# Does a Na2SO3 solution oxidize to Na2SO4 in an oxygen rich atmosphere?

Let's assume I have a pure solution of water and $\ce{NaOH}$. When this solution comes in contact with an atmoshpere containing inert gases and $\ce{SO2}$, then $\ce{NaHSO3}$ and $\ce{Na2SO3}$ will be formed into the solution.

The question is, if the atmosphere now contains also $\ce{O2}$, would these reactions happen?

$$\ce{2Na2SO3 + O2 <=> 2Na2SO4 + 2e-}\\ \ce{2NaHSO3 + O2 <=> 2NaHSO4 + 2e-}$$

What would happen to the freed electrons?

• That's quite an atmosphere you have there... Well, the redox reaction can and will happen; as to the freed electrons, there would be none. Oct 16, 2015 at 7:55
• Well actually I do not have such an atmosphere, it was just to simplify the question compared to the original problem. How can I figure out the equilibrium constants for these reaction? I cannot find the H and S of $\ce{NaHSO3}$ anywhere. Could you suggest a software to simulate these equilibrium constants? That would help a lot with my work. Oct 16, 2015 at 7:59
• I don't have these constants or $\Delta G$'s at hand, so keep searching. As for software, it must ultimately rely on the same reference data. Oct 16, 2015 at 8:12
• Would you not have 1/2 O2 + 1/2 H2O + 2 e- -> 2 OH-? You can look up the relevant reduction potentials to find the equilibrium constant. Granted it's not exactly standard conditions but it's a start. Oct 16, 2015 at 9:05
• You can also try to find values in the CRC Handbook of Chemistry and Physics. I think the suggestion by @orthocresol is right, this is certainly a reaction that will happen in an aqueous solution. If you have oxygen and sulfur dioxide in the atmosphere, you will also have to consider $\ce{2SO2 + O2 <=> 2SO3}$. Oct 26, 2015 at 3:31

I'm not sure what is the stable form at room temperature, since we have both sodium sulfite ($\ce{Na2SO3}$) and sodium sulfate ($\ce{Na2SO4}$) in our lab and they're both stable, in air. I do know that the sulfite will oxidise to sulfate with the addition of bleach.
That said, your redox reaction is not quite accurate. In sulfite, sulfur is in the $\ce{S^4+}$ oxidation state. In sulfate, it is $\ce{S^6+}$. What happens in the reaction is that the sulfur gives away two electrons, which then go to an oxygen that changes from $\ce{O^0}$ to $\ce{O^2-}$.
One way of writing it would be $\ce{S^4+ + 1/2O2^0 -> S^6+O2^2-}$. Total charge is preserved and there are no free electrons.