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Ag/AgCl reaction picture (right side)

For the Ag/AgCl half reaction, where did the free electron that AgCl captured come from? Did it come from the Ag wire? If so, why would Ag wire give an electron to AgCl, both atoms are Ag and have the same electronegativities, right? Also, what's the purpose of the KCl(aq) solution, because I don't see where any of it comes into play.

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The $\ce{Ag/AgCl/Cl-}$ electrode is basically ordinary $\ce{Ag/Ag+}$ electrode. The difference is that the concentration of $\ce{Ag+}$ is kept very low by presence of $\ce{AgCl(s)}$ and excess of $\ce{Cl-}$ ( from $\ce{KCl}$ ) due $\ce{AgCl}$ solubility product $K_\mathrm{sp}=\ce{[Ag+][Cl-]}$.

By other words, electrons come from the half reaction $\ce{Ag(s) -> Ag+(aq) + e-}$,
linked to the equilibrium reaction $\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$.

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Yosimba2000 asks about the origin of the free electron that AgCl captured. It does not come from some part of this Ag/AgCl electrode. It comes from another electrode. Matter of fact an electrode does not work alone. It has to be coupled with another electrode, in order to produce a cell, whose performance is measured by a potential difference. This other electrode can be any reducing agent like Zinc or Hydrogen dipped in aqueous solution of its ion. The electron is created by an equation like $\ce{Zn-> Zn^{2+} + 2 e-}$ or $\ce{H2 -> 2H^+ + 2 e-}$

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  • $\begingroup$ I am not so sure he asks about this. $\endgroup$
    – Poutnik
    Nov 11 '21 at 18:08
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    $\begingroup$ Well, I am pretty sure he is looking for such an explanation. Let's wait his reaction $\endgroup$
    – Maurice
    Nov 11 '21 at 20:17
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    $\begingroup$ As I read the OP’s question, I wondered of they were not realizing that a reduction half-reaction was not the same as a properly balanced “full” reaction with no electrons loitering about on either side of the reaction. So I upvoted this answer because I think it addresses the OP’s question. I also upvoted Poutnik’s answer because the two answers wrap it up well. $\endgroup$
    – Ed V
    Nov 11 '21 at 22:15

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