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Harmless
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Electrolysis of Sulfuric Acid with Copper electrodes
Does anyone know what will happen?
I thought that the copper would dissolve creating copper sulfate, but what would happen with the sulfuric Acid?
does anyone know the full reaction?
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12AX7
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Anode: Cu - 2e- > Cu(2+)
Cathode: 2H+ + SO4(2-) + 2e- > H2 + SO4(2-)
Copper metal dissolves as cupric ions (since cuprous are generally unstable, particularly in sulfuric solution), while hydrogen ions are discharged as
hydrogen gas.
With an inert anode, the sulfuric acid (or sulfate ion if using a reasonably alkaline salt) is oxidized to persulfuric acid (S2O8(2-) or SO5(2-) as
the case may be). When warm, it decomposes on formation to yield some amount of oxygen. The cathode reaction remains the same, or may include a
persulfate > sulfate reduction depending on conditions.
Tim
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woelen
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I agree with 12AX7, but making persulfate requires quite specific conditions (which??). I've tried this reaction quite a few times with graphite anode
and I almost exclusively get oxygen at the anode:
2H2O - 4e --> 4H(+) + O2
When this is balanced with the cathode reaction, you simply have decomposition of water into the elements.
On the other hand, I also have read many times about the formation of the peroxodisulfate.
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Chris The Great
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I believe that with inert electrodes this will produce substantial amounts of ozone at the right temperature and concentrations... I can't remember
the details but I read them here in some of the ozone threads from way back...
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bio2
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Here you go! Forget right now where it came from.
Potassium peroxydisulfate,
K2S208, also known commonly as
potassium persulfate, is one of the easiest salts to prepare by
electrolytic oxidation, because it is only slightly soluble and
separates from solution in good yield. The electrolyte is a
solution of potassium acid sulfate. KHS04, saturated at about
5 °, 'that is, containing about 40 grams of salt to 100 grams of
water.
The anode is a short piece of platinum wire, 1 cm of
22 gauge (0.644 mm diameter) being satisfactory, and the cathode
is a small strip of platinum foil. The cathode should be placed
an inch or two above the anode. Reduction of persulfate at the
cathode is inappreciable, and it is not necessary to separate anode
and cathode by a porous pot. Good yields of persulfate are
obtained by using a high current density and a low temperature.
Addition of a little fluoride raises the yield somewhat.
Take about 150 ml of potassium acid sulphate in, preferably,
a tall narrow "electrolytic" beaker, and surround the beaker by
an ice bath. Add about 0.1 gram of sodium fluoride if desired,
though it is not necessary. Insert the electrodes, and pass a
current of 1 to 1.5 amp for 2 hours. Six volts will be sufficient
potential and will not require a rheostat, since the current can
be adjusted by moving the cathode up and down. Keep a record
of the time and the current, and maintain the current constant
to 0.1 amp, so that you can calculate the number of coulombs
passed and thence the theoretical maximum yield for the reaction
2HSO7 -o SO - 2H + 2e.
Potassium peroxidisulfate separates as a crystalline precipitate
after the first 10 minutes. At the end of the electrolysis, filter
the precipitate on a small sintered glass filter or a Gooch crucible,
wash with alcohol and ether, and dry. The yield is about 5
grams. From the yield, calculate the current efficiency, that is,
the observed yield divided by the maximum theoretical yield
for the quantity of electricity passed. The current efficiency
should be about 50 per cent.
The peroxydisulfate can be produced faster by passing a higher
current, with a longer wire for the anode, if the solution is kept
cold--a condition that can be maintained only with very efficient
stirring.
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