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The blue bottle experiment is a bottle half full of a colorless solution (dextrose, KOH, and methylene blue, in the version that I saw) and air that, when shaken, turns blue through the methylene blue oxidizing. Over time, the blue color fades while the methylene blue is reduced and the solution goes back to colorless. This can be repeated.

My question is, can this be repeated indefinitely, or is there some chemical in the solution that is being eaten up? Is the mechanical energy of shaking enough to sustain the reaction indefinitely? If it can be used up, what exactly is being used up?

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    $\begingroup$ Nothing last forever. $\endgroup$ – Nilay Ghosh Dec 10 '17 at 4:14
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    $\begingroup$ @NilayGhosh How do you know?:) $\endgroup$ – andselisk Dec 10 '17 at 4:16
  • $\begingroup$ @andselisk 2nd law of thermodynamics $\endgroup$ – Nilay Ghosh Dec 10 '17 at 4:22
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    $\begingroup$ @andselisk I edited my post. Did I make it more clear what I'm actually asking? (which is whether or not the blue bottle experiment can be reused indefinitely) $\endgroup$ – Daffy Dec 10 '17 at 4:37
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    $\begingroup$ @Zhe But the act of shaking the bottle oxidizes it again. From what I understand, oxidization and reduction are inverses. Doesn't this mean that after some time, the solution returns to its original, non-shaken state? Isn't it simply flipping back and forth between oxidization states? Or am I just misunderstanding terminology? $\endgroup$ – Daffy Dec 10 '17 at 4:38
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The blue bottle experiment, and other similar reactions like the "chemical traffic light", are not infinitely repeatable. They are driven by the oxidization of glucose (or some other suitable reducing agent, such as ascorbic acid) by atmospheric oxygen, and will eventually stop working once the bottle runs out of either glucose or oxygen. However, each cycle only consumes a rather small amount of glucose and oxygen, so the reaction can be repeated quite many times.

Also, the blue indicator dye (methylene blue) itself is not used up in the reaction. Even if the reaction does "run out of fuel", it can be made to work again just by adding more glucose to the solution and/or simply by opening the bottle to let more oxygen in. However, since the glucose is not getting fully oxidized all the way to water and carbon dioxide in this reaction, the various incomplete oxidization products will gradually accumulate in the solution.

I'd expect that, if you kept adding more and more glucose to the solution, eventually you'd end up with a sticky sugary alkaline goo full of partly oxidized glucose products that was too viscous to be swirled around the flask any more. At that point, you'd really have no choice but to dispose of the experiment and start over with fresh ingredients.

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  • $\begingroup$ I could understand the chemical traffic light not being infinitely repeatable, since there's no input of energy. The video I watched on the blue bottle experiment didn't specify any of the chemicals being eaten up in the reactions. He only talked about the oxidization/reduction of the methylene blue. So I figured the energy was being supplied through shaking. Thanks for the informative answer :) $\endgroup$ – Daffy Dec 11 '17 at 0:44
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It can be repeated as long as there is oxygen in the bottle.

This is due to the fact that the oxidation and reduction that is happening will always consume the respective chemicals and only the methylene blue is being transformed back and forth.

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    $\begingroup$ I am not sure. I think the dextrose is being consumed by oxidation as well. Moreover if it it being converted to carbon dioxide this will react in acid-base manner with the KOH. Only the methylene blue component seems to last "forever". $\endgroup$ – Oscar Lanzi Dec 10 '17 at 10:54
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    $\begingroup$ Yes you are absolutely right... That's what I meant only that I couldn't remember what reacted how, so I only wrote about the oxygen $\endgroup$ – Raven Dec 10 '17 at 11:06
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    $\begingroup$ That is not true. I've done this reaction dozens of times, and eventually the colors are different and then it finally stops working. $\endgroup$ – Zhe Dec 10 '17 at 14:05
  • $\begingroup$ And in how far does this contradict my answer? $\endgroup$ – Raven Dec 10 '17 at 14:18

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