Why does the Blue Bottle Demostration stop eventually?

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The Blue Bottle Demonstration involves a redox reaction where glucose is added to a NaOH solution with indigo carmine, resulting in color changes. The process eventually stops due to the depletion of oxygen, which is necessary for the re-oxidation of indigo carmine to its original color. When the solution is vigorously shaken, it may become stuck in a blue state because all available oxygen has been consumed, preventing further oxidation. Indigo carmine undergoes two oxidation steps: first to red and then to green, with the second reaction involving oxygen. Understanding the specifics of these redox reactions and their reversibility is crucial for completing the lab report.
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I'm trying to finish a lab report on the Blue Bottle Demostration where glucose is added to a solution of NaOH. Then Indigo carmine is added so the solution would go from red to yellow to green (My teacher called it the stop light reaction :D)
I need to figure out why does the process eventually stop.
Here is what I think:
Since O2 in the flask is what re-oxidizes the indigo carmine to reform the initial colour at equilibrium, without O2, the reaction will eventually stop.
My question is, what happened to the O2 that oxidizes the indigo carmine? does it just stay there?
 
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What I meant was that when we did the lab in class, when we shook the solution very vigorously, the solution turned blue and never turned back. If all O2 oxidized are then reduced, why did the reaction get stuck?
One more question:
I read on the internet that indigo carmine was oxidized twice, once to give the red colour, and once more to give the green colour. How?
Thanks!
 
I'm not too familiar with this lab, but the reaction with glucose is the first oxidation/reduction reaction with the indicator, the next reaction involves the indicator with oxygen gas, that is the electrons are allocated to the oxygen. I'm not quite sure how reversible this particular oxidation/reduction reaction is. You should access the reaction products for this second reaction.

I'm guessing that the redox reaction in this case are successive, remember that two electrons are involved in the reaction.
 
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