Sciencemadness Discussion Board - Copper Sulfate

Copper Sulfate

Al Koholic - 4-12-2003 at 22:38

I'm attempting to produce a good amount of copper sulfate for some copper plating experiments I want to try in the very near future. What I have been doing lately is running the 5V, 20A DC line from a computer power supply (AT) right into a pyrex beaker containing 800ml of 30% H2SO4 and 2 copper electrodes.

It was working great for a while and I was getting erosion of the copper anode and the solution was turning blue. The way I figure it, each molecule of H2 that leaves at the cathode is another Cu++ ion for my solution.

Anyway, after a few hours of operation I started to get a nice reddish buildup on the cathode as well. I think this may be copper (II) oxide although I am not sure. It did not have good adhesion to the surface of the electrode. However, I reasoned that since hydrogen was still coming off, I MUST be putting copper cations into solution still to balance the charge.

Now I can barely sustain a reduction of hydrogen for more than 2 minutes because the cathode gets coated with this red coating and just quits on me. What I think may be happening is that because I am at such a high voltage I am possibly plating copper back out onto the cathode as well now which might be responsible in some way for the coating that builds up. Since no more hydrogen is being evolved, am I simply moving copper from one electrode to the other? I think I may be...

This brings up my real question. After doing some reading on galvanic cells and electrolytic cells and consulting the handbook of chem and phys, I noted that Cu++ + 2e- => Cu is a 0.34 volt potential...ie: spontaneous. Sooooo I need to drive the reaction the other way with a voltage source greater than 0.34V. Now I am beginning to think that the buildup is a result of having both electrodes in the same bath...should I separate them with a salt bridge?

Also, I am now also going to modify a power supply (an old one with the big ass transformer and cap) by placing an AC dimmer switch between the power connection and the wall socket, then remove the voltage control board from the circuit of the supply so that power goes right from the transformer through a rectifier and into the cap providing me with essentially variable DC, controlled by the dimmer. This would allow me to run at say...0.5V and hopefully avoid providing force for other side reactions.

Saerynide - 5-12-2003 at 02:47

I was also planning on making copper sulfate. Could I make copper sulfate by using a copper anode and a carbon or stainless steel cathode and using a solution of magnesium sulfate? I was hoping to get Mg(OH)2 and CuSO4.

chemoleo - 5-12-2003 at 05:50

Hmm, a few things I dont understand.
If you were indeed just transferring copper from one electrode to the other, then I wouldnt understand why the electrode would 'quit on you', as freshly plated copper also conducts electricity... so the reaction (even if you just electrolyse water) should continue until the electrode is corroded completely.
You want the copper sulphate for plating (sorry to disappoint you it doesnt work very well)... so I guess you could use CuCl2, too? In that case I seem to remember that Cu dissolves in HCL, with H2O2 - but dont take my word for it. Someone else in the forum will remember details for sure.

Copper plating.

Tacho - 5-12-2003 at 08:56

Chemoleo is right. Smooth, shiny copper plating using copper sulfate is not easy. I tried every possible combination. Only made it right once, using low voltage, very low current and graphite substrate.

I believe you have to use cyanide (poison!)salt compositions to have a beautifull finish.

over engineering

Mr. Wizard - 5-12-2003 at 10:18

If you want to eat up the copper with the sulfuric acid, why don't you just use alternating current? You don't need any fancy transformers, just put a safe load in series with the electrolysis cell. Start off with a 60 watt light bulb, which will give you about .5 amp of current through the solution. You will have to be careful that you don't get across the two electrodes, and get electrocuted when it is out of the solution, as the voltage will rise to line voltage when the resistance of the cell goes up. You will have to replace the acid as it is used up, and you willl have to replace the copper as it goes into solution. The solution (electrolyte) will also get warm if you use too much current. You can regulate the current with different size loads. Another safer method would be to use a stepdown transformer with AC output and it would we less likely to shock you. Have fun.

Al Koholic - 5-12-2003 at 10:43

This is all what I though would happen when I first started too....I assumed total corrosion of electrodes which I weighed so I would know when the reaction was done by the weight of the remaining copper.

The red coating on the cathode still confuses me and I can only think it means I am using too high a voltage. Something I also noticed was that yes, before I implemented fan cooling of the solution it would get quite warm after some time. Once it had warmed up sufficiently it began to produce gas at the anode as well!!! Must have been oxygen due to splitting water. At this point the reddish copper oxide coating began to deposit even more rapidly. Perhaps I need to keep the temp down.

Also I thought I would let you all know about my experiments with just this acid solution and plating. I managed to nicely plate a butterfly knife just by doing an acid dip and then a CuSO4/acid dip. The coating was not removed by vigorous cleaning with steel wool. It has begun to corrode because of all the use I give it but that was expected....adhesion still seems to be good.

