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Author: Subject: Separation of a US nickel
cyanureeves
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blogfast25 you wrote: precipitate the supernatant as mixed hydroxides? with what? this solution would have all of the nickel since copper solubility maxed out reducing the percentage of copper left in solution with nickel,right? the precipitate should then be heated with NaOH and leached out. leached out? how? filtered?poured out? will the nickel remain as insoluble hydroxide?
m1tanker78
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 Quote: Originally posted by blogfast25 Now that's what I call copper! So you're selectively plating out the copper on a steel cathode, while the nickel enters into solution, as NiSO4 right? What's the electrolyte... dilute H2SO4? And what voltage and amps are you reading? Very interesting, Tank, much more so than the 'magnetic effort'...

The initial electrolyte is HCl in distilled water. I set the power supply to provide 777mV. I add just enough HCl to pull 500mA initially. As time goes on, the current draw increases to as much as 2.5A. Copper(II) is selectively reduced to Copper(I) at the cathode. Nickel cannot exist as Nickel(I) in this setting which is precisely what's exploited. The copper can be removed by gently lifting the cathode out of the cell and scraping it into a separate container. This form of copper is very spongy so you'll undoubtedly drag out some acidic nickel solution with it.

p.s.: Blogfast, the 'magnetic effort' was something I had to get off my chest. Done.

Tank
m1tanker78
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I don't know when I'll be able to pick up where I left off on this experiment so I'm going to report what I have. I had to interrupt the electrolysis; I was aiming for 20 to 24 hours (or until the anode eroded away).

*The total run time was about 14 hours.
*Electrolyte was dilute hydrochloric acid.
*Voltage was 0.777V throughout.
*Current varied from 0.5A to 2.5A.
*Actual average power consumption (measured at the wall) was 51W x 14 hrs = ~$0.06. *17 coins x$0.05 = 85 cents.
*The anode weighed 84.2g before and weighed 46g after (38.2g dissolved).
*I scraped the copper off the cathode (into distilled water) about 8 times throughout the experiment.

Here's a look at the solution:

And the copper (don't know the precise composition) removed periodically:

The copper is brick red when fresh and turns a lighter color as the dragged-in acid begins to slowly attack it. The water slowly takes on a blue color as the copper dissolves. The copper is extremely porous and now has a glittery appearance - probably micro crystals of metallic copper mixed in with Cu+??

The total cost of this experiment was less than a buck - if I don't count my time and labor.

Tank
cyanureeves
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to not have to be at the mercy of online nickel compound sellers: priceless!
m1tanker78
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 Quote: Originally posted by cyanureeves to not have to be at the mercy of online nickel compound sellers: priceless!

I have to disagree with you, Reeves. The fun factor and learning experience is priceless. I should probably take this opportunity to learn about Moles, Coulombs and knock some rust off of standard red/ox potentials. THAT would be priceless!

Tank
Sedit
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Have you checked the precipitated Copper with a magnet to ensure that not to much Nickel is coming along for the ride?

Knowledge is useless to useless people...

"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story before."~Maynard James Keenan
blogfast25
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 Quote: Originally posted by cyanureeves blogfast25 you wrote: precipitate the supernatant as mixed hydroxides? with what? this solution would have all of the nickel since copper solubility maxed out reducing the percentage of copper left in solution with nickel,right? the precipitate should then be heated with NaOH and leached out. leached out? how? filtered?poured out? will the nickel remain as insoluble hydroxide?

The solution at that point still contains small amounts of copper sulphate.

Precipitate the lot as Ni(OH)2/Cu(OH)2 with NaOH or KOH and filter. Treat the filter cake with strong NaOH or KOH (hot, preferably): as Cu(OH)2 is mildly amphoteric (it forms cobalt blue cuprate anions = Cu(OH)<sub>4</sub><sup>2-</sup> it will enter solution, leaving behind the pure Ni(OH)2 (which isn't amphoteric). Wash profusely with clean hot water and the 'pure' Ni(OH)2 is yours. This method is especially workable if the copper is the minority constituent.

Tank:

Thanks for the data. Surprisingly low currents there but all in all a nice method for separating Cu and Ni. Nice work and well done!

In future and to keep run time lower, try larger electrodes, put closer together. Larger electrodes allow higher currents with acceptable current densities. Shorter distances between electrodes allows higher currents w/o overheating (ohmic power) because of lower overall cell resistance. In industrial electrolysis, electrodes are often frighteningly close together: time is money!

[Edited on 26-10-2011 by blogfast25]

[Edited on 26-10-2011 by blogfast25]

blogfast25
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 Quote: Originally posted by Sedit Have you checked the precipitated Copper with a magnet to ensure that not to much Nickel is coming along for the ride?

Nickel is less ferromagnetic than Fe: small amounts would go undetected by magnetism. Also, if the applied voltage is correct, the separation should be near 100 %.

