Rrerku

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?

$$
beginarraycc
hline
cetextOxidized species <=> textReduced species & E^⦵(puV) \
hline
beginalign
ceH2O(l) + e- &<=> 0.5 H2(g) + OH-(aq) \
ceH+(aq) + e- &<=> 0.5 H2(g) \
ce0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq) \
ce0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)
endalign
&
beginarrayr
-0.83 \
0.00 \
+0.40 \
+1.23
endarray
\
hline
endarray
$$

(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 $ceH+$ occurs.

As $ceOH-$ 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.
$ceOH-$ or anions of weak acids like $ceClO-$ do not survive in acids. Acids do not survive in hydroxides.

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

$$beginalign
ceO2(g) + 4H+(aq) + 4e- &<=> 2 H2O(l)\
ce2H+(aq) + 2e- &<=> H2(g)
endalign$$

For the alkaline electrolysis, similarly, use the reactions where $ceOH$- occurs.

$$beginalign
ce2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)\
ceO2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)
endalign$$