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Author: Subject: Copper Sulfate
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[*] posted on 21-12-2003 at 18:07


first, DON'T EVER USE A LIGHT DIMMER ON TRANSFORMERS!!! At best you'll get a spectacular shower from the dimmer because they are not designed to drive highly inductive loads and the currents forced back into it will surely destroy the semiconductors.

now, if you have nitric acid, lucky you! what you do is that you dissolve copper in nitric acid, slowly add sodium hydroxide untill all the remaining acid is neutralised, and all the copper nitrate precipitates as blue copper hydroxide. then you heat the solution to boiling, the copper hydroxide will dehydrate to give black copper oxide, let this settle, then decant, filter the oxide, wash with plenty of water, then dissolve this in sulphuric acid... this way you end up with at least 99% copper sulphate, and i'm sure you can get 99.9% if you washed the copper oxide well, and started with pure reactants.
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[*] posted on 22-12-2003 at 05:24


Mumbles
Drop a teaspoon full of bicarbonate of soda into a cup of freshly boiled water and see if you believe me. It decomposes much more readily in solution.

[Edited on 22-12-2003 by unionised]
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[*] posted on 23-12-2003 at 18:19


Quote:
Originally posted by hodges
Quote:
Originally posted by Theoretic
The reaction with sodium bicarbonate isn't the method here, it fizzes with CuSO4 as well. There's a substitution reaction, and the "copper bicarbonate" decomposes to copper carbonate, carbon dioxide and water (only alkali metal and ammonium bicarbonates are stable in a solid state). I tried it. It worked. :D:D:D


Okay, it looks like I did not have a good test then. I know when I mix MgSO4 with NaHCO3 there is no visible reaction. But aparently there is a reaction with CuSO4. I still have the solution - when I get my pH meter that I ordered from E-Bay I will measure the pH and compare it to the normal pH for a CuSO4 solution. This should tell whether I actually have a weak H2SO4 solution or just an even weaker (since no color visible) CuSO4 solution.

Hodges


I received my pH tester and found the pH of the "H2SO4" I produced through electrolosys of weak CuSO4 solution is 0.8. Although there is no longer any green color, some copper remains because if I put in Zn or Al I see some Cu plate out. Also when I titrated with NH4OH I ended up with a rather dark yellow solution. This may be due to impurities in the CuSO4 or contamination left on the carbon rod from the battery it was removed from.

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[*] posted on 29-12-2003 at 00:32


A pH of 0.8?? And you said you used a weak solution?!

Am I missing something here? :o

I just thought of something else. Why electrolyze MgSO4 to make copper sulfate which will be electrolyzed again to get H2SO4? Is it possible to make H2SO4 by just electrolyzing MgSO4 using the flowerpot method?

[Edited on 29-12-2003 by Saerynide]
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[*] posted on 29-12-2003 at 14:29


Quote:
Originally posted by Saerynide
A pH of 0.8?? And you said you used a weak solution?!

Am I missing something here? :o

I just thought of something else. Why electrolyze MgSO4 to make copper sulfate which will be electrolyzed again to get H2SO4? Is it possible to make H2SO4 by just electrolyzing MgSO4 using the flowerpot method?

[Edited on 29-12-2003 by Saerynide]


Well it was a relatively weak solution compared to the amount of CuSO4 that could be dissolved in H2O. I don't remember my chemistry real well, but I seem to recall that a 1N solution of a strong acid has a pH of 0. If the pH was 1, it would imply 0.1N which would be 0.05M. And it was not far from pH of 1 so it was probably something like 0.06M, which would be pretty weak. Someone correct me if I'm remembering incorrectly how pH varies with molarity.

Yes presumably H2SO4 could be made from MgSO4 as long as the solutions of the two electrodes did not mix (flower pot method as you suggested would probably work). But using CuSO4 you can just use two carbon rods and no flower pot or other membrane is necessry.

I was basically just playing around - was not necessarily thinking this would be the most practical way to make H2SO4. I have a reasonably good source of H2SO4 but was interested in the electrochemistry of the reactions.

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[*] posted on 31-12-2003 at 04:24


Yeah pH does vary with molarity. I didnt think a solution of H2SO4 with such a low molarity could be so acidic :o

Anyways, I tried using a paper towel cause I didnt have the time to go out to find a flower pot yet (apparently, places here dont sell non-glazed flower pots very often :mad: ). It didnt work cause my paper towel kept on drying out :mad:

[Edited on 31-12-2003 by Saerynide]
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[*] posted on 2-1-2004 at 07:48
Commercially available CuSO4


Source of information:
http://www.finishing.com/0000-0199/064.html

Quote (from the last entry):
'The composition of K77 root Killer is >99% Copper Sulfate'

There is also a recipe for copper-plating without exotic chemicals.