Haggis - 5-12-2003 at 13:55

I know the main idea of this is making copper sulfate, but at least in America, it is common in the plumbing isle. It is used as a root killer for septic tanks. "Drain Line Root Killer" or the like. It is, however, a pentahydrate, or hexahydrate, I cannot remember. It is available though. Resume conversation.

Saerynide - 6-12-2003 at 00:20

So what about the magnesium sulfate idea? Does that work, since magnesium hydroxide isnt very soluble?

unionised - 6-12-2003 at 03:30

I'd have to look up the solubillities, but I don't think copper hydroxide is much more soluble than Mg(OH)2. On heating it converts to CuO which is very insoluble. At best, you will have to keep the solution cold.

Why go to all this trouble? Cu disolves in hot H2SO4 anyway. (Do this outside)

Saerynide - 6-12-2003 at 05:27

I dont got expensive-fancy-online-ordered chemicals. I have to make do with whats easily obtainable at home

Polverone - 6-12-2003 at 12:00

But if you're in the US, you don't need to order H2SO4 online. Go to a hardware store and look at the drain openers until you find one that has the container sealed inside a plastic bag. Look at the warning labels. It's probably concentrated H2SO4.

Edit: Ah, if your signup IP correctly indicates your location, then you're certainly not going to be able to get chemicals like you can in the US! I don't think your electrochemical method will work. However, copper + an acid + H2O2 will cause the metal to dissolve pretty easily and yield the corresponding metal salt. You could use vinegar and H2O2 to make copper (II) acetate, then precipitate it as carbonate (with addition of sodium carbonate). Now you can react the basic copper carbonate with even weak or dilute acids to get the corresponding copper salt. Can you find dilute H2SO4? It's used for adjusting pool pH, or (somewhat more concentrated) replacing the electrolyte in lead/acid storage batteries.

Actually, if you can get one of these forms of H2SO4, you might be able to use it directly with H2O2 and copper metal.

[Edited on 12-6-2003 by Polverone]

unionised - 6-12-2003 at 15:23

Most places in the world have car batteries that contain sulphuric acid. Diulte sulphuric acid can be concentrated quite effectively by boiling it to drive off the water (You need an inert container to do this). If you do that with copper present you will get copper sulphate (and SO2, unfortunately).
You could add hydrogen peroxide to avoid the waste of acid but copper (and quite a lot of other things) catalyse the decomposition of peroxide so the yield based on peroxide will be low, and that assumes you can get it.
I don't know where you live but if it happens to be warm and dry you can even concentrate acid by just leaving it for the water to evaporate at room temperature. At 50% RH you get up to about 45% W/W. Better if it's hotter or drier.

Saerynide - 6-12-2003 at 20:54

I wish I was still in Canada where you could get pretty much anything from Home Depot... you cant get anything here

And you know whats stupid? Half the drain cleaners here dont even got warning labels/ingredients lists at all! They dont even say the're caustic and that therefore you should not stick you hand in them

No wonder why they never work when Im trying to unclog my drain :mad:

[Edited on 4-1-2004 by Saerynide]

Al Koholic - 6-12-2003 at 22:25

As far as I'm concerned...conc. sulfuric is just too valuble to me to be messing around with in the production of CuSO4. I'm sure if you have unending supplies you would disagree but I have just lost my source of good concentrated acid. Now I can only get battery electrolyte OTC and thats not good for reacting with Cu.

Anyway, electrolysis is just the way to go for making something like this IMHO. It would be nice if I could just make my own H2SO4 from something cheap and simple like sani-flush (NaHSO4 + heat --> Na2S2O7 + heat --> Na2SO4 + SO3 --> SO3 + H2O --> H2SO4). This is off topic and another project entirely...

unionised - 7-12-2003 at 04:20

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.

Saerynide - 7-12-2003 at 07:13

What kind of alcohol? Or does any kind work?

Electrolysis

chloric1 - 7-12-2003 at 11:25

Saerynide-- I have experimented with MgSO4 electrolysis and I think you could make some CuSO4. The technique is to get a small(4 oz) flower pot and plug the hole with plumbers putty. This will be your diaphragm. Use a plastic container with a lid. Cut a hole above the flower pot and insert a copper anode. Then insert a lead cathode adajacent to the pot. The lid should fit snuggly over the top of the flower pot. Use a high voltage 12 volts or more because you will have a voltage drop across the walls of the pot. You are using the pores of the ceramic to separate compartments. the flower pot should contain copper sulfate in solution and the cathode compartment will have a pasty Mg(OH)2 deposit. thsi method will also work with NaCl and Fe anodes to make FeCl2!! USe Nickel or stainless cathode for NaCl electrolyte!!

