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Electrolysis of gold problem

  1. Jun 7, 2005 #1
    In the electrolysis of Au3+ (aq) solution, gold is deposited. How much gold is deposited in 6 hours by a constant current of 0.540 A?

    I know you need to find the moles of e- by using It/F. But I'm not sure where to go from there.

    Any help would be great.

  2. jcsd
  3. Jun 7, 2005 #2


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    Use Faraday's law,compute the mass of Au by making those multiplications/divisions and then compare to the fninal result.

    The formulation's kinda vague,so you can leave the final answer in Kg...

  4. Jun 7, 2005 #3


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    Consider the half-reaction

    Au3+(aq) + 3e- -------------> Au(s)

    Now 1 mol of Au3+ requires 3 Faradays (3*96500 C) to deposit 1 mol of Au

    Therefore, if the amount of charge passed is 6*3600*.540 C, how many moles of Au will be deposited?
  5. Jun 7, 2005 #4
    Okie dokie, thanks.
  6. Jun 7, 2005 #5
    I just used n = It/zF (where z= amout of "excess" electrons). Then multiplied the answer by the molar mass of gold. That gave me 7.94g.
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