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Faraday's Electrolysis Forumlae question

  1. Jul 12, 2004 #1
    i have a practical assesment i need to complete at school and i come across this problem....i am to electroplate a 20 cent coin with a layer copper, i have all the experiment setup but i'm a bit confused abit about the faraday formula...
    one of the calculations i need to work out is the THEORETICAL amount of copper that is plated out.......
    do i work out this answer via the faraday formula?, cause i am confused...
    faraday's forumla is [no. moles electrons = (current x time)/96500)]...correct me if i'm wrong......
    i cannot see how i can work out the theoretical amount that should be plated out from the copper :confused: ......please help me. It only talks about the amount of electrons.......or does it have something to do with Cu --> Cu2+ + 2e-
    please help me out here, because this assesment is due soon, and i need to have it all clear (not a open book thing)lol :smile: .......
  2. jcsd
  3. Jul 12, 2004 #2
    well, current x time will give you charge, charge over Faraday's Constant will give you moles of e-, but anyway, that's what you have up there

    yes, you could work it that way IF they give you the mass you need to plate the coin successfully, OR you could go through a bunch of rigmarole to look up the unit cell dimensions of copper, then figure out how many moles would give you at least a monolayer or two, and so on, but I doubt the problem is that involved

    did they give you "the coin starts off this many grams and after plating should weigh that many grams?" If so:

    Just use the formula to figure out how many moles of electrons would have to pass to reduce x moles of Cu(II) ( 2 times x ) and that's it (besides converting back to grams if you need to).
  4. Jul 13, 2004 #3
    Well if you have to electroplate with copper, you will have to use something like copper sulfate and deposit copper on the coin starting from copper ions. You need (solvated) copper ions for this and so, the equation

    Cu^{++}(aq) + 2e^{-} -> Cu(s)

    represents in totality what you need to do. From this, you see that you need "2 moles of electrons" (thats how these sums work and it seems odd at first sight) to effect the process of depositing 1 mole of copper from 1 mole of copper ions. So, you need 96480C times 2, or 96500C times 2(whatever you prefer) to do this. But let us say that you have a current source that gives out a (constant) current of I amperes.

    So in time t, it will give a charge equal to I*t coulombs. So the amount of copper (in moles) deposited through this source in a time of t seconds is given by

    \frac{1}{(2 * 96500) C} * [ (I*t) C ] moles

    You can convert this to mass by multiplying with the molecular weight of copper. That should see you through I think.

    Hope that helps.
    Last edited: Jul 13, 2004
  5. Jul 13, 2004 #4
    but according to what ur saying.....if i do it that way, it only gives me the number of moles of electrons. how to i find the mass of copper using the number of moles of electrons given.
    i cant see the relationship between the copper and the electrons.
    or maybe i'm a bit dud i i dont understanding what your trying to tell me....lol :)
    is it possible for you to explain to me a bit clearer....sorry about it
  6. Jul 13, 2004 #5
    wait wait wait.......what your trying to say is that i need to work out the number of moles of electrons required. because the copper to electron ratio is a 1:2 ratio, i just divide the no. of moles of electrons by 2 and that gives me the no. of moles of copper and i get the mass from there, by timesing by the molar mass of Cu.....
    is this right, or have i got the wrong interpretation?
  7. Jul 13, 2004 #6
    What do you mean by "timesing"? I am sorry, but I mean multiply by the molar mass. Here's the list of steps...

    1. First write the equation for the electrochemical process.
    2. Next identify the stoichiometric coefficients of the relevant species. In our case, we need 2 moles of electrons to deposit one mole of copper given one mole of copper (cuprous) ions.

    NOTE: 1 mole of electrons corresponds to a coulomb charge of 96500 C or 96480 C.

    3. Now, use the unitary method or ratios, whatever you're comfortable with. Its somewhat like this:

    2 * 96500 C of charge is required to deposit 1 mole of Cu.
    x C of charge will deposit [tex]\frac{1 mole}{2*96500C}*xC = \frac{x}{2*96500}[/tex] mole or moles of Cu.

    4. Now multiply this with the molar mass of copper (63.54 g/mole) to get the mass of copper deposited.

    Hope that helps...

    Last edited: Jul 13, 2004
  8. Jul 13, 2004 #7
    To add to what I said sometime back, x is simply the charge that your current source gives in a time interval t. So if the current source is a constant current source rated at I amperes and the time for which the process is conducted is t seconds, then x = I * t coulombs. Equate this to the ratio for x, so the mass deposited in terms of I and t is given by the equation

    mass = \frac{I * t}{2*96500}*63.54 g
  9. Jul 16, 2004 #8
    OH OK i got it now......
    all clear now.........after i saw that last equationi got it.
    its A.L.L clear now
    thanks a million maverick280857

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