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Final state of concentration cell?

  1. Jul 5, 2013 #1

    somasimple

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    Hi,
    here is an example of a concentration cell
    http://www.jce.divched.org/jcedlib/qbank/collection/conceptests/echem.html [Broken]
    1/ The final state is assumed when concentrations are equal in each compartment, right?
    2/ Graphic evolution of the voltage is a decreasing curve, right?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 5, 2013 #2

    Borek

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    Yes2.
     
  4. Jul 5, 2013 #3
    1) technically, it's when the activities are the same. If the ionic atmospheres in each half cell are different, activity of the ion in question is modified and the final state may have somewhat different concentrations in each half cell (assuming copper electrodes in both for example)
    2) applying ohm's law, current will be linearly related to voltage in the cell assuming constant load. From the nernst equation the voltage is proportional to the log of one concentration divided by the other. For initial half cell concentrations C1 and C2, the electric current will add the stoichiometric amount of ions to one cell solution and subtract an equal amount from the other. cell voltage ~ log((C1-integral(current, time 0 to t))/(C2+integral(current, time 0 to t)))

    I wrote a short loop to see what the functional dependence of current is (setting resistance to 1 ohm and neglecting faraday constant etc) and it turns out the functional dependence is an exponential decay, which is not surprising.

    For[t = 0, t < 200, t++,
    v[[t + 1]] =
    Log[(1 - Total[Take[v,t]*.005])/(2 + Total[Take[v,t]*.005])]]

    1 and 2 were taken to be the initial half cell concentrations and the Total[] is just an integral over time of how many ions have been removed or added to solution. Mathematica code btw.
     
  5. Jul 5, 2013 #4

    somasimple

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    Thanks you²

    Another question:
    Let's take 3 vats:
    1/ The first (A) contains a copper sulfate solute at 0.1 M and a copper electrode.
    2/ The second(B) contains a copper sulfate solute at 0.1 M and a copper electrode.
    3/ The third (C) contains a copper sulfate solute at 1 M and a copper electrode.

    They are connected by salt bridges A to C and B to C.
    And with voltmeters by A to C and a second voltmeter B to C.

    Is the voltage of each voltmeter a fraction (1/2) of the equilibrium voltage of my first example (the voltage is divided by the number of vats that contain the lower concentrations)?
    How is it possible to rearrange the Nersnt equation?
     
    Last edited: Jul 5, 2013
  6. Jul 5, 2013 #5

    Borek

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    No chemistry here, this is a simple circuit question.

    attachment.php?attachmentid=60085&stc=1&d=1373019733.png

    Note: could be I got signs wrong (battery symbols may need to be reversed), but it doesn't matter much.
     

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  7. Jul 5, 2013 #6

    somasimple

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    I understand the diagram but I seem to fail why it is not chemistry:
    Compartment C is the one where is the chemical reduction, right?
    Compartment A and B have chemical oxidation, right?

    Note: B is not connected to A (as in your picture)
     

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    Last edited: Jul 5, 2013
  8. Jul 5, 2013 #7

    somasimple

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    You're right it doesn't matter at all.
    It may matter if the experiment is modified like the picture?
    Electroneutrality must be balanced between C //A and B (because electrons flow to C from A and B)?
     

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    Last edited: Jul 5, 2013
  9. Jul 5, 2013 #8

    Borek

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    Sorry, I mislabeled the points.

    attachment.php?attachmentid=60088&stc=1&d=1373028373.png

    No much other choice - you need to reduce concentration of 1M solution and increase concentrations of 0.1M solutions to reach the equilibrium.
     

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  10. Jul 5, 2013 #9

    somasimple

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    So, if we remove the salt bridges, there is an electroneutrality violation that stops very quickly the chemical reactions, and, equalization of concentrations is made impossible?
    Since 2 vats (A, B) give 1 electron to a single vat (C) that must react to this violation but can't compensate it, the voltage measured between AC or BC must be lower than a quite same situation where B is removed (like the picture below) ?
     

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