[k_amy21]

Hello,

I am curious about one thing: Sodium bisulfite (SBS) is commonly used as oxygen scavenger. Therefore, I would expect no oxygen to be present inside a sodium bisulfite solution. Well... it seems that my SBS solution contains ca. 0.5 mg/L oxygen, measured with an oxygen probe. How come there is oxygen? I would expect that all oxygen is consumed by the reaction with bisulfite. Is it just a delay in the reaction? Or there is another explanation? I would appreciate if someone can clarify this to me.

Thank you in advance.


Regards
Amy

[Borek]

Have you tried to estimate equilibrium concentration of oxygen from redox potentials?

[k_amy21]

Borek, thank you for your quick reply and suggestion. If I understand well, using the redox potential for each half cell reaction (HSO3-/HSO4- and O2/O2-), I would calculate standard cell potential E, which would allow me to calculate equilibrium constant K (E=0.0592/n x log K). The value of the K would give me an indication of the reaction rate between HSO3- and O2. Is this what you were thinking?

Unfortunately, I am not able to find redox potential for the HSO3-/HSO4- half reaction. But I did find a paper about reaction rate of HSO3- and O2 in seawater vs. in fresh water. Seems the latter is much slower due to less dissolved species and hence, lower conductivity. This is exactly my case; SBS is dissolved in deionized water. Therefore, some oxygen would easily be measured inside this solution when left in the open air. Is this correct way of thinking?

[Borek]

If I understand well, using the redox potential for each half cell reaction (HSO3-/HSO4- and O2/O2-), I would calculate standard cell potential E, which would allow me to calculate equilibrium constant K (E=0.0592/n x log K).

Yes.

The value of the K would give me an indication of the reaction rate between HSO3- and O2. Is this what you were thinking?

No, you are mistaking equilibrium with kinetics, these are two different things.

Unfortunately, I am not able to find redox potential for the HSO3-/HSO4- half reaction.

You should be able to find redox potential for SO₄^2-/SO₂(aq) - the latter is more or less what bisulfite solution is, should be good enough for a rough estimate.

But in a way you are right that kinetics can play a role here too, especially if the solution is open to the air.

[k_amy21]

I found the redox potential for SO₄^2-/SO₂ and O₂/O^2- and calculated the K to be 3 x 10³⁹. Borek, if you could explains what this value means now I would appreciate it. 

[Borek]

calculated the K to be 3 x 10³⁹

Looks way too high for me.

Broadly speaking when you have equilibrium constant you can estimate equilibrium concentrations. You probably know starting concentration of bisulfite, you probably know pH (if not, do some reasonable guesses), you can assume some fraction of bisulfite got oxidized (to make things easier: assume exactly half). Just solve for the oxygen concentration, plug and chug.

[k_amy21]

Ooh ok, I think I got it now; how to calculate the concentration of a specie based on the K and concentrations of other species in the reaction. Thanks a lot for the tips and leading mean to the "solution". I'll check once again the calculation of K, I might have messed up something knowing my clumsiness.

[Borek]

Standard redox potential for SO₂/SO₄^2- seems to be around 0.17 V, reduction of oxygen takes place at 0.40 V, so the E_cell is in the 0.2V range, that should yield log(K) in the 0.2/0.059 range (a lot of handwaving here, but I am talking just about estimating an order of magnitude, 10³⁹ is for cells with EMF over 2V).