Question about aqueous electrolytic cell

[waytogo1]

Hi everyone,
I was doing a problem on electrolytic cell in aqueous solution, and I cannot seem to understand this. Please help!

In the problem,
the aqueous solution has H2O (obviously) with Cu2+ and SO42- and there are two Pt electrodes in the solution with a power source attached to them. There is also a table of reduction potential:
O2 + 2H2O + 4e– → 4OH– (E=0.40)
SO42– + 4H+ + 2e– → H2SO3 + H2O (E=0.20)

And according to the answer to the problem, the anode produces O2 gas.
I don't get this.... doesn't anode prefer oxidizing H2SO3 and not OH?
O2 + 2H2O + 4e– → 4OH– has a higher reduction potential than the other equation... which means that it is more likely to be reduced and less likely to be oxidized right? So shouldn't we be seeing more SO42- produced rather than O2 gas?
thanks!

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[Hadi7183]

Based on what you have mentioned, it's an electrolytic cell, so the NON-spontaneous reaction is forced to go through.

 

[waytogo1]

But whether or not it is spontaneous or non-spontaneous, reduction still happens at cathode and oxidation still happens at anode...
And in the solution sheet for this question, it explains that Cu2+ is reduced to Cu (s) at the cathode, which has reduction potential of 0.34. This makes sense because 2H+ + 2e– → H2 E=0. So, Cu2+ does have a higher reduction potential so it should be reduced here...
So why doesn't the same logic apply to anode? Shouldn't anode oxidize something that has a higher oxidation potential (lower reduction potential)? In this case, shouldn't SO42– + 4H+ + 2e– → H2SO3 + H2O happen instead of producing O2 at the anode?

 

[waytogo1]

I've attached the diagram for this problem...
I'm trying to figure out what happens at electrode #2... and why it produces O2.
thanks!

 

[Hadi7183]

Electrodes #2 and #3 are parts of a bridge to form a closed circuit between two physically separated half-cells. You need to focus on #1 and # 4. In an electrolytic cell, anode is the (+) electrode and cathode is the (-) electrode. In this case, #4 is the anode and # 1 is the cathode. As I said, in an electrolytic cell, the power supply forces the non-spontaneous reaction to happen, therefore:

- At #4 (anode), you have this oxidation reaction: 4OH– → O2 + 2H2O + 4e– (O2 bubbles out). Note that this is the only reaction possible in the left half-cell anyways.
- At #1 (cathode), you have this reduction reaction: Cu2+ + 2e- → Cu (this reaction is favored, b/c it has a higher reduction potential, compared to that of SO42-)

I hope this clear things up!