Neil
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This is great... now for a condensed write up for the pre-pub folder...
m1tanker78
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 Quote: Originally posted by Sedit Have you checked the precipitated Copper with a magnet to ensure that not to much Nickel is coming along for the ride?

Yes, I've poked around with a strong magnet in a ziplock back. Nothing was picked up but just to be clear, some Nickel(II) Chloride solution is dragged in from the cell at each cathode scraping. Most of it should dissipate after rinsing the copper 3 or 4 times with distilled water. It also helps to break-up the clumps before rinsing.

 Quote: Shorter distances between electrodes allows higher currents w/o overheating (ohmic power) because of lower overall cell resistance. In industrial electrolysis, electrodes are often frighteningly close together: time is money!

Blogfast: I agree with you about the electrode spacing but I'll point out that you'd run into a big problem by closing the gap. It won't take long for the copper deposit to bridge over to the anode and create a short circuit.

On that note, one of my next improvements would be to find a suitable 'bag' to hold the cathode. This would greatly facilitate the removal of the copper. As it is, it's hard to avoid dropping a little of it back in the cell each time the cathode is removed unless you scrape every 20 to 40 minutes.

Tank
blogfast25
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Tank:

Or some membane of sorts, to prevent shorting...

Chemistry Alchemist
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Does the Copper/Nickel form an alloy that is soluble in HCl? i dissolved some coins in HCl, the colour started off as green (Nickel II Chloride) but after a while of dissolving still it turned a brown colour... i pored off about 50 mls and added Aluminum until it stopped fizzing, this produced a red precipitate that looks alot like Copper, would Nickel also be precipitated as well? could i add the precipitate to more HCl to dissolve the Nickel and leave the Copper as it is?

cyanureeves
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wont aluminum drop all metals from solution?adding hcl would alloy the sponge back into solution unless your solution remained nickel green but i bet it went clear.why dont you add hydroxide to the chloride solution like blogfast25 suggests then strong heat the dropped hydroxides with more hydroxide and leach?plante1999 and i have successfully separated chromium from iron this way.maybe nickel or copper will remain as insoluble hydroxide.
Chemistry Alchemist
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Yeah Aluminum will precipitate all metals but Copper is not reactive to HCl so why would i have copper chloride in the solution?

cyanureeves
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yikes! i understand your initial question. i will go back to the bleachers now!
blogfast25
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 Quote: Originally posted by cyanureeves maybe nickel or copper will remain as insoluble hydroxide.

If there's enough alkali, Cu(OH)2 will enter solution as Cu(OH)<sub>4</sub><sup>2-</sup>, cuprate, a cobalt blue solution. Copper is slightly amphoteric. Nickel isn't at all, it stays as Ni(OH)2.

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could i just add more HCl to the Copper/nickel powder to dissolve the nickel and leave teh copper powder?

m1tanker78
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 Quote: Originally posted by Chemistry Alchemist could i just add more HCl to the Copper/nickel powder to dissolve the nickel and leave teh copper powder?

That won't work because both metals will enter solution as +2. You can dissolve the metal powders and try a mild reducing agent. I used electrolytic Hydrogen to selectively reduce Cu(II) to the insoluble Cu(I). I'm still working on the half cell reactions and so forth. Other reducing agents have been proposed and suggested upthread.

Tank
m1tanker78
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Here are some progress images on the copper side of the electrolytic process. Bear in mind that I didn't take much care in rinsing or otherwise preventing oxidation. I rinsed once with tap water (not recommended) and once with distilled water (better).

Final settling before decanting off as much of the water as possible. I wasn't able to figure out what the crystals were. Crystals of alkali carbonates (from the tap water) seem out of place in this setting:

Copper 'mud' after decanting off water:

Lack of care on the rinsing step (IMO) quickly manifests itself upon exposure to air:

I suppose it isn't too late to give it a proper rinse but I'll leave it as is for now. Most of the drab green areas seen in the above pics now appear (edit) lime green after several days.

Tank

[Edited on 11-1-2011 by m1tanker78]
blogfast25
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Very nice! Now wash, dry and melt down ;-) !

Sedit
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I am finding the Sulfate salt of the dissolved coins WILL precipitate with the addition of Ammonium hydroxide ONLY the Nickle compounds.

A wash of Ni and Cu sulfates took up ONLY the Cu and left the Ni as a sludge.

I want to personally ask Woelen to perform a simple inorganic experiment for us all if he may since I have great faith in his skills, materials and attention to detail. I wish to know the solubility of CuSO4 in concentrated H2SO4... also the solubility of NiSO4 in concentrated sulfuric acid.....

In the end I would like to see these two saturated solutions mixed together. I have a huge suspicion that the Copper sulfate will drop out due to A common ion effect leaving the Nickle behind much more pure then before still in solution.

