phunethicalaiming
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1st Post - Electrolysis of MgSO4 & Isolation of Products
1st Post! 
I know this is a simple experiment, I just have a couple questions regarding this simple experiment that caught me offguard while I was showing my
kids (3 & 6). I know most people that demonstrate this only carry out the reaction to show that the solution turns blue and that a precipatate is
formed.
I wanted to go further and isolate the CuSO4 and Mg(OH)2. I've been applying current to the solution using a 9V, and want to ensure that all MgSO4
is reacted upon, as that is the only way I figure to have a relatively pure CuSO4 product.
1) Will electrolysis stop when MgSO4 is completely reacted?
2) Filtering the solution leaves a blue solid behind, which I must believe is Mg(OH)2. However, Mg(OH)2 is supposed to be white. Am I wrong, or is
just that the water passing through the filter somehow leaves behing CuSO4 to color the Mg(OH)2 blue? This one is throwing me. I add dH20 to the
filter and it absorbs a significant amount of CuSO4 everytime but the process seems endless and the blue color remains. On a side note, the bluish
color of the solution always seems so much more brilliant when looking at the "white" precipatate.. Somebody enlighten me
3) When evaporating off the H20 under heat, I just slightly burned some of the crystals, giving them a tannish color, rather than the bluish white.
What effect does that have? Is this oxidation? Is it just a physical change?
All help is appreciated!
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zoombafu
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First of all what exactly are you doing? Im assuming that you have an aqueous solution of magnesium sulfate, and you are running a current through it
using copper electrodes.
If this is the case then magnesium hydroxide will precipitate out, and the solution will turn a blue, from the dissolved copper II sulfate being
formed.
To get all of the magnesium hydroxide you just need to make sure that there is enough copper for it to react with, and keep the reaction running long
enough.
A current will continue to run through out the solution, because the copper cations and the sulfate anions will remain in the solution which will
conduct electricity, so you can't tell if the reaction is over by checking the flow of the circuit.
The precipitate should be MG(OH)2, so keep on washing it to remove any copper II sulfate. (use warm water because it dissolves copper II sulfate more)
To obtain copper II sulfate first filter out all of the magnesium hydroxide, then evaporate, or heat the solution. You can then recrystallize it to
obtain a purer product.
I'm not sure if you 'burnt' it, but if you are trying to obtain a pure product for future reactions it would probably be best not to use the possibly
'burnt' substance.
I hope this helps
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bbartlog
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Without a divided cell, this will not work as described. Mg(OH)2 reacts with CuSO4 to form Cu(OH)2 and MgSO4 (reforming your original solution). As
copper dissolves at the anode and forms CuSO4 in the cell, you will get increasing amounts of Cu(OH)2 (blue) in with your Mg(OH)2, and further you
will never remove all of the Mg from solution for the same reason.
This is why your precipitate is blue. It also explains the discoloration on heating: Cu(OH)2 is unstable and starts to decompose to CuO at 80C (says
wiki... I think some say even lower).
[Edited on 5-12-2011 by bbartlog]
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