kt5000
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Electrolysis of copper in weak sulfuric acid
I have some weak sulfuric acid on hand, around 33%. I tried doing the electrolysis with it and two copper electrodes. I was hoping to see this
reaction:
Cu(s) + H2SO4(aq) ==> CuSO4(blue solid) + H2(g)
Didn't happen. We noticed a strong HCl smell which seemed to be heavier than the surrounding air. None of the reagents contained Cl. The solid
precipitate was also red, not the blue CuSO4 we expected. Here's what I think happened:
4Cu(s) + H2SO4(aq) ==> H2S(g) + 4CuO(red solid)
Question is, why that reaction rather than the CuSO4?
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bfesser
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This has been discussed before, please <img src="./images/xpblue/top_search.gif" /> <a href="search.php">Search</a> before posting.
Also, try using <img src="./images/xpblue/bb_subscript.gif" /> for subscript in your formulæ (e.g.
H<sub>2</sub>SO<sub>4</sub> . Happy searching.
Not confirming/denying your suspected reaction; but it's important to note that <a href="http://en.wikipedia.org/wiki/Copper(I)_oxide"
target="_blank">copper(I) oxide</a> <img src="../scipics/_wiki.png" /> (Cu<sub>2</sub>O) is red (or yellow), while <a
href="http://en.wikipedia.org/wiki/Copper(II)_oxide" target="_blank">copper(II) oxide</a> <img src="../scipics/_wiki.png" /> (CuO) is
black.
[Edited on 7/27/13 by bfesser]
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kt5000
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Apologies.. I try to UTFSE. I'm new to the chem tinkering and it's sometimes hard to figure out how to describe what I'm looking for.
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ElectroWin
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check the MSDS for the H2SO4 product you have; I am guessing you mistook the smell. it's SO2 gas, coming off the anode. you will find the reaction
there is as follows:
Cu + H2SO4 --> CuO + SO2 + H2O
the CuO just formed then dissolves in H2SO4:
CuO + H2SO4 --> CuSO4 + H2O
at low pH this should be a green coloured solution.
if you got red precipitate, it probably formed at the cathode as Cu2O.
what was the current density on the electrodes?
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kt5000
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Thanks.. Sounds like exactly what I have. The red precipitate is there and I later noticed the green tint to the water.
Is current density simply the number of amps divided by the surface area of the electrode? I used 12-AWG copper wire, about 10mm long, so
approximately 67.77 mm2 of area and 1.5 amps of current. So around 22mA / mm2.
Does the current density affect what reaction occurs, or just the rate?
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ElectroWin
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your calculation of current density seems correct; the value works out to be quite high, though.
22 mA/mm^2 = 2.2 A/cm^2 = 22 kA / m^2
yes, current density affects what reactions occur. so does pH, temperature, electrolyte concentration.
i suggest making electrode surface area much larger (by 100x and more).
the anode will corrode if it gets more than 10 A/m^2; the cathode will either collect that red sludge at the high densities youre giving it, or it may
plate metallic copper, iirc around 400 A/m^2 if pH < 6.
[Edited on 2013-7-27 by ElectroWin]
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kt5000
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I ran a batch this afternoon (outdoors). It did corrode the anode badly and had a bunch of red sludge on the cathode. No copper plating on the
cathode. Sounds like I need to find a proper current meter so I can get a better handle on the current density. That 1.5 amps was based on the power
supply and may be way off.
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