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Working of Voltaic cell

  1. Sep 20, 2004 #1
    Can anyone tell me how the Voltaic cell works?
    In my text book(grade 10 India), a diagram with a beaker containing Dilute Surlphuric acid , 1 zinc plate(with negative signs all over it), 1 copper plate(positive signs all over it) and the 2 plates are connectedto a bulb.
    I want to know why the Zn plate becomes - (Another doubt. metals are electropositive? )and Cu plate is +? :cool:
     
  2. jcsd
  3. Sep 21, 2004 #2

    chem_tr

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    Electrochemical voltage rank order

    Hello,

    A voltaic cell contains one eager electron donor and one eager electron acceptor. In here, zinc is an electron donor and copper is the electron acceptor, since [tex]Zn \rightleftharpoons Zn^{2+}+2e^-; E_0=-0,76 V[/tex], where [tex]Cu \rightleftharpoons Cu^{2+}+2e^-; E_0=+0,35 V[/tex].

    We learn that zinc is eager to give electrons more than copper, as evidenced by the E0 values; metallic zinc spontaneously gives two electrons to reach a more stable state, where you have to give some energy to convert metallic copper to its divalent ion. That's why metallic copper does not react with non-oxidizing acids like HCl (remember that E0 of hydrogen is accepted to be 0).

    Here, sulfuric acid is the electrolyte, carrying electrons from zinc to copper.

    Regards, chem_tr
     
  4. Sep 23, 2004 #3

    ShawnD

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    I'm not exactly following what you're saying.

    First of all, these voltages seem backwards. You've written zinc oxidizing as being a non-spontaneous reaction, but it actually is spontaneous. You've written copper oxidizing as being spontaneous, but it's non spontaneous.
    Secondly, when you say the copper is the electron acceptor, are you saying the copper is the oxidizer? The way I see it, the acid is the oxidizer (+0.17V to reduce), the zinc is the reducer (+0.76V to oxidize), and the copper just happens to be there. IMO, the copper could be replaced with carbon, silver, gold, or platinum and the reaction would go exactly the same.
     
    Last edited: Sep 23, 2004
  5. Sep 23, 2004 #4

    chem_tr

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    You seem to be right. The acid oxidizes zinc more readily than copper; so it is very easy to produce Zn2+ rather than Cu2+.

    I've taken these from a chemistry textbook, and these values are all calculated according to the assumption that hydrogen produces 0 volts, as you well know. So let's speculate that zinc is very eager to be oxidized with hydrogen ions than copper. The negative value indicates that this energy is given to the environment; but a positive value shows that some energy must be given to achieve the oxidation. If we write copper's oxidation in the reverse way, I mean, copper's reduction, then we'll see that 1,11 volts of energy is gained with this redox reaction.

    I hope these will settle the topic.

    Regards,
    chem_tr
     
  6. Sep 23, 2004 #5
    I cant comprehend anything what you are saying. Can you say in simple 10th grade chemistry terms.
     
  7. Sep 23, 2004 #6

    chem_tr

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    I'll try to be as simple as possible.

    In our universe, all metals have different properties, some resemble each other more than others, so several grouping systems are discovered. One of these groupings is done according to their electron-donating power (eagerness to oxidation). If a metal is eager to be oxidized, then its standard oxidation potential (determined with relation to hydrogen's potential accepted as 0) will be negative, which means that energy is spontaneously given if that metal is oxidized.

    Please note that the standard reduction potential has the same absolute value, with inverse sign.

    If the standard oxidation potential is positive, then the reduction potential is negative, so it means that the metal is more eager to be reduced than to be oxidized.

    Now let me tell you something about what you see in the diagram from your textbook. Sulfuric acid is both electrolyte and oxidizer, and causes zinc to be oxidized to zinc(II) ions, giving away the unwanted two electrons, thus zinc is (-) pole, a.k.a. cathode.

    Copper is also oxidized to copper(II) ions with the action of sulfuric acid, though simple acids like HCl cannot evolve hydrogen from copper (copper is more passive than hydrogen). Oxidative acids can, however, cause copper to be converted into Cu(II) ions. Copper ions are then reduced with the electrons sent from zinc, because it is spontaneous for copper. That's why copper is (+) pole, a.k.a. anode.

    I hope you understood now why there are two different signs on zinc and copper.
     
  8. Sep 23, 2004 #7
    But if electrons are lost by zinc then it should be + and if copper gains electrons it should be -
     
  9. Sep 23, 2004 #8

    chem_tr

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    Well, it is confusing indeed. I definitely know that Zn/Cu/H2SO4 system produces metallic copper on zinc plate, this is consistent with their electrochemical voltage order. The designations (-) and (+) probably refer to their voltage signs; since Zn gives -0,76 volts, it is of high electron potential (negative pole); Cu has +0,35 volts to be of low electron potential (positive pole). The textbook you have, I think, explained the issue like that.
     
  10. Sep 24, 2004 #9
    I think that's correct. Thanks
     
  11. Jul 26, 2009 #10
    yah yah!! m also having the same doubt....INFACT i hav joined this site cuz of this doubt only .......

    PLEASE explain me this.....
     
  12. Jul 26, 2009 #11
    hOw copper is produced in on zinc plate.....See in my book it is given that Zn ---> Zn2+ + 2e-
    and copper will get those electrons and get reduced to copper....sO there will be no electrons ion zinc rod and it shUD be positively charged but...this is not happening....ZINC is getting negatively charged.(according to book...).....i hav not understood please explain simply
     
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