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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?

Taiki
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  • 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. – MaxW Oct 27 '15 at 21:37
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    @MaxW You mean that on a standard hydrogen electrode $\ce{H (g)}$ is ionised into $\ce{H+ (aq) + e-}$ right away? – Taiki Oct 27 '15 at 22:05
  • Yes. The point is that the reaction is taking place at the surface (boundary) not "inside" the solution. Think of the reverse reaction. You're electrolizing H+ ions to get H2(g). The reaction takes place at the surface of the electrode not in the bulk of the solution. – MaxW Oct 27 '15 at 22:09
  • @MaxW OK. Thanks. And then by removing the proton hydration enthalpy from question 2 I'll get the right number for enthalpy change now? – Taiki Oct 27 '15 at 22:39
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    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. – orthocresol Oct 27 '15 at 23:39
  • @Taiki - Sorry, I just don't remember how to work the rest of this off the top of my head. It's been 40 years since I took P-Chem. This is just sort of funny reaction. – MaxW Oct 27 '15 at 23:57
  • @orthocresol - Re remark "You don't even have to consider the reaction in the context of an electrochemical cell. " I surely don't understand this. This is definitely a half-cell reaction. Obviously there has to be oxidation too. You can't just generate H+ i0ns to get a positively charged solution. – MaxW Oct 28 '15 at 00:04
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    @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. – orthocresol Oct 28 '15 at 00:09
  • @orthocresol - Well yes using the "standard hydrogen electrode" as the stake makes sense for relative electrochemical measurements since the hydrogen half-cell would be reproducible. When you add reduction and oxidation half-cells whatever bias there is cancels. But the half-reaction itself must have some absolute value. I doubt that the "real" value could be zero. – MaxW Oct 28 '15 at 00:51
  • The point being that if you tried to use a copper electrode as the standard you could probably reproduce the solution to the required degree of accuracy. But the electrode itself would probably be very hard. How pure is "pure" copper? Having done a lot of XRF work no metal is "pure." With XRF you can easily find impurities in anything. You can even tag a lot of things to the geographic source by the distribution of trace impurities. – MaxW Oct 28 '15 at 01:19
  • Occurs to me that in order to have some sort of "absolute" value then you need to know H2(g) pressure and concentration of H+ ions and Temperature. As I remember standard hydrogen cell uses 0.1N acid and H2(g) at STP. – MaxW Oct 28 '15 at 01:22
  • @orthocresol The discussion between you two is interesting. So what I'm doing is calculating the 'absolute' enthalpy change of the reaction occurring at the standard hydrogen electrode? And conventionally such change is defined to be zero? – Taiki Oct 28 '15 at 11:52
  • Actually, I thought about it and we are not really disagreeing about anything. :) It is just that in order to define a "standard" reaction enthalpy, we need to define the pressure of the H2 gas and the concentration of the H+ ions, which is of course taken to be the standard state - 1 bar for H2 and 1 M for H+. Max was talking about the reaction in the context of an electrochemical cell, but my only point was that as long as you specify the initial and final states sufficiently clearly, the enthalpy change doesn't depend on how the reaction is carried out – orthocresol Oct 28 '15 at 12:49
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    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. – orthocresol Oct 28 '15 at 12:51
  • @orthocresol Now it's all clear. Thank you very much! – Taiki Oct 28 '15 at 12:59

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