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Can anyone provide the answer to this cells in series question? watch

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    Hi all

    This sounds like an easy query but its deceptive and I've seen this question come
    up on several other forums and caused quite a debate among people of all science
    knowledge levels. To the extent that I have yet to see an answer that truly made sense....

    It is commonly known that when cells are in series, the voltages are added to make the
    combined total. But why exactly is this?

    Lets say you have 3 x 1.5 V dry cells stacked anode to cathode as expected. This as we
    know will provide a combined potential difference of 4.5 V.

    When dealing with a single cell, the situation is relatively simple - electrons flow from the
    anode (eg zinc) to the cathode (via wire) where they pass through the metal (copper/brass) cap and into the carbon electrode inside the cell. The de-energised electrons in the electrode then react with the surrounding electrolytic paste inside the cell in order to avoid the build up of negative charge (which would prevent further flow of current) and to avoid the increase of pressure that would occur due to the formation of gaseous species (typically ammonia and hydrogen).

    How does this work when dealing with multiple cells?

    For there to be a combined 4.5 V this must mean that each coulomb of charge leaving the
    anode now has 3x the energy as it did before (4.5 V = 4.5 J / 1 C).

    But how does this voltage build up from the top cell's anode to the bottom cell's anode which is connected to the conducting wire? As the cells are in series each anode is in direct
    contact with a cathode (apart from the last one), but how do the energised electrons pass
    through each cell in order to build up the voltage? They cant pass through the carbon
    electrode because they would react with positively charged species such as NH4+ in the
    surrounding electrolytic paste and also the carbon electrode isn't in direct contact with both
    the cathode and anode.

    I am going to post this in the chemistry forum as well to see if there any nice electro-chem people around!
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    That's a great question. I don't know the exact answer but perhaps it lies in this article

    http://www.mpoweruk.com/chemistries.htm

    "The electrical (pump) pressure or potential difference between the + and - terminals is called voltage or electromotive force (EMF)."

    If you apply a charge to one of the ends by placing another battery next to it, the potential difference between the + and - terminals will be more. In other words an additional battery serves to drive up the rate of chemical reaction across both batteries by virtue of the fact that there is an additional potential difference to speed things up.

    Hope this helps.
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    (Original post by Science hopeful)
    Hi all

    This sounds like an easy query but its deceptive and I've seen this question come
    up on several other forums and caused quite a debate among people of all science
    knowledge levels. To the extent that I have yet to see an answer that truly made sense....

    It is commonly known that when cells are in series, the voltages are added to make the
    combined total. But why exactly is this?

    Lets say you have 3 x 1.5 V dry cells stacked anode to cathode as expected. This as we
    know will provide a combined potential difference of 4.5 V.

    When dealing with a single cell, the situation is relatively simple - electrons flow from the
    anode (eg zinc) to the cathode (via wire) where they pass through the metal (copper/brass) cap and into the carbon electrode inside the cell. The de-energised electrons in the electrode then react with the surrounding electrolytic paste inside the cell in order to avoid the build up of negative charge (which would prevent further flow of current) and to avoid the increase of pressure that would occur due to the formation of gaseous species (typically ammonia and hydrogen).

    How does this work when dealing with multiple cells?

    For there to be a combined 4.5 V this must mean that each coulomb of charge leaving the
    anode now has 3x the energy as it did before (4.5 V = 4.5 J / 1 C).

    But how does this voltage build up from the top cell's anode to the bottom cell's anode which is connected to the conducting wire? As the cells are in series each anode is in direct
    contact with a cathode (apart from the last one), but how do the energised electrons pass
    through each cell in order to build up the voltage? They cant pass through the carbon
    electrode because they would react with positively charged species such as NH4+ in the
    surrounding electrolytic paste and also the carbon electrode isn't in direct contact with both
    the cathode and anode.

    I am going to post this in the chemistry forum as well to see if there any nice electro-chem people around!
    Well I don't know what level of study you're at or what sort of answers you've already had and are unsatisfied with.

    I think you might be having problems with thinking about electrons as whizzing round and round a circuit many times at a very high speed (which seems sort of reasonable given the way students are introduced to electricity). The electrons in conductors are in reality moving very slowly, it's called drift velocity, but the current goes at the speed of light because the start their slow movement almost all at the same instant. you can look at the drift velocity of charge carriers, namely electrons with the formula
    I=nvqA

    I current
    n number density, the number of free charge carriers per unit volume of material
    v drift velocity
    q charge per charge carrier
    A cross sectional area of material

    realistic values for copper wire which has n about 8.5x1028 would be < 1.0x10-3 ms-1

    in a cell you've got a chemical reaction that's releasing electrons and a chemical reaction that's accepting electrons both at the same rate, the electrons aren't passing through cells, they stop when they've neutralised a +ve ion.
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    (Original post by Joinedup)
    Well I don't know what level of study you're at or what sort of answers you've already had and are unsatisfied with.

    I think you might be having problems with thinking about electrons as whizzing round and round a circuit many times at a very high speed (which seems sort of reasonable given the way students are introduced to electricity). The electrons in conductors are in reality moving very slowly, it's called drift velocity, but the current goes at the speed of light because the start their slow movement almost all at the same instant. you can look at the drift velocity of charge carriers, namely electrons with the formula
    I=nvqA

    I current
    n number density, the number of free charge carriers per unit volume of material
    v drift velocity
    q charge per charge carrier
    A cross sectional area of material

    realistic values for copper wire which has n about 8.5x1028 would be < 1.0x10-3 ms-1

    in a cell you've got a chemical reaction that's releasing electrons and a chemical reaction that's accepting electrons both at the same rate, the electrons aren't passing through cells, they stop when they've neutralised a +ve ion.
    Thanks for this..... I'm going to do some YouTubing on drift velocity!

    I guess my big struggle is understanding, if each electron returning to the cell is neutralised in the electrolytic paste, how in a series of cells is the voltage added? I do understand that the electrons themselves do not pass through the cells.

    For 3 cells in series, the difference in p.d between the cathode of the 1st cell and the anode of the 3rd cell is 4.5 V, so by the power of maths that means each coulomb entering the circuit has 4.5 J of energy (as opposed to 1.5 J)..... how physically does the energy triple?
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    (Original post by Science hopeful)
    Thanks for this..... I'm going to do some YouTubing on drift velocity!

    I guess my big struggle is understanding, if each electron returning to the cell is neutralised in the electrolytic paste, how in a series of cells is the voltage added? I do understand that the electrons themselves do not pass through the cells.

    For 3 cells in series, the difference in p.d between the cathode of the 1st cell and the anode of the 3rd cell is 4.5 V, so by the power of maths that means each coulomb entering the circuit has 4.5 J of energy (as opposed to 1.5 J)..... how physically does the energy triple?
    well you're stacking the cells end to end in series and each one has a pd of 1.5V across its terminals... it's a consequence of the conservation of energy that the pd's add up...

    perhaps analagous to raising a constant mass through a uniform gravitational potential gradient in 3 equal distance steps.
 
 
 
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