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Current in lemon battery

  1. Mar 8, 2012 #1

    ojs

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    Hi, I was doing some tests with lemon batteries today and found that it produced little current (not enough to light a small LED at least and I serially connected 4 lemons getting above 3V) which is explained in some places in the following way:

    2 electrons are produced by the zinc (if your using zinc and copper) and they are absorbed by the H near the cathode so there are few extra electrons to go through the circuit.

    But is this quite right? I mean isn't there enough electrons to use all around us? I always had the image of extra electrons in metals and that all we had to do was supply voltage and voila, current was produced. My image was that we didn't need any one specific electron producer. So shouldn't the copper wires connecting the lemons to the led contain enough free electrons so that when 3V are supplied then the electrons start to move?

    Of course the wire is electrically neutral so if my understanding is correct then an equal amount of positive charge would have to be in place in the wire, that could be protons or positively ionized atoms, I have no solid idea here.

    If someone could clarify my confusion here that would be much appreciated.
     
  2. jcsd
  3. Mar 9, 2012 #2

    NascentOxygen

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    Staff: Mentor

    More than likely your single lemon cell did not produce sufficient voltage to be able to light the LED.
    Potentially, just enough to light some LEDs. Probably dimly because the voltage will fall as current flows.
    One molecule of the zinc reacts and liberates two electrons at the cathode. At the same time, at the anode two electrons react with H+ to form an atom of hydrogen, then two of these atoms form a molecule of hydrogen gas. The flow of the electrons through the external circuit constitutes a flow of electricity.
    Plenty of electrons in metals, but they need something to push extra electrons in one end before an equal number of electrons will emerge from the other to flow though a motor or light and perform useful work for us. :wink:
     
    Last edited: Mar 9, 2012
  4. Mar 9, 2012 #3

    ojs

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    Ok, so the voltage drops when I finish the circuit with the LED because it can't produce the electrons needed to power the circuit?

    What is confusing me is the initial voltage, that should be enough of this "something to push extra electrons" (I mean the voltage is created by charge, so there should be enough charge to create a current). But are you saying that when I close the circuit with the LED then the charge dissipates and the lemon battery isn't strong enough to produce more electrons to keep up the current and so not enough charge gets created at the cathode and anode and then the voltage drops?

    Unfortunately I forgot to measure the voltage across the LED when the circuit was closed, only measured the voltage when the circuit was open.
     
  5. Mar 9, 2012 #4

    NascentOxygen

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    Because it can't produce sufficient voltage when you draw current. The reaction proceeds faster as you allow electrons to travel through the wire, but the products of the chemical reaction now begin to suffocate the reaction in the vicinity of the anode and cathode.
    It does give that impression, but then fails to live up to its promise.
    Not really. The voltage is created by the particular metals and acid solution. The greater the area of metals you can get into the lemon, the greater the current it can deliver.
    If you do the experiment at night, with the room darkened you may see the LED glowing dimly.
     
  6. Mar 9, 2012 #5

    NascentOxygen

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    If you squeezed the juice into a cup, and dangled the metals into the jar of juice, then you could rapidly stir it and this might allow the reaction to proceed at full pace without dropping off when you draw current. Fresh electrolyte would be constantly bathing each electrode.

    I should have said "At the same time, at the anode two electrons react with two H+ ions to form a pair of hydrogen atoms (in early times termed nascent hydrogen) these then combining to form a molecule of hydrogen gas."
     
  7. Mar 9, 2012 #6

    cmb

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    What you are doing in pushing dissimilar metals into your lemon is to activate a chemical reaction. It is not a particularly strong chemical reaction, and the reactants need time to diffuse through to react with your electrodes. I suspect you are, indeed, generating the voltage (as the voltage is defined by the chemistry of the reaction, and you have enough here) but that the set up does not allow for a sufficiently rapid chemical reaction from which enough electrical energy can be generated to light your led.

    You could try connecting your lemon cells to a capacitor, to which is connected a resistor and serial switch to your led. Leave the switch open whilst enough lemon-reactions occur to charge the capacitor, then open the switch to your led.

    Also - make sure you have connected the polarity correctly to your led (it is a diode after all)!!... lemons don't usually come with + and - terminals!!!
     
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