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?