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Can any two electrodes from the table make voltaic cell?

  1. Aug 2, 2015 #1
    Hello everybody,

    I would like to ask you a question about voltaic cells. Can any two electrodes from the table attached to this post make a voltaic cell? There will always be a potential difference no matter which half reactions you select. From my point of you, any two electrodes (or half reactions) can make a voltaic cell. What do you say?

    Thanks for help,
    CroSinus SEP.jpg
     
  2. jcsd
  3. Aug 2, 2015 #2

    Borek

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    Staff: Mentor

    In principle - yes.

    Many combinations are not practical though.
     
  4. Aug 4, 2015 #3
    Thanks for answering my post.
    CS

    But one thing is not clear to me. If we take into consideration formula:

    delta G = - z*F*E
    the spontaneous half reactions imply negative ΔG. In other words only chemical half reactions for which ΔG is negative are spontaneous. But not all half reactions from the table given above are spontaneous.

    How can two non spontaneous chemical reactions make a voltaic cell?
     
    Last edited: Aug 4, 2015
  5. Aug 4, 2015 #4

    Borek

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    Staff: Mentor

    You are mistaking half reactions and reactions.
     
  6. Aug 4, 2015 #5
    I don't know. In my opinion, if I reconsider the problem once again, a voltaic cell will function only if two spontaneous half reactions take place at the electrodes. One of the half reactions must be spontaneous oxidation, and another spontaneous reduction.

    Am I right? I would appreciate your comments about the problem.

    Sincerely,
    CS
     
  7. Aug 4, 2015 #6

    Borek

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    Staff: Mentor

    No, half reactions are not spontaneous. Whole reaction (which combines two half reactions) can be.
     
  8. Aug 6, 2015 #7
    You are right. Like in the Daniell cell. The oxidation of zinc electrode is spontaneous while the oxidation of copper electrode is not. The reduction of Cu2+ ions is spontaneous instead.

    But what if I take Li/Li+ and K/K+ electrodes. Would such a cell be possible?
     
  9. Aug 6, 2015 #8

    Borek

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    Staff: Mentor

    Possible? Definitely. Good luck designing it :wink:
     
  10. Aug 13, 2015 #9
    You are right, it will be difficult to make such a cell. Maybe we should use a melt instead of a solution. But it is interesting to discuss the issue for the sake of theory.
     
  11. Aug 13, 2015 #10
    This is a classical problem in electrochemisty. You can find other tables of standard potentials, E°s, of half cell reactions having the same numerical value but opposite sign, for the reactions in your table . For example for the Li/Li+ -3.040V. Thus, every reaction in your table and that table will have opposite signs for the values of E°s. Therefore, the question arises whether the potential of LI/Li+ is +3.040V or -3.040V? The American tables give the positive value and the British tables give the negative value. Much can be said in support of each. This issue was resolved by an international convention, which uses the negative value for this half cell E°. Without going into too much of a detail, I shall be brief.

    The way to understand the relation between E° and the ΔG° and the corresponding spontaneity is this way: Choose a reaction from the table. Combine it with H/H+ reaction. You will get a voltaic cell. A spontaneous reaction, in principle, is possible. Take for example, the Li/Li+ and couple it with SHE. You will get a cell with a E° value of 3.040V. In this cell Li goes to Li+ spontaneously and H+ goes to H2 spontaneously.

    Now take, Cu/Cu2+ reaction. Combine it with H/H+ reaction. You will get a voltaic cell. A spontaneous reaction, in principle, is possible. You will get a cell with a E° value of 0.337V. In this cell H2 goes to H+ spontaneously and Cu2+ goes to Cu spontaneously.
    Now you can think of combining Li/Li+ and Cu/Cu2+ reactions (or any two reactions for that matter). You get a cell with a E° value of the difference of the two E° values, 3.377V. In this cell Li goes to Li+ spontaneously and Cu2+ goes to Cu spontaneously.

    Finally, a positive value of E° for a half cell reaction in a standard table of E° values does not indicate that it is a spontaneous reaction, in as much as a negative value of E° for a half cell reaction in that table of E° values does not indicate a nonspontaneous reaction.

    A half cell reaction does not occur by itself. So, when combined with another half cell reaction, a cell is formed and the two half cell reactions occur spontaneously (as you guessed correctly) to give a spontaneous cell reaction with a value of E° for the cell which is the difference between the two E°s of the individual half cells chosen.

    P. Radhakrishnamurty
     
  12. Aug 15, 2015 #11
    Thank you very much for a detailed answer and explanation.

    Is there an easy way to say which electrode is going to be an anode and which cathode when I randomly choose two half reactions from the table to form a voltaic cell.

    Thanks a lot,
    CroSinus
     
  13. Aug 15, 2015 #12
    In the table you have chosen, the upper couple that you choose acts as anode (the negative terminal) and the lower couple the cathode (the positive terminal) of the battery.

    P. Radhakrishnamurty
     
  14. Aug 15, 2015 #13
    OK, thanks for the answer. What will happen if I take both electrodes from the same side of the table. For example, from the upper part.

    CS
     
  15. Aug 15, 2015 #14
    When you take two couples at random (both can be from the same side of zero (not of the table!)) one will be upper and the other lower.
    You can go about doing what you want, with that understanding.

    P. Radhakrishnamurty
     
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