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, 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]) + ionization enthalpy (first ionisation energy: 1312 kJ/mol [ref]) + proton hydration enthalpy (-1150 kJ/mol [ref]). 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?

  • $\begingroup$ The equation would be correct assuming that you using the "standard hydrogen electrode" so that the oxidation reaction is at the interface and solvation of the H2(g) is not required. $\endgroup$ – MaxW Oct 27 '15 at 21:37
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    $\begingroup$ @MaxW You mean that on a standard hydrogen electrode $\ce{H (g)}$ is ionised into $\ce{H+ (aq) + e-}$ right away? $\endgroup$ – Taiki Oct 27 '15 at 22:05
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    $\begingroup$ You don't need to remove the hydration enthalpy. I think your analysis is perfectly fine as it is. You don't even have to consider the reaction in the context of an electrochemical cell. I could invent a fantasy planet 4000 light-years away from Earth, where aliens carry out this reaction on a daily process via miniscule tweezers that pluck out the electron from the nucleus, and the enthalpy change of the reaction (at a given $T$ and $p$) would still be the same. $\endgroup$ – orthocresol Oct 27 '15 at 23:39
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    $\begingroup$ @MaxW yes, and that is part of the reason why $\Delta H$ for the reaction above is conventionally specified as 0 (because whatever difference there is will be cancelled out by the oxidation half-equation). But if you are just interested in the reaction as written (as OP is), I doubt it has to be physically carried out. Basically I am saying you can do all these calculations on paper and that half-reaction would still be endothermic regardless of what the oxidation half-equation is. $\endgroup$ – orthocresol Oct 28 '15 at 0:09
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    $\begingroup$ Also, you're absolutely right about the "absolute" enthalpy change. Conventionally it's defined to be zero because the formation of the other counterion will "cancel out" the absolute enthalpy change of your reaction. $\endgroup$ – orthocresol Oct 28 '15 at 12:51

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