Wattage to produce 6L of air from Water in less than a second? So I understand the question is probably vague to a scientist, but this is all the specifics I could mustard up with my non-scientific background.

So I understand the basics of the process and have used it occationally in a garage environment to fiddle around, but never really thought about the applications until a particular Indiegogo campaign claimed to have created artificial gills small enough to hold in your mouth and generating 45 minutes of breathable air.
The campaign have gotten a wide spread of media attention and it got me thinking, how much wattage would be required to produce 6L of air over the course of a second which happens to be a rough average of a adults lung capacity.

I understand that submarines use this process to create breathable air, but as someone mentioned they have a nuclear reactor onboard and "it would not be wise to attach that to your face".

Would shortening a Lithium Ion/Polymer battery be sufficient to create such a quick reaction? Or are we talking megawatts in order to produce such a volume of air in that time span?

Update to the numbers used:

Pointed out by @DavePhD, you don't use 6L per breath, rather 6L on average per minute.
Taking into account that this would be a physical exercise under water, you'd normally use anything between 14-17L per minute, on average 16L.

Also noted is that I should have used P instead of the term wattage, thanks @Loong
I guess it would have to be the amount energy released by the reaction $$\ce{2H2 + O2 -&gt; 2H2O}$$ where you have four liters of hydrogen gas and two liters of oxygen gas, all divided by one second (or less, increasing the wattage as you decrease the time). 

You asked about "air" — but you can't get normal air from electrolysis of water. 

So, the heat of combustion for hydrogen is $286\ \mathrm{kJ/mol}$. $$286\ \frac{\mathrm{kJ}}{\mathrm{mol}} \times \frac{4\ \mathrm L}{22.4\ \frac{\mathrm L}{\mathrm{mol}}} = 51.1\ \mathrm{kJ}$$ and that is assuming atmospheric pressure which is certainly low, but how low depends on your depth below the surface of the water. So with the "barely under the surface" energy of $51.1\ \mathrm{kJ}$, we can then divide that by seconds: $$\frac{51.1\ \mathrm{kJ}}{1\ \mathrm s} = 51.1\ \mathrm{kW}$$ so you would need a bare minimum of $51.1\ \mathrm{kW}$ of power to convert water into a 67:33 mixture of $\ce{H2:O2}$. That seems like a lot of energy in a small amount of time to me: think about a six-liter balloon with the perfect combustion ratio of hydrogen and oxygen. That reaction is a lot faster than one second, but it's still a lot of energy. Tough to get into a small, portable battery and have it fill a capacitor with that amount of e energy every few seconds.

I'll be honest, I'm just doing these calculations on-the-fly, and I may have made a mistake or five, but it seems reasonable to me. If anyone takes any issue with my answer, please, feel free to edit/give suggestions.

Edit: This Wikipedia page has some information on hydrogen-oxygen diving air. They use a 96:4 ratio. Not sure why, but if that's the best ratio, this would take significantly more energy per breath, since a lot of oxygen would just get dumped.