[toddy0101]

Hello guys,
I am having trouble getting the right answer for the following problem:
the problem reads, " Predict the half-reactions occurring at the anode and the cathode for the electrolysis of aqueous Na2SO4."
I was able to to get the answer for the reaction that happens at the cathode. For the anode reaction I got, H2SO3 + H2O --> SO4 2- + 4H+ + 2e- E= .20 v and 2 H2O -> O2 + 4H+ + 4e- E=.82V [(H+)] = 10 ^-7 M.
 I concluded that since E for the SO4 2- reaction is more negative than E for the water  this is will be reaction that takes place at the anode. The answer however  is  H2Ol -> O2 + 4H+ + 4e- E=.82V [(H+)] = 10 ^-7 M. I don't understand why this is the answer.

Thanks a lot ,

[Hunter2]

I think your equation is wrong. Sodium sulfate is an inert salt, an aquaeous electrolysis guides to hydrogen on cathode and oxygen on anode.