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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?

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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 $\ce{2H2O + 2e- -> H2 + 2OH-}$.

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