In the electrolysis of water, reduction half equation is $$\ce{2H2O + 2e- -> H2 + 2OH-}$$ with a standard $E$ value of $-0.83\ \mathrm V$.
If $\ce{H+}$ is present in water, why can $$\ce{2H+ + 2e- -> H2}$$ not occur, with potential $0\ \mathrm V$, which is more favourable?
See, this is pretty much the same reaction. The potential is not always exactly what they say in the reference book. It depends on concentrations. The potential of $\ce{2H+ + 2e- -> H2}$ is $0 V$ only in standard conditions, that is, when the concentration of $\ce{H+}$ is 1 mol/L. When it is lower, the potential is different (see Nernst equation). Indeed, $\ce{H+}$ is present in pure water, but the concentration is quite low ($10^{-7}$ mol/L). If you plug that concentration into the Nernst equation, you'll get the same value that is listed for $\ce{2H2O + 2e- -> H2 + 2OH-}$.
