What voltage should copper electrolysis be run at?

Posted by admin on May 16th, 2010 and filed under copper | 2 Comments »

I’m trying to purify copper by running a direct current through a copper sulfate solution. I’ve been trying it at 6 volts, but this results in a brown sludge being deposited at the cathode, instead of metallic copper. Does anybody know what voltage is normally used for this operation?

Current density is more important than the specific voltage. A typical "Cell Voltage" would be 0.15V to 0.30V and the overall cathode current density should be about 15 mA/cm2. (note: anode passivity occurs at current densities above 25 mA/cm2.) Also, dissolved copper ion concentration and sulfuric acid concentration are important. You want to have about 120 g/L Copper Sulfate (about 0.7 molar) and 215 g/L Sulfuric Acid (about 2 molar).

What were the first copper tools created?

Posted by admin on May 13th, 2010 and filed under copper | 1 Comment »

After the discovery of copper working and before the discovery of bronze, the History channel said that copper was used to make tools and decorations. They showed a lot of copper decorations, but no tools. So, what tools DID early man make from copper? Could they make swords, even if they had to be sharpened often? Could they make axes? Hammers? What tools could they make?

P.S. It was hard to decide what topic to put this question in.

it seems to me that the VERY first things that would be made would be the most simple. a small blade for instance; a knife. possibly an arrowhead; simple tools for hunting/preparing food. tools of war, like the sword you mentioned, would be made later

We recently let a copper bottom pan over heat on the glass cooktop of our range. How do we remove copper?

Posted by admin on May 11th, 2010 and filed under copper | 2 Comments »

The top is clean except for a copper semi-circle left from melted copper from the pan. Is there an effective way to remove the copper residue?

Contact the cooktop manufacturer. I never heard of this happening before, pan must have been defective.

What causes copper to be oxidized when it is mixed with hydrochloric acid?

Posted by admin on May 8th, 2010 and filed under copper | 2 Comments »

I preformed an experiment in chemistry in which we placed a piece of copper into HCL and left it there for a week. When I went back to look at the experiment, the copper had been oxidized. What caused this? And what is the chemical equation? I know that hydrogen will not oxidize copper due to their placement on the activity series.
Thank you for your help.

IF any copper oxide was formed while the copper was setting in hydrochloric acid, the copper oxide would dissolve. Copper may not dissolve in hydrochloric acid. (see 1) But copper oxide is a basicoxideand it would dissolve in hydrochloric acid. Per this reaction: CuO + 2 HCl = CuCl2 + H2O

Copper Annealing

Posted by admin on May 8th, 2010 and filed under copper | 1 Comment »

Copper Annealing. When copper is worked, meaning bending, folding, hammering, etc., the copper becomes more and more rigid. Continued work will cause the copper to fracture. To avoid this, it is time to anneal the copper.

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Oxidation of Copper

Posted by admin on May 7th, 2010 and filed under copper | 25 Comments »

A chemical demonstration of the oxidation of copper metal by nitric acid. Carleton University, Ottawa, Canada.

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How much copper can you get from 10g of copper carbonate?

Posted by admin on May 7th, 2010 and filed under copper | 6 Comments »

Hi,

How much copper can you get from 10g of copper carbonate?
Also, what are the methods to extract the copper from the copper carbonate?

I think I know a method, which is to heat the copper carbonate to get copper oxide, and then use a reducing agent to take the oxygen away. Any others? Also, confirming my method will be a great help.

Thanks buds.
All :- Thanks for your help :D .

well first you need to know the mw of Cu, C, and O which is:
Cu= 63.5460 g/mol.
O= 15.9994 g/mol
C= 12.0107 g/mol

CuCO3 is what we are looking at (copper carbonate)
63.5460+12.0107+15.9994(3)=123.5549 g/mol

(63.5460/123.5549)*100= 51.43 percent composition

.5143*10g=5.143g

Your method for extraction appears to be correct. You can also use strong acids as well, I believe.