What is the purpose of balancing a redox half equation using H+ and H2O and electrons?
One reason to track solvated hydrogen (or hydroxide) ions and water in a redox reaction is that the hydrogen ions often get transferred too, and you need to know how that takes place to render the products.  We should expect hydrogen-ion transfers as part of the process in a protocol solvent such as water or a concentrated protic acid, as the electrons are charged and some compensating charge-transfer process must take place to keep the charges in balance.
Copper in concentrated nitric acid, where nitrogen is reduced (primarily) to the +4 oxidation state, is a well-worn example.  The oxidation half-reaction is, of course, copper being oxidized to $\ce{Cu(II)}$.  So we look at the reduction half-reaction.  In its &quot;raw&quot; form we would have
$\ce{HNO_3 -&gt; NO2}$
We balance the elements other than hydrogen and oxygen, which is already done, and then compare the combined oxidation states of those elements in the reactants and products to find that the nitric acid must have taken one elecron:
$\ce{HNO_3 + e^- -&gt; NO2}$
Now the charges are not balanced, and so we invoke that compensating hydrogen ion transfer.  In acid this is set up by adding (solvated) hydrogen ions to balance the charges:
$\ce{HNO_3 + \color{blue}{H^+} + e^- -&gt; NO2}$
Water is used to balance the hydrogen which, if we did things correctly, balances the oxygen as well giving our familiar result
$\ce{HNO_3 + H^+ + e^- -&gt; NO2 +H2O}$
and upon combining with the copper oxidation:
$\ce{Cu + 2 HNO_3 + 2 H^+ -&gt; Cu^{2+} + 2 NO2 + 2 H2O}$
in which we see that in the final reaction, hydrogen ions are indeed transferred.  So we needed to account for them.