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Topic: Galvanic coatings  (Read 4693 times)

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Offline jray

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Galvanic coatings
« on: January 08, 2008, 08:56:09 AM »
And the last problem i want you to advise:

1.9 g of a metal was electrodeposited on the cathode, when electrolysis of the salt of this metal (III) was carried out for 1.5 hour and electric current was 1.8A. Identify the metal.

I couldn't understand one thing, that this is practical mass and i need to find theoretical, but there's no current efficiency and i don't know how to find it. If i had that i could find it, by the formula:
m=E*J*t/96500 and when i calculate E of the metal i can identify it by multiplying by it's valance. So could you help my finding theoretical mass?

Offline Borek

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Re: Galvanic coatings
« Reply #1 on: January 08, 2008, 09:17:53 AM »
Assume 100%.
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Offline jray

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Re: Galvanic coatings
« Reply #2 on: January 08, 2008, 09:19:28 AM »
So it can't be 1.9 - 100 or am i missing something
                    x   - 100

Offline ARGOS++

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Re: Galvanic coatings
« Reply #3 on: January 08, 2008, 07:00:55 PM »

Dear Jray ;

I believe your problem requires a little different recipe to be solved then yours.

Assuming 100% current efficiency you have to find out (calculate) how many moles the 1.9 g deposited metal are to determine its Atom Mass (AM), and then consult the Periodic System to identify your unknown ( = find the element with the nearest AM).
I found a final difference of 1.3% in AM by calculation.

So the recipe looks like:
  • 1.)   Determine how many mole electrons you spent during 5400 sec (= 1.5h).
    (Remember: 1 A =  6.241 * 1018 electrons / sec; and 1 mole = 6.02214 * 1023 )
  • 2.)   Divide the number of mole electrons you found by 3, because:
             Me+3  + 3e-  ----->  MeĀ°
    Now you have found the number of moles that deposit 1.9g of your unknown metal.
  • 3.)   I think the small rest to find its AM can now be left to you.

I hope this small recipe is clear enough to be easy followed.


Good Luck!
                    ARGOS++


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