Electrolysis of water: Which equations to use? (IB Chem) There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells? 

$$
\begin{array}{cc}
\hline
\ce{\text{Oxidized species} &lt;=&gt; \text{Reduced species}} &amp; E^⦵(\pu{V}) \\
\hline
  \begin{align}
    \ce{H2O(l) + e- &amp;&lt;=&gt; 0.5 H2(g) + OH-(aq)} \\
    \ce{H+(aq) + e- &amp;&lt;=&gt; 0.5 H2(g)} \\
    \ce{0.5 O2(g) + H2O(l) + 2 e- &amp;&lt;=&gt; 2 OH-(aq)} \\
    \ce{0.5 O2(g) + 2 H+(aq) + 2 e- &amp;&lt;=&gt; H2O(l)}
  \end{align}
&amp;
  \begin{array}{r}
   -0.83 \\
    0.00 \\
   +0.40 \\
   +1.23
  \end{array}
\\
\hline
\end{array}
$$

(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)
For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs.    

As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.    

Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. 

But note that using reactions with half of a molecule is not necessery.

$$\begin{align}
\ce{O2(g) + 4H+(aq) + 4e-  &amp;&lt;=&gt; 2 H2O(l)}\\
\ce{2H+(aq) + 2e- &amp;&lt;=&gt; H2(g)}
\end{align}$$

For the alkaline electrolysis, similarly, use the reactions where $\ce{OH}$- occurs.

$$\begin{align}
\ce{2 H2O(l) + 2e- &amp;&lt;=&gt; H2(g) + 2 OH^-(aq)}\\
\ce{O2(g) + 2 H2O(l) + 4e- &amp;&lt;=&gt; 4 OH^-(aq)}
\end{align}$$