Flood Control - electrolysis - limit to rate of conversion?
Floodwaters are immense in quantity.  Let's estimate how much energy would be required to convert that quantity of water to $\ce{H2}$ and $\ce{O2}$.


The Mississipi river has a (non-flood) flow rate of about 593,000 cubic feet per second, or 17000 cubic meters per second.  That's $1.7 \times 10^7$ liters per second, or $1.7 \times 10^{10}$ grams per second, or roughly $10^9$ moles per second.
Faraday's constant is the conversion factor between moles and Coulombs.  It's value is about 96000 ampere-seconds per mol.  Let's approximate it as $10^5$ ampere-seconds per mole.  
It takes four moles of electrons per mole of water to complete the electrolysis.  
Multiply the numbers together, and you get that to electrolyze all the waters of the Mississippi river, you'd need $4 \times 10^{14}$ amperes of current.  That's about the amount of current present in a billion bolts of lightning.  Yikes!  
Exactly how much voltage would be required depends on your design of electrolytic cell you use for this giga-bolt scale.  Let's say it's 5 volts.  That means the power you need for the electrolysis is $2\times10^{15}$ watts.  That's 2000 terawatts.  But the world's current total energy consumption is about 12 terawatts...