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For a science experiment you need to electroplate

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    For a science experiment you need to electroplate a 130 nm thick zinc coating onto both sides of a very thin, 3.0 cm*3.0 cm copper sheet. You know that the charge carriers in the ionic solution are divalent (charge 2e) zinc ions. The density of zinc is 7140 kg/m^3
    If the electroplating apparatus operates at 1.3 mA, how long will it take the zinc to reach the desired thickness?


    2. Relevant equations

    not sure

    3. The attempt at a solution
    I have no idea how to solve this problem
     
  2. jcsd
  3. Apr 4, 2009 #2

    LowlyPion

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    Re: Electroplating

    Figure that you are going to need to supply 2 electrons for each zinc atom you are to plate onto the copper.

    Figure the volume of the plating as total surface area times the depth.

    They give you the density so figure how many atoms need to be plated. That times 2 electrons then has to pass as the current doesn't it?
    And they give you the rate that electrons can be supplied, so ...
     
  4. Apr 4, 2009 #3
    Re: Electroplating

    so we know area and thickness of the plate and the density of the zinc, and the current.
    so we can get the volume of the plate and using density we can figure out the mass.
    volume = 3*10^-2* 3*10^-2* 130*10^-9 m = 1.17*10^-10 m^3
    and the mass = 1.17*10^-12 m^3 * 7140 kg/m^3 = 8.35*10^-7 kg
    what is the next step? how do i figure out number of atoms?
     
  5. Apr 4, 2009 #4

    LowlyPion

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    Re: Electroplating

    Looking up zinc I see the standard atomic weight is 65.38 g/mol.
    http://en.wikipedia.org/wiki/Zinc

    How many atoms in a mole?

    And 2 electrons needed per atom and ...
     
  6. Apr 4, 2009 #5
    Re: Electroplating

    so in 1 mole the weight of zinc is 65.38 g
    so in 8.35*10^-7 kg 1.3*10^-8 mole
    right?
     
  7. Apr 4, 2009 #6

    LowlyPion

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    Re: Electroplating

    And now how many atoms per mole?

    Does Avogadro ring a bell?
     
  8. Apr 4, 2009 #7
    Re: Electroplating

    i dont know. its complicated for me.
    in 1 mole we have 6.022*10^23 atoms
    so in 1.3*10^-8 mole we have 1.3*10^-8 * 6.022*10^23 = 7.83*10^15 atoms
    so number of electrons = 7.83*10^15 *2 = 1.57*10^16

    then how do i figure out the time?
     
  9. Apr 4, 2009 #8

    LowlyPion

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    Re: Electroplating

    Well what is the current? 1.3 ma right? So how many electrons per second is that?

    1 ampere is 1 Coulomb / sec

    and 1 Coulomb is equal to 6.241 *1018 electrons.
     
  10. Apr 4, 2009 #9
    Re: Electroplating

    1 Coulomb is equal to 6.241*10^18 electrons.
    so 1.57*10^16 electron = 2.5*10^-3 C
    I = Q/t
    so t = Q/I
    = 2.5*10^-3/1.3*10^-3
    = 1.92 s

    is this right?
     
  11. Apr 4, 2009 #10

    LowlyPion

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    Re: Electroplating

    Oh, and you want me to check numbers too?

    Well, now that I look ... there are 2 sides to be plated. So that one is not quite right.
     
  12. Apr 4, 2009 #11

    LowlyPion

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    Re: Electroplating

    Looks like it propagates the factor of 2 all the way. I get 3.785s.
     
  13. Apr 4, 2009 #12
    Re: Electroplating

    ok so for both side the total time would be 1.92*2 = 3.84 s
    I tried plugging that in but that didnt work.
    any idea what Im doing wrong here?
     
  14. Apr 4, 2009 #13

    LowlyPion

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    Re: Electroplating

    Not sure why. Not without redoing the whole daisy chain.
     
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