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Help to understand reversibility in galvanic cell

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data:-I am confused about some lines in my textbook about galvanic cell


    2. Relevant equations:-The lines are as follows............
    The cell is connected to an external source of potential that opposes and exactly balance the cell potential . If the external potential is reduced infinitesimally ,the cell reaction occurs in it's spontaneous direction.An opposite infinitesimal change in external potential will cause the reaction to proceed in the reverse direction.The cell reaction is then said to occur under reversible conditions.Thus,the cell can be made to behave in a reversible manner by balancing their potentials by an external potential so that no current flows.
    The reversible electrical work done in a galvanic cell by cell reaction is equal to decrease in gibb's energy.Hence,
    Electrical work= - deltaG



    3. The attempt at a solution:-Let us take if I have a single displacement, reversible reaction, AB + C <--> AC + B. Then I increase the concentration of B on the right, what happens? The AC molecule splits, and forms more AB to re-balance the equation. This is the same thing happening in the galvanic cell. Lets assume the cell has a potential of 1.1V and I apply a current of exactly 1.1V to the circuit, the electrodes will no longer undergo a redox reaction, as there is no potential gradient that would cause the electrons to move from one electrode to the other. But if I drop the external voltage to 1V, then the cell will begin undergoing a redox reaction, albeit at a slower rate to bring the voltage back to 1.1. If I apply a greater voltage of say 1.2V, then electrons are going to go the opposite direction, and oxidize the cathode and reduce the anode.
    Actually someone has explained me in this manner,i want to verify this.Is this correct?
    Please help.
    Regards.
     
  2. jcsd
  3. Jan 31, 2015 #2

    mfb

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    Sounds fine.
    Note that you don't have to care about the internal chemical details if you want to analze circuits.
     
  4. Feb 4, 2015 #3
    How one can apply current of 1.1 v as volt is unit of potential.
     
  5. Feb 4, 2015 #4
    Are we charging galvanic cell here,so voltage of 1.1 is applied from external source.To charge it, a power/voltage source is connected positive terminal to positive terminal of the battery, and negative terminal to negative terminal of the battery (parallel connection),right?
     
    Last edited: Feb 4, 2015
  6. Feb 4, 2015 #5
    What does this mean?
     
  7. Feb 4, 2015 #6

    Bystander

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    It's a typo.
    Correct.
     
  8. Feb 4, 2015 #7

    Bystander

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    Higher voltage goes one way, lower, the other.
     
  9. Feb 4, 2015 #8
    Why when there is parallel connection we subtract the two potentials?
     
  10. Feb 4, 2015 #9
    Sorry,still didn't understand.
     
  11. Feb 4, 2015 #10

    Bystander

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    This gives you the magnitude of the emf driving the reaction. Zero, it goes nowhere. Positive it goes one way, negative the other. Whichever you choose to define as forward or reverse as far as calling positive or negative forward or reverse (European or American conventions).
     
  12. Feb 4, 2015 #11
    I understood the whole thing which i had written in my attempt to a solution ,but how this explains reversibility in galvanic cell?
     
  13. Feb 4, 2015 #12

    Bystander

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    When the textbook is discussing "reversibility," it's attempting to state that whatever electrical work is done on (or by) the cell in either direction can be completely recovered only when the cell is run reversibly, and that the cell can be run reversibly only by making infinitesimal changes in the potential it is working against, or that is working on it. What it means is that any real use of the cell, work done charging it, or work done by it during discharge is NOT reversible.
     
  14. Feb 4, 2015 #13
    Increasing or decreasing the potentials by 1 volt,will it be considered as infinitesimal change?
     
  15. Feb 4, 2015 #14
    One more thing about reversibility is
    In my textbook it is written that for the reaction to be reversible the system should be in mechanical equilibrium with it's surroundings.(the external pressure should be equal to pressure of the gas ,this balance of pressure is called as mechanical equilibrium)
    I don't understand if there is no pressure difference between system and surrounding,how work will be done by/on the system?
     
  16. Feb 4, 2015 #15

    Bystander

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    Not infinitesimal. This is the same "infinitesimal" as is used in your math courses.
    There is mechanical work, and there is also electrical work, "Eit."
     
  17. Feb 4, 2015 #16
    Can I draw a conclusion from this that emf resist changes as we have learnt in physics class.
     
  18. Feb 4, 2015 #17
    How this answers my post 14.
     
  19. Feb 4, 2015 #18

    Bystander

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    Yes.
    Electrical work is voltage times current times time.
     
  20. Feb 4, 2015 #19
    But in my textbook this mechanical equilibrium is given in context of piston cylinder .How electricity can come in this type of problems.
     
  21. Feb 4, 2015 #20

    Bystander

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    Textbook writers are lazy --- they like to stick to the historical development of thermodynamics, pV work in steam engines, and very often (almost always) fail to generalize the "work term" in the first law to include ALL work, chemical, mechanical as pV, mechanical as surface tension times area, electrical, magnetic, .....
     
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