blip - 7-12-2003 at 12:34

When I was experimenting with making Cu(OH)2 from electrolysis, I found that an NaCl electrolyte would cause the red covering but Na2CO3 wouldn't. I may try chloric1's method on some salts when I'm bored.

hodges - 17-12-2003 at 18:43

Quote:

Originally posted by chloric1
Saerynide-- I have experimented with MgSO4 electrolysis and I think you could make some CuSO4. The technique is to get a small(4 oz) flower pot and plug the hole with plumbers putty. This will be your diaphragm. Use a plastic container with a lid. Cut a hole above the flower pot and insert a copper anode. Then insert a lead cathode adajacent to the pot. The lid should fit snuggly over the top of the flower pot. Use a high voltage 12 volts or more because you will have a voltage drop across the walls of the pot. You are using the pores of the ceramic to separate compartments. the flower pot should contain copper sulfate in solution and the cathode compartment will have a pasty Mg(OH)2 deposit. thsi method will also work with NaCl and Fe anodes to make FeCl2!!

I've done this even simpler. I had two bowls of MgSO4 solution, with a paper towel bridging them. I had copper electrodes on both sides. Anode side the solution gradually turned blue-green. Cathode side I did get some flaky magnesium hydroxide, and also some thick white powder which i had to scrape off the electrode from time to time (MgO?). About the time I calculated that enough amp-hours had passed through the solution to replace all the magnesium ions with copper ions, I noticed the solutions started trying to mix on their own (i.e. the green color started to flow through the paper towel towards the cathode, and some of the white precipitate started to flow into the anode side. I stopped the current flow at this point. Not sure of the ultimate purity but the anode side ended up with a beautiful blue-green color and when the water evaporated left blue-green crystals.

Later I took some of the CuSO4 I had made and electrolyzed using a carbon rod at the anode. I ended up with copper at the cathode. It was spongy and did not stick very well, but I verified that it conducted electricity. Once the blue color was gone I assumed I had a weak H2SO4 solution. I verified this by using reacting it with NaHCO3 and noting the fizzing.

Hodges

Saerynide - 17-12-2003 at 22:17

Sweet!!! I always wondered why no one talks about making H2SO4 by electrolysing CuSO4, so I thought it wasnt possible, but I guess I was wrong

Thanks so much for posting that.

unionised - 18-12-2003 at 12:28

You can make H2SO4 that way, I used to electrolyse copper sulphate till most of the blue colour had gone. Shake it up with lead shavings and filter it (Pb +CuSO4--> PbSO4 +Cu) to remove the copper then boil it to remove the water. Guess what! I never made a lot of acid.
BTW the trick with NaHSO4 uses dry ethyl alcohol. I think the azeotrope would do.

Theoretic - 19-12-2003 at 04:24

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.

hodges - 20-12-2003 at 17:35

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.

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

unionised - 21-12-2003 at 05:27

Sodium carbonate will fizz with acid but not with copper sulphate (unless you heat it rather strongly).
If you don't happen to have sodium carbonate you can make it by boiling sodum bicarbonate solution.

Mumbles - 21-12-2003 at 09:44

Are you positive that boiling a solution of Sodium Bicarbonate will yield Sodium Carbonate? I have seen the decomposition temperature referenced at 270oC numerous times. Last time I checked boiling water gets no where near this, unless you have some sort of super autoclave.

http://www.hummelcroton.com/nahco3_m.html

http://www.tabex.com/Tabex%20MSDS/Total%20Alkalinity%20up.pd...

[Edit] Ok, on further searching I found that 270 was the complete decomposition temperature. It starts to decompose around 93. So yes, it would would form some Carbonate, but no where near complete conversion.

[Edited on 12-21-2003 by Mumbles]

pyroscikim - 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.

unionised - 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]

hodges - 23-12-2003 at 18:19

Quote:

Originally posted by hodges

Quote:

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.

Hodges

Saerynide - 29-12-2003 at 00:32

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

Am I missing something here?

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]

hodges - 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?

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.

Hodges

Saerynide - 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

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]

Commercially available CuSO4

overseer - 2-1-2004 at 07:48

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]

Saerynide - 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?

Al Koholic - 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! :-)

hodges - 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?

Hodges

Saerynide - 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

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]

hodges - 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.

Hodges

Saerynide - 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?

hodges - 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.

Hodges

Saerynide - 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??

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]

hodges - 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:

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?

Hodges

Al Koholic - 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! :-)

Saerynide - 8-1-2004 at 00:59

Quote:

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]

I am a fish - 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

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.

hodges - 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.

Hodges

Saerynide - 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]

hodges - 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]

Saerynide - 10-1-2004 at 08:43

Lol, we are like just repeating each other's words, but in different ways

axehandle - 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.