[Edited on 2-1-2004 by overseer]
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[*] posted on 3-1-2004 at 06:19


I tried bridging with a different type of paper towel this time and it didnt dry out. But I've encountered some even worser problems.

1: This red stuff coats the copper anode and stops the reaction. Looks like copper oxide to me. It doesnt stick on very hard, but not exactly adhesionless either. I have to rub it off with paper towels.

2: I dont seem to be getting CuSO4 at the anode. I get a clearish liquid with Cu(OH)2 precipitate, while at the cathode, I dont see much Mg(OH)2 precipitate, but instead the solution is light blue.

Does anyone have any clue whats going wrong? :(
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[*] posted on 4-1-2004 at 15:38


Yeah, the red stuff is Cu2O. I wasn't using a salt bridge cell so I don't know about the other problem you seem to have.

I do want to add that I have this nice butterfly knife that was kinda the point behind getting CuSO4 in the first place (because I wanted to plate it) that I actually did try to plate via acid dip method. I took some of my mostly H2SO4 + a little CuSO4 solution and dripped it on some iron, got a nice precip and because the iron had been bead blasted I actually had decent adhesion. Anyway, confident that simple displacement would work, I dipped the knife. The handles plated soooooo well. I think the metal must be zinc because I think you can plate Cu onto Zn with a simple acid dip. I could not even touch the coating with steel wool, it was on there. The blade on the other hand, being stainless steel did not plate so well. It has a dull Cu finish with decent adhesion but not nearly as good as the handles which look like pure, bright copper. Anyway, I'm going to keep working on it because theres some things I'd like to try with plating non-conductive surfaces but I'll get back to this when I'm through experimenting! :-)
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[*] posted on 4-1-2004 at 17:20


Quote:
Originally posted by Saerynide
1: This red stuff coats the copper anode and stops the reaction. Looks like copper oxide to me. It doesnt stick on very hard, but not exactly adhesionless either. I have to rub it off with paper towels.

2: I dont seem to be getting CuSO4 at the anode. I get a clearish liquid with Cu(OH)2 precipitate, while at the cathode, I dont see much Mg(OH)2 precipitate, but instead the solution is light blue.

Does anyone have any clue whats going wrong? :(


I think you may have cathode and anode mixed up. Cathode is negative side of battery, anode is positive. The positive side should be where the CuSO4 forms, and it sounds like that is what is happening from your description. On the negative side, if you are seeing a red precipitate, this is probably Cu or some oxide of Cu. That would indicate that copper ions are getting into that side of the reaction through your porous membrane. With a good membrane you should only see Mg(OH)2 (which you do have to scrape off from time to time though). I used two separate bowls, bridged with an absorbant brand of paper towel. I didn't end up with any significant Cu ions crossing through the paper towel until all the Mg ions were used up, at which point I just stopped the reaction. I'm wondering if you tried to do it with just one bowl, with the two halves separated by a paper towel? Or maybe you let the reaction go long enough that in fact all the Mg ions were used up and Cu ions started coming across?

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[*] posted on 5-1-2004 at 01:27


Im positive that I did not mix up the electrodes cause the steel cathode didnt corrode like how it would if it was an anode. And, yeah, I used 2 separate bowls. I dont see how one could possibly separate a solution with a paper towel in one bowl :D

But I think your suggestion that the Mg ions getting used up is probably the case. But I cant understand why that would happen. Wouldnt it just float over and start plating the knife instead of turning the solution blue?

Edit: If I was doing this with just MgSO4 to make H2SO4 and Mg(OH)2 with 2 separate bowls, the two solutions wouldnt start mixing and neutralizing each other, would they? I dont see how it could cause theres only 1 kind of anion and cation in there. The cell would just continue to break down water, right?

[Edited on 5-1-2004 by Saerynide]
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[*] posted on 5-1-2004 at 16:32


Quote:
Originally posted by Saerynide
If I was doing this with just MgSO4 to make H2SO4 and Mg(OH)2 with 2 separate bowls, the two solutions wouldnt start mixing and neutralizing each other, would they? I dont see how it could cause theres only 1 kind of anion and cation in there. The cell would just continue to break down water, right?

