2
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$ Oct 16, 2015 at 7:55
  • $\begingroup$ 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. $\endgroup$
    – Pilitio
    Oct 16, 2015 at 7:59
  • $\begingroup$ 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. $\endgroup$ Oct 16, 2015 at 8:12
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 16, 2015 at 9:05
  • 1
    $\begingroup$ 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 $\ce{2SO2 + O2 <=> 2SO3}$. $\endgroup$ Oct 26, 2015 at 3:31

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.