Electrolysis of water - why not pure water?

In the electrolysis of water, $\ce{H2O}$ molecules are being reduced at the cathode (forming $\ce{H2}$ and $\ce{OH-}$) and oxidized at the anode (forming $\ce{O2}$ and $\ce{H\_3O+}$).


[It is said](http://www1.lsbu.ac.uk/water/electrolysis.html#comp) that:



> 
> The electrolysis of water **usually involves dilute**, or moderately concentrated, **salt solutions** in order to reduce the power loss driving the current through the solution, but the presence of salt is not a requirement for electrolysis.
> 
> 
> 


What is the reason for "power loss"? How does the presence of foreign ions change that?


The electrolysis seems possible to happen in pure water alone (as the quote also says).




The terrible conductivity of water can explain this.


Pure water is a really bad electric medium, since it has a low tendency to ionize by itself (we know the low concentrations of ions involved by looking at the Kw).


This means that the voltage needed to get a current flow from anode to cathode would be really high, with consequences both in term of safety and energetic waste.


A salt dissolved provides ions, which make the diluted solution a good electric medium, and allow a current to easily flow between anode and cathode.


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Pure water is a very good insulator. Due to it's very low self ionization, there is not a sufficient amount of ions to carry current unless a significant overpotential is applied in order to overcome this kinetic barrier. 


Sea water, which has a salinity of about 3.5%, is about one million times as conductive as pure water. According to [this Wikipedia article](https://en.wikipedia.org/wiki/Electrolysis_of_water): 



> 
> Electrolysis of pure water requires excess energy in the form of
>  overpotential to overcome various activation barriers. Without the
>  excess energy the electrolysis of pure water occurs very slowly or not
>  at all. This is in part due to the limited self-ionization of water. 
> 
> 
> 


The article sited by the OP states that the resistivity of pure water is 18 MΩ cm while that of sea water is 20 Ω cm. This implies that a relatively huge overpotential is required to electrolyze pure water. At these high potentials more high-energy side products like ozone are likely to be produced, as well as excess heat. This results in power lost to the production of both chemical and heat energy.


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