I won't do any rib poking, but there is some flawed thinking here. Certainly not off topic though.
Nomenclature: First, when you pass a current through water, you are not breaking the hydrogen bonds, which are intermolecular forces, but (a) breaking the bond between hydrogen and oxygen and (b) forming molecular oxygen and molecular hydrogen, $\ce{2H2O -> 2H2 + O2}$, and this reaction can be broken down into two half reactions: $$\ce{4H2O + 4 e- -> 2H2 + 4OH-}$$ $$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Potentials: The 1.2 V you refer to likely arises from the standard reduction potential of oxygen, which is the reverse of the second reaction above. I won't go in to the thermodynamics here, but suffice it to say that a brief look at a standard reduction potential table indicates that the reduction of water (first equation) has a potential of -0.83 V and the oxidation of water has a potential of -1.23 V for a grand total of ~ -2 V. (Note, these values are pH dependent and standard reduction potential tables report values at a pH of 0.)
Electrochemistry When you apply an overpotential to a system (the potential beyond what is needed to get the oxidation/reduction reactions to go forward) then two things will happen (a) the reaction will go faster and (b) heat will be generated. This heat would be analogous to heat due to friction in a physical system and is considered wasted energy.
Thermochemistry Finally, we get to the answer to your question, which is "probably not, but the way you asked it, yes" (hunh?) Recall the nomenclature issue above. Adding heat to water will break the hydrogen bonds in water eventually leading to the water changing from a liquid to a gaseous state. However, adding heat to water would not result in the oxidation/reduction reactions outlined above because an important ingredient is missing: electrons.
