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This might be elementary, but why aren't electrons included in the equilibrium constant for a half-reaction?

For example, for

$$\ce{O2 (g) + 4H+ (aq) + 4e- -> 2H2O(l)}$$

we have

$$Q = \frac{(a_\ce{H2O})^2}{(a_\ce{O2})^2(a_\ce{H+})^4}$$

without the electron.

This means that the activity for electron is 1? Why is that so? Is it because of any definition just like solid and pure liquid?

Atkins$^1$ says that because the electrons are 'stateless'. I am not sure what that mean.


[1] Peter Atkins, Julio de Paula, James Keeler. Physical Chemistry (11th Edition). OUP. 2018. Page 218.

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    $\begingroup$ You've clearly used some MathJax before; but I think you can use it more efficiently. For one, fractions with nested subscripts/superscripts should really be written out in display style $$...$$, otherwise it's extremely difficult to read. Secondly, you can use $\ce{H2O}$ $\ce{H2O}$ to typeset chemical formulae more easily, rather than using \text{} on every element symbol. Consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? for a refresher, or introduction, if you've not seen these before. $\endgroup$
    – orthocresol
    Jul 13 at 18:37
  • $\begingroup$ Atkins at times generates useless points and unnecessarily complicates matter by using overly fancy language (esoteric). Anyway, he is getting pretty old and most likely his new co-authors added this. Yes, you can assume unity for electron activity. $\endgroup$
    – M. Farooq
    Jul 14 at 2:53
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    $\begingroup$ @M.Farooq I can't get why that textbook had much success. It often rises Qs without giving solid As. In addition, it uses so many difficult to read characters and, especially, breaks the flow in too many boxes, highlights, activities and the like. $\endgroup$
    – Alchimista
    Jul 14 at 9:43
  • $\begingroup$ Hmm I am not completely sure, but my guess is that the electron is generally provided by an electrode. In that case, the electrode surface, volume etc. are constant throughout the reaction, so its ability to transfer electrons is also constant, so whatever the factor is, we can neglect it by setting it to 1. $\endgroup$
    – S R Maiti
    Jul 14 at 11:24
  • $\begingroup$ @Alchimista, Atkin's was never my favorite for the reasons you state above. $\endgroup$
    – M. Farooq
    Jul 14 at 13:24

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