I can not seem to keep CuSO4 in the solution of sulfates very well.... it keeps precipitating out and this could be a great find IMO.

PS: I cant edit pictures very well at all and many are quite large, I miss posting pictures... If I posted some very large ones is there any Mods willing to resize and fix them for me so I can actively post my progress in various areas until my computer is fixed?

[Edited on 2-11-2011 by Sedit]

Knowledge is useless to useless people...

"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story before."~Maynard James Keenan
blogfast25
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 Quote: Originally posted by Sedit I am finding the Sulfate salt of the dissolved coins WILL precipitate with the addition of Ammonium hydroxide ONLY the Nickle compounds. [Edited on 2-11-2011 by Sedit]

What ratio Cu/Ni are you using with the ammonia? If the copper ammonia complex is much stronger (larger Kf) than the nickel complex, then the ammonia would complex the copper preferentially. It then becomes a matter of stoichiometry. You could test it by adding Cu2+ to a nickel ammonia complex: if Kf,Cu >> Kf,Ni then the copper should 'steal' the ammonia... A bit like some complexometric titrations. It should then be possible to precipitate the Nickel as hydroxide or maybe carbonate.

You could check the common ion effect also with Na2SO4 or K2SO4. If it works for separating Cu and Ni, that would be interesting.

One way of uploading photos more or less hassle free is to use a Blogger account (FREE! GRATIS for LIFE!): uploading pictures to a free Blogger blog is child's play (you don't even need to publish a single post). Then you copy the URL assigned to the picture into your posts at SM. It's what I do. Easy peasy.

[Edited on 2-11-2011 by blogfast25]

Sedit
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I would like to say the ratio is 25%/75% Ni/Cu sulfates but that can't be since when dissolving the coins in Sulfuric acid the Copper sulfate always precipitates out to some extent meaning there has to be a higher concentration of Nickle sulfate in the resulting solution then would be expected from the coins.

I had some mixed crystals from them sitting around and when I added Ammonia to it the Copper sulfate which made up the bulk of these crystals went into solution as the complex and a green precipitate sludge settled out on the bottom just as before with previous experiments using the solution instead of the crystals.

I am confident that Ammonia can be used to separate the two, the only main downside is the fineness of the Nickle precipitate and the repeated washings needed to make it pure. If this electrochemical means presented by Tank is as complete and as simple as it seems then it would trump my method of using the Ammonia a thousand times over.

It's only academic curiosity now after seeing Tanks experiments and nothing more since my way is messy and liable to great loss. Selective precipitation through the common ion effect however could be of great value if I find the reaction goes to a completion that is worthy of being used. I remember one time where the ammonia did not precipitate the Nickle as expected and I believe that was when I was using the HCl salt IIRC, this would mean that Ammonia sulfate plays a role in the precipitation of the Nickle hydroxides to some extent.

Knowledge is useless to useless people...

"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story before."~Maynard James Keenan
m1tanker78
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Sedit, I used strong aqueous ammonia when I experimented with your proposed method of separation of the metals. I wonder if that has anything to do with our observations being out of alignment??

I thought copper sulfate precipitated because of a lack of water in the concentrated H2SO4. Same thing happened when I used H2SO4 + KNO3 to dissolve some nickels. Regardless, that constitutes a significant separation of the metals which is the title and goal in this thread!

The ammonia method needs more experimentation. The sulfate method dissolves and segregates the copper and is practically hands-free. The electrolytic method also dissolves and segregates the copper but requires attention and input ~hourly. Then there's the need to build up the anode.

All afford room for improvement.

Tank

blogfast25
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 Quote: Originally posted by m1tanker78 Sedit, I used strong aqueous ammonia when I experimented with your proposed method of separation of the metals. I wonder if that has anything to do with our observations being out of alignment??

IF (I'm still not wholly convinced) it's possible to separate the two with ammonia, then using the right amount of ammonia would be crucial.

From this table:

http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm

... can be gleaned that the K<sub>f</sub> (complexation constant) of the copper (II) tetrammonium complex is about 1.1 x 10<sup>13</sup> and for the nickel (II) hexammonium complex it's 5.5 x 10<sup>8</sup>. That's several orders of magnitude of difference, which would seem to strongly suggest that if the right amount is used (basically C<sub>NH3</sub> about 4 x C<sub>Cu2+</sub>, the ammonia would preferably complex the copper but not the nickel. But use too much ammonia and part or all of the nickel will be complexed too.

Note: the complexation constant for the copper - ammonia complex is:

K<sub>f </sub> = [Cu(Amm)<sub>4</sub><sup>2+</sup>] / ( [Cu<sup>2+</sup>] x [Amm]<sup>4</sup> )

[Edited on 3-11-2011 by blogfast25]

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