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Electrolysis. |
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| Jun7-05, 01:27 AM | #1 |
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Electrolysis.
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. Thanks. |
| Jun7-05, 01:37 AM | #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... Daniel. |
| Jun7-05, 01:55 AM | #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? |
| Jun7-05, 01:58 AM | #4 |
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Electrolysis.
Okie dokie, thanks.
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| Jun7-05, 02:00 AM | #5 |
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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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