They should not mix (IMHO) unless by capilary action. Presumably you would get Mg(OH)2 at the cathode and H2SO4 at the anode. However, once all the sulfate ions are gone from the cathode side, the solution may not conduct much anymore (not sure how good a conductor Mg(OH)2 is - probably not too good given that it doesn't dissolve very well). The anode would still conduct fine with the H2SO4 though. If H2SO4 is the desired product you should be able to replace the solution in the cathode side from time to time to continue the reaction.

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[*] posted on 6-1-2004 at 03:22


Crap.... I poured most of the excess MgSO4 into the bowl with the anode.... Wouldnt it still work cause the Mg ions would go the the cathode?
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[*] posted on 6-1-2004 at 18:34


Quote:
Originally posted by Saerynide
Crap.... I poured most of the excess MgSO4 into the bowl with the anode.... Wouldnt it still work cause the Mg ions would go the the cathode?


Yes, but what about when all the SO4 ions are gone on the cathode side? At that point you will have just Mg(OH)2 on the cathode side and that will not conduct electricity very well. You can go ahead and use your setup, just replace the cathode solution from time to time if H2SO4 is your desired product.

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[*] posted on 7-1-2004 at 03:25


But at that point, wouldnt I be left with two bowls containing H2SO4 and Mg(OH)2?

Edit: It has been four days since my H2SO4 cell has been running. My god, it sure takes forever to separate 0.5 mol of epsom salts :mad: But anyways, the Mg(OH)2 is coming out ok, but the H2SO4 has turned a VERY strange color. It had started to yellow after a few hours, but now its an orange yellow, like amber. I was thinking it might be the carbon, but would that turn orange?? :o There is also the possibility that some of my solder wires dissolved in it, but neither tin or lead sulphate is yellow/orange... Any ideas?

[Edited on 7-1-2004 by Saerynide]
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[*] posted on 7-1-2004 at 15:02


Quote:
Originally posted by Saerynide
But at that point, wouldnt I be left with two bowls containing H2SO4 and Mg(OH)2?

I think you will probably use up all the MgSO4 at the cathode first. This would likely cause the circuit to pretty much stop conducting, before all the Mg ions are gone from the anode. Thus you would get a mixture of MgSO4 and H2SO4 at the anode.
Quote:

It has been four days since my H2SO4 cell has been running. My god, it sure takes forever to separate 0.5 mol of epsom salts :mad:

1 Faraday per mole times charge. 1 Faraday is somewhere around 27 ampere-hours if I recall correctly. So 0.5 moles * 2 = 27 ampere hours. If it has been running for four days, that would be about 100 hours and would imply your current is 1/4 amp or less. Do you have a way of measuring current? If you are drawing more current than that then the reaction is probably complete and you would then be taking apart the water.
Quote:

But anyways, the Mg(OH)2 is coming out ok, but the H2SO4 has turned a VERY strange color. It had started to yellow after a few hours, but now its an orange yellow, like amber. I was thinking it might be the carbon, but would that turn orange?? :o There is also the possibility that some of my solder wires dissolved in it, but neither tin or lead sulphate is yellow/orange... Any ideas?

Not sure what that would be. You are usign carbon rods from a dry cell? Maybe there is some chemical left in these from the battery. Orange could be iron - do you have any iron nearby that could have gotten into the solution?

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smile.gif posted on 7-1-2004 at 16:01


1 Faraday is the charge on a mole of electrons and is equal to 96,485 coulombs of charge.

27 A*h is 27hr*3600s/hr = 97,200 coulombs....it seems you remember pretty damn correctly! :-)
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[*] posted on 8-1-2004 at 00:59


Quote:
Not sure what that would be. You are usign carbon rods from a dry cell? Maybe there is some chemical left in these from the battery. Orange could be iron - do you have any iron nearby that could have gotten into the solution?


I dont see how any iron couldve gotten in. My cathodes are stainless steel, but theyre in the other bowl and there is not a spot of corrosion on them. Plus, cathodes dont corrode... do they??

Oh well, iron or not, Im not letting my four days of waiting go to waste. To purify that remaining solution of iron, magnesium sulphate and sulphuric acid, this is what I plan to do:

Electrolyze the solution again, but connecting it to the cathode side. This should precipitate remaining Mg2+ and bring remaining SO4 2- to the anode, right?

Then I would be left with a solution of only iron and sulphuric acid. I will then electrolyze the solution again for a long time to make sure all iron is plated out as iron metal or iron (II) oxide.

Does this work?

[Edited on 8-1-2004 by Saerynide]
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[*] posted on 8-1-2004 at 11:46


Quote:
Originally posted by Saerynide
Yeah pH does vary with molarity. I didnt think a solution of H2SO4 with such a low molarity could be so acidic :o


Remember that the pH scale is logarithmic. Reducing the pH by one unit corresponds to increasing the hydrogen ion concentration by ten times. When you consider that some acids have a pH of -2, a pH of 0.8 isn't particularly low.




