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November 01, 2025, 12:19:58 pm

Author Topic: Help...Faraday's laws  (Read 1896 times)  Share 

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shadezofemerald

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Help...Faraday's laws
« on: October 01, 2008, 01:40:03 pm »
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A couple of questions i didn't know how to work out:

1) A charge of 0.400 faradays was passed through 1.00 L of 1.00 M copper II sulfate solution using carbon electrodes. What's the concentration of Cu ions in solution after electrolysis?

2) Three electrolytic cells containing silver nitrate solution, Copper II sulfate  and Chromium III Sulfate solution, respectively, where connected in series so that the same amount of electric charge passed through each cell. Metal was deposited at the cathode of each cell. If 10g of silver was obtained from one cell, what mass of metal would be obtained from each of the other cells?

3) Electrolysis of concentrated NaCl solution produced 15.0 L of Chlorine gas, measured at SLC. What mass of silver would be produced if the same amount of electricity were passed through a silver-plating cell?

 :o



ENTER 2008: 87.05

1st Preference 2009: Commerce/Arts @ Deakin Uni- Melbourne

Collin Li

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Re: Help...Faraday's laws
« Reply #1 on: October 01, 2008, 01:55:44 pm »
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Question 1

0.400 faradays is synonymous with saying .

Another way you can think of it is: , which then implies:

You should be able to identify that copper will be deposited as a result of electrolysis, at the expense of copper ions. Using the moles of electrons, we can work out the amount of copper ions consumed:



Originally there was 1 mole of copper ions (using ), so now there is: of copper ions remaining.

So the final concentration is:

Collin Li

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Re: Help...Faraday's laws
« Reply #2 on: October 01, 2008, 02:01:34 pm »
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Question 2 and 3 are done best with knowledge of Faraday's second law: the amount of product you get, for a given amount of charge, is inversely proportional to its charge.

That is, if I put enough charge to produce 6 moles of silver from , then:

- From I can produce 3 moles of copper (half)

- From I can produce 2 moles of chromium (a third)