It more efficient to generate heat burning electrolyzed hydrogen, or through an electric resistance? It's convenient and simple to use electricity to generate heat, doubtlessly, say in en electric kettle or boiler or heater, but is it more efficient to generate heat burning electrolyzed hydrogen, or through the resistance (kettle  / boiler / heater)? Assume the same amount of electricity was used.
 A: According to this wikipedia article, electrolysis has an efficiency between 70-80%, so not all of the electric energy used for electrolysis would go into dissociating water molecules, but some would be lost as heat.
Therefore, the energy that can be obtained from burning the hydrogen obtained from electrolysis would be less than the electric energy spent for electrolysis.
A: Sorry for my poor english. My native language is french.
I think there is something missing in your question. The current in the resistance is not enough to find the energy. You also need voltage. 1 A at 1 V is different from 1 A at 1000 V. So I assume the same voltage : the minimum voltage to produce a certain amount of hydrogen and oxygen.
It suppose that we produce the hydrogen and oxygen reversibly (very slowly, at the minimum necessary voltage). In this case, the minimum electrical work is equal to the variation of Gibbs free energy : $W=∆G$
Since I assume the same voltage and the same current, this energy would be that transformed into heat if it was supplied it to a resistance.
Now, the question is whether the combustion of hydrogen produces more or less heat than the energy required to produce it. The thermal energy released is this time the variation of enthalpy $W'=∆H$.
The relation between the two is $∆G=∆H-T∆S$ with evidently $∆S>0$ for liquid water electrolysis. So we find $W<W'$ : If I am not mistaken, you gain more heat when you produce hydrogen and then burn it.
The difference does not seem negligible since we have a Gibbs standard free energy of 237 KJ / mol and a standard enthalpy of 285 KJ / mol.
Obviously all this is theoretical and does not take into account the efficiency lower than 1.