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[*] posted on 8-1-2004 at 18:51


Quote:
Originally posted by Saerynide
Oh well, iron or not, Im not letting my four days of waiting go to waste. To purify that remaining solution of iron, magnesium sulphate and sulphuric acid, this is what I plan to do:

Electrolyze the solution again, but connecting it to the cathode side. This should precipitate remaining Mg2+ and bring remaining SO4 2- to the anode, right?

Then I would be left with a solution of only iron and sulphuric acid. I will then electrolyze the solution again for a long time to make sure all iron is plated out as iron metal or iron (II) oxide.

Does this work?


No, if you reverse the polarity then any Mg(OH)2 that now forms will react with your H2SO4 to produce MgSO4 again!

If there is iron or another similar metal dissolved in your H2SO4, though, you can get it out using your second idea. Electrolyze the H2SO4 side only (both electrodes in the same container). Make sure you use a carbon rod for the anode or else if you use a metal such as copper or iron it will go into your solution. The best thing for the cathode would also be a carbon rod - but alternatively you could use copper such as a wire since H2SO4 does not react with copper except in strong concentrations. Any iron in the solution should plate out or precipitate as the oxide on the cathode. Be sure you remove the electrodes as soon as you disconnect the current or else it will dissolve again in the H2SO4.

Once you have your H2SO4 you can test its strength by dropping a couple drops of it on zinc. Based on the speed at which hydrogen is evolved, you can get an idea of the concentration. You could also try aluminum foil but there may be a delay of up to a few minutes before that reaction starts due to the oxides that form on the surface of bare aluminum.

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[*] posted on 9-1-2004 at 03:00


Quote:
[quoteNo, if you reverse the polarity then any Mg(OH)2 that now forms will react with your H2SO4 to produce MgSO4 again!


I should've been more clear. I didnt mean reversing the polarity of the electrodes. I meant that I'll place the cathode into the H2SO4 + MgSO4 solution, and place the carbon anode into a new bowl with a bit of water. Come to think of it, I should plate out the iron first...

This is why I think it would work:

The Mg ions will form Mg(OH)2 as the same amount of SO4 ions will migrate over the bridge to the water until all the Mg ions are used up from the cathode side. There would then be H2SO4 and Mg(OH)2 left at the cathode because I originally had excess SO4 ions in there. These will react back to form MgSO4 at the cathode. Electrolysis will split the MgSO4 again, which should continue to bring SO4 ions to the anode until the Mg ions are again used up. This process will repeat until eventually, only Mg ions are left and the solution stops conducting.

Does that make sense or was it a load of crap? :(

[Edited on 9-1-2004 by Saerynide]
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[*] posted on 10-1-2004 at 07:53


Quote:
I'll place the cathode into the H2SO4 + MgSO4 solution, and place the carbon anode into a new bowl with a bit of water.

The Mg ions will form Mg(OH)2 as the same amount of SO4 ions will migrate over the bridge to the water until all the Mg ions are used up from the cathode side. There would then be H2SO4 and Mg(OH)2 left at the cathode because I originally had excess SO4 ions in there. These will react back to form MgSO4 at the cathode. Electrolysis will split the MgSO4 again, which should continue to bring SO4 ions to the anode until the Mg ions are again used up. This process will repeat until eventually, only Mg ions are left and the solution stops conducting.


I didn't quite follow all that (I'm easily confused), but one thing I'm thinking is that your water solution at the anode will not conduct much electricity. Unless the water is impure, in which case it will conduct but you will have ions moving between the bowls.

I think you can still use your solution - just keep it the anode and replace the solution on the cathode side with a new MgSO4 solution. That will give you more sulfate ions and allow the magnesium ions to move to the cathode where they will form magnesium hydroxide. When this is done then you can electrolyse the H2SO4 solution in a single bowl with carbon electrodes for both anode and cathode to plate out the iron.

Hodges

[Edited on 1/10/04 by hodges]
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[*] posted on 10-1-2004 at 08:43


Lol, we are like just repeating each other's words, but in different ways :D
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[*] posted on 28-1-2004 at 12:26


"NaHSO4 mixed with alcohol gives a solution of H2SO4 in alcohol and Na2SO4 as a precipitate.
Boil, or better, distil off the alcohol and get the acid."

Does this really work? It sounds too good to be true, considering that I can buy NaHSO4 by the tons at any pool supplier.




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