I'm reading Detchko Pavlov's Lead-Acid Batteries: Science and Technology (maybe an old version than the link) and in the first chapter the book mentioned "it is currently unknown why electrolysis of water happens so slowly".
If we think about it, isn't that strange? We have a pair of electrodes in a pot of acidic water, with a potential of 2+ volts, but nothing happens? And I do remember asking the same question when I was in middle school but really didn't get a convincing answer.
Do we know the answer now?
Here's the text of the book. Mine is a Chinese translation version so I will try to translate it back.
As is shown by the E/pH diagram of Figure 2.1, an lead-acid battery in open-circuit is thermal-dynamically unstable. The self-discharge reaction between the electrodes will electrolyse water into $\ce{H2}$ and $\ce{O2}$. At the same time, the following reactions happens on the electrode plates:
On the positive plate:
$$ \ce{PbO2 + H2SO4 -> PbSO4 + H2O + \frac{1}{2} O2} $$
On the negative plate:
$$ \ce{Pb + H2SO4 -> PbSO4 + H2} $$
If the battery is constructed with high-purity lead and sulfuric acid, the above reactions happens at a very slow rate. To this day, we have not fully figured out what basic reactions so strongly inhibits the precipitation of $\ce{H2}$ and $\ce{O2}$.
Also in this document
What is the exact shape of this outgassing-rate/voltage curve and how is it derived?