iron-carbon (steel) phase diagram. Again, the phases are labeled.



In a Pourbaix diagram, the axes are pH and applied electric potential $(E_0)$ in volts instead of temperature and pressure (pure substance) or temperature, pressure (sometimes omitted), and composition for mixtures. The lines represent boundaries, but the "phases" are now the predominant iron-containing ion in solution. These diagrams seem to be primarily for aqueous systems, and so there is never an absence of $\ce{H+}$ ions. They may not be needed in the redox equation, but they are in solution. Otherwise, how would we have different pH along the horizontal axis? 

Below is the Pourbaix diagram for aqueous solutions of iron. Notice that not all lines are horizontal or vertical. Each line represents a set of conditions ($E_0$ and $\text{pH}$) where two ionic species are in dynamic equilibrium.



For example, at $E_0=1.0\ \text{V}$ and $\text{pH}=2$, the predominating species is $\ce{Fe^{3+}(aq)}$. If we increase the $\text{pH}$ to $5$, the predominating iron species becomes $\ce{Fe2O3}\cdot\ce{nH2O}(s)$, rust. The reaction might be (we increase the pH by added $\ce{OH-}$):

$$\ce{2Fe^{3+}(aq) +6OH- (aq) -&gt; Fe2O3(s) + 3H2O(l)}$$

Note that this reaction is redox neutral - all elements remain in the same oxidation state. Vertical lines represent redox neutral transitions/reactions.

If we start at the same place $E_0=1.0\ \text{V}$ and $\text{pH}=2$, and change $E_0$ from $1.0 \ \text{V}$ to $0.4 \ \text{V}$, the predominating species becomes $\ce{Fe^{2+}(aq)}$. When we lower $E_0$, it is like adding electrons to the system (it is really encouraging electrons to flow through the system in a specific way), we get this reaction, which is pH-neutral. 

$$\ce{Fe^{3+}(aq) + e- -&gt; Fe^{2+}(aq)}$$

As you noted, there is no $\ce{H+(aq)}$ in this reaction. Horizontal lines are pH-neutral transitions/reactions.

A diagonal or curved line is neither pH or redox neutral. For example, let's go from $E_0=0\ \text{V}$ and $\text{pH}=6 \ (\ce{Fe^{2+}(aq)}$ to $E_0=0\ \text{V}$ and $\text{pH}=10 \ (\ce{Fe2O3\cdot nH2O(s)}$ crossing a diagonal line. The reaction now involves a change in oxidation state and in pH. Since we are traveling horizontally, the electrons lost by $\ce{Fe^{2+}}$ must go somewhere. 

$$\ce{2Fe^{2+}(aq) + 4HO- (aq) -&gt; Fe2O3(s) + 2H2(g) + H2O(l)}$$