Electrolysis of water: Which equations to use? (IB Chem)

Question

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?

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

(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.)

Answer 1

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.

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

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

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