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Electrochemistry question

  1. Aug 27, 2016 #1
    Hi everybody!
    While I was studying the Pourbaix diagram of chlorine ( and its disproportionation), I got stuck in a conceptual problem about the potential E. The diagram, as I understood, it's supposed to represent the equilibrium between the various species, but knowing that at equilibrium the ΔG of the reaction is zero, E have to be zero as well. But I kinda feel that this concept isn't applicable to half-reactions (like this case, in which the equilibrium is between Cl- / Cl2 , Cl2/HClO and so on). If it's not applicable, then why is it so? And then, what does the equilibrium mean in this case?
    Thanks in advance
     
  2. jcsd
  3. Aug 27, 2016 #2

    Borek

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    Staff: Mentor

    Hard to speak about an equilibrium of a half reaction - for an equilibrium you need a whole cell (two half reactions).
     
  4. Aug 27, 2016 #3
    I thought about that too, but I still don't understand a couple things about the Pourbaix diagram. Like, I don't know how to visualize the equilibrium in the solution, for example, on my book it is said that on the line that represents the coexistence of Cl2 and HClO the following expression is valid:

    E= 1.6 - 0.06 pH that is derived by E= 1.6 + 0.03 log ( [HClO]^2 * [H+]^2 / [Cl2] )

    So from this expression is it implicated that I can change the concentration of the species as I want and the equilibrium will exist with those concentrations, or will the concentrations of HClO and Cl2 be fixed?
     
  5. Aug 28, 2016 #4

    Borek

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    Staff: Mentor

    Pourbaix diagram tells you which species are dominating the solution. It is not like others are not present.

    If you add anything to the solution that has somehow (it doesn't matter how for this discussion, let's say "by external means") forced its pH and E, this added species will react (with whatever the "external force" needs to supply) till the solution is dominated by what the diagram predicts. In normal situation adding something can easily change the pH and E, moving the system into another area on the diagram.
     
  6. Aug 28, 2016 #5
    Ok, now I think I understand, I didn't see that way. Thank you for your help!
     
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