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Advanced Lemon battery!

  1. Aug 21, 2008 #1
    Hey!

    Well, we did the lemon battery in school. Re-did it at home. Played with my flashing LED's... And then realized I have to make a documentation on it. (boring part)

    Anyway, the simple one page thing has grown to 5 pages. Teacher better like it. hehe.

    2H+(g) + 2e- → H2 (g) is the cathode half reaction. But I wanted to go a step further and make the molecular reaction. Can someone help me by explaining where the H's come from? What kind of an acid is in lemons? Would this differ from bananas?

    Also in the redox table for a Zinc-Copper cell, you have the following:
    E°cell=E°cathode-E°anode
    =0.34-(-0.76)
    =1.1V @ Standard conditions.

    In this one you still use copper, but hydrogen is being reduced. Does that mean you have to take it for Hydrogen or copper? Would it then be:
    E°cell=E°cathode-E°anode
    =0.00-(-0.76)
    =0.76V @ Standard conditions.

    Thanks. :D
     
  2. jcsd
  3. Aug 21, 2008 #2

    HallsofIvy

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    The acid in lmeons is citric acid. Yes, that differs from bananas!
     
  4. Aug 23, 2008 #3
    Hey,

    Thanks for your reply. But I still don't know all the other questions...

    This is what I did find out. The Cu is still the cathode. But instead its not really part of the reaction in some way. I'm guessing thou that the Cu might be eaten away the acid maybe?

    All I know is that it is the Cathode. So the volts would be 1.1. :D

    Next, I did research on citric acid. It has 4 OH groups, therefor it is able to release 4H's right?
    C6H8O7 + 4e- --> C6H4O7- + 2H2
    Is this right? Should the copper also come in somewhere?

    I'm trying to find a redox table with citric acid on, seeing that the one in my book doesn't have it...

    Thanks. :)
     
  5. Aug 24, 2008 #4
    ok. I know this isnt right. But maybe it gives someone an idea to helo me.

    Well I have been playing around with the exuations. Got out as something like this.

    C6H8O7 + Cu + 4E- --> 2H2 + 2H2O +2CO2

    Now the last bit I wasn't sure of. So I just tried to make something of it.

    C6H8O7 + 7Cu + 4E- --> 2H2 + 2H2O +2CO2 + C4Cu7O

    Dont know if you evin get something like that.

    Anyway. Please guys. I seriously need this. Its due for tomorrow.

    Any ideas yet?
     
  6. Aug 29, 2008 #5
    Hey.

    Is my question really stupid? Really hard? Under the wrong heading?

    My project is already in so I can't get marks on it anymore. If that bothered anyone. But I still need to know.

    Im as curious as hell and I don't know where to find out.

    Please help me out.

    Thanks.

    Dark_Fire
     
  7. Aug 29, 2008 #6

    chemisttree

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    Gold Member

    Can't help you until you tell me how you conducted your lemon battery experiment. I don't think you were generating hydrogen, BTW. I believe that the acid was only the charge carrier (electrolyte) and didn't actually enter into the redox reactions. You were probably just using a couple of electrodes (zinc and copper?) inserted into a lemon. The lemon just acted as an electrolyte.

    Citric acid has three acidic protons.
     
  8. Jul 20, 2009 #7
    I am having trouble getting my LED to light at all. I have gradually increased the number of lemons in my cell and now have 8 with a voltage reading of 5.2. The LED is working as I have tested it with a regular battery, but the LED still won't light when connected to the lemon battery. HELP!!
     
  9. Jul 21, 2009 #8
    Hmm. If you can measure a definite voltage in your lemon cells it ought to work. I can't tell what might be wrong unless you have a more detailed description. You're sure you aren't connecting the LED in the wrong way? o.0
     
  10. Jul 21, 2009 #9

    Borek

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    Just because you are measuring voltage high enough doesn't necesarilly mean you will be able to light a diode. Voltage drops under load if the internal resistance of the battery is too high.

    No idea what is the current needed to light a diode, so I can be off in this specific case - but in general I am right :wink:
     
  11. Jul 21, 2009 #10

    Redbelly98

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    drashling, welcome to Physics Forums :smile:

    Have you tried reversing the polarity of the LED? Perhaps it is in the wrong direction.

    And I have no idea how much current a lemon battery can generate. To answer your question, typical LED's use 10 to 20 mA of current, but even 2 or 3 mA should cause a visible glow.
     
  12. Jul 21, 2009 #11
    Hi thanks for your replies,
    I think it may be a problem with current. I have finally gotten a very faint glow from the diode using new lemons. Does anyone know what current is needed to light a diode?
    Thanks for your help.
     
  13. Jul 21, 2009 #12
    Sorry RedBelly98 just reread your post and realised you had answered my question already. I will keep trying and keep you posted.
     
  14. Jul 21, 2009 #13

    negitron

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    What sort of LED are you using? Typical red and green LEDs based on a gallium-arsenide chemistry have forward voltage drops of around 1.7 volts and maximum forward currents in the 20-30 mA range, these should produce a readily visible light with as little as 3-5 mA. However, other chemistries such as used to produce blue, violet and white LEDs can require higher forward voltages, in the 3-5 V range.

    In any case, regardless of the open-circuit voltage of your battery, you want to check the voltage across the LED while it's connected to be sure you're meeting the minimum turn-on voltage level.
     
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