joseph1990
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Electrolysis Na2SO4
So there really isn't too much information out there on Na2SO4 in terms of reduction potential. I have an aqueous solution of .1M Na2SO4 and 2 copper
electrodes.
At the Anode the reaction is
2 SO42- (aq) S2O82-(aq) + 2 e- o Eox =-2.00 V
2 H2O(l) O2(g) + 4 H+ (aq) + 4 e- o Eox = -1.23 V
So we basically have the water being oxidized because of the lower ox number
At the cathode
Cu2+(aq) + 2 e- Cu(s) o Ered = 0.34 V
2 H2O(l) + 2 e- H2(g) + 2 OH-(aq) o Ered = -0.83 V
And here at the cathode we have the copper being reduced.
Am I correct to assume that 12.09 grams of copper will precipitate in the solution after 2 hours at 7.5 amps?
So it seems as though most of the videos where people are doing electrolysis are making a mistake by saturating the solution since excess na2so4 will
be left in the solution, so how much na2so4 is necessary for the 12.09 grams of copper to precipitate? Or in other words, how fast is Na2SO4 being
used up in the cell?
Thanks for your help.
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joseph1990
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Crap, I think that equation is the dissolution of cuso4. One mistake equals new maze.
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woelen
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At the anode you don't get one of the two reactions in your post. The easiest reaction is oxidation of copper metal to copper(II) ions. You need
graphite, platinum or MMO if you want the above reactions to occur.
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joseph1990
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Quote: Originally posted by woelen  | | At the anode you don't get one of the two reactions in your post. The easiest reaction is oxidation of copper metal to copper(II) ions. You need
graphite, platinum or MMO if you want the above reactions to occur. |
So hypothetically if I were to use the copper electrode how much na2so4 will be used up? I'm assuming 1 gram won't be enough to deposit the required
reduction of copper(II) ions.
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woelen
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No Na2SO4 will be used up, it only serves as a means of making the liquid conductive. At the anode, copper ions go into solution. At the cathode,
water is decomposed (to hydrogen and hydroxide ions) and also copper ions are precipitated if at sufficient concentration. The sodium ions and sulfate
ions remain in solution.
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chornedsnorkack
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How can the reaction work?
What you need is Cu at cathode.
You need a soluble Cu salt for that.
Which you do not have.
As the reaction begins, you only have Na2SO4 at cathode.
Meaning that the only reaction possible at cathode is reduction of H2 - no Cu at cathode yet. As the sulphate ions travel to anode to dissolve Cu
there, the only possible counteranions at cathode will be OH-.
As the Cu cations dissolved at anode migrate towards cathode, they will meet OH- ions formed by the cathode.
And precipitate as insoluble and therefore non-conducting Cu(OH)2.
The net reaction will therefore be Cu (dissolved from anode)+2H2O->H2 (reduced at cathode)+Cu(OH)2 (precipitated some distance between electrodes)
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