Calculating enthalpy change

My friends and I were talking casually about electrolysis, fuel cells, and whatnots when the following question arose:


Is $\ce{H2 (g) -> 2H+ (aq) + 2e-}$ endothermic or exothermic?


I have the following questions (by the way I am a physicist):


1. Is the reaction sensible? (I think it probably is, but no one is sure.)
2. I said the reaction was endothermic, with the following reasoning. According to [this table](https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29), the standard enthalpy change of formation $\Delta H\_\text{f}º$ of the reactant is zero, so if $\Delta H\_\text{f}º$ of the products are positive then the reaction is endothermic.


Electrons are fundamental particles so their $\Delta H\_\text{f}º$ is zero. As for $\ce{H+ (aq)}$, its $\Delta H\_\text{f}º$ is the atomization enthalpy of hydrogen (218 kJ/mol [[ref]](https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29)) + ionization enthalpy (first ionisation energy: 1312 kJ/mol [[ref]](https://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page))) + proton hydration enthalpy (-1150 kJ/mol [[ref]](http://pubs.acs.org/doi/pdf/10.1021/jp982638r)). Therefore,


$2 \Delta H\_\text{f}º[\ce{H+ (aq)}]$ = 2 (218 + 1312 - 1150) kJ/mol = 760 kJ/mol


which is a positive number, so the reaction is indeed endothermic. Am I right?


