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1/2 O2 + H2O + 2e- --> 2OH- /// Is there a name for this equation? And how is it triggered?

It's an equation from an electrolyte (KOH) and air. How do I learn more about this equation and how to make it happen? (It's important in batteries...)... thanks.

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I think it's an oxidation-reduction half reaction. It is not something you observe, it is just a technique to solve problems.

I could be wrong... I don't get that stuff you said about batteries.

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Anode: Zn + 4OH? ? Zn(OH)42? + 2e? (E0 = ?1.25 V)
Fluid: Zn(OH)42? ? ZnO + H2O + 2OH?
Cathode: 1/2 O2 + H2O + 2e? ? 2OH? (E0 = 0.34 V pH?11)
Overall: 2Zn + O2 ? 2ZnO (E0 = 1.59 V)

eh..... the formatting ......... no sub / super scripts.....

let me give you think link....

https://en.wikipedia.org/wiki/Zinc%E2%80%93air_battery#Reaction_formulas
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Okay, so what is happening is that you put zinc in an electrolyte (Pottassium Hydroxide I think), along with what is called a "Gas Diffusion Electrode" (don't be intimidated by the name of that, please, lol).

Then what happens is some how the oxygen from the air reacts with... something... I guess the water? But we have to ask, where does the water come from?

Anode: Zn + 4OH? ? Zn(OH)42? + 2e? (E0 = ?1.25 V)
Fluid: Zn(OH)42? ? ZnO + H2O + 2OH?
Cathode: 1/2 O2 + H2O + 2e? ? 2OH? (E0 = 0.34 V pH?11)
Overall: 2Zn + O2 ? 2ZnO (E0 = 1.59 V)

The water comes from the Zincate (the Zn(OH)2?4) turning into Zinc Oxide (ZnO) and water and Hydroxyl ions (OH-)

Keep in mind! I am NOT a chemist or chemistry major! So if I can follow this, it should be possible for you too!!!

Then we ask, how does the Zincate appear? Well, the zincate appears in the anode reaction, which comes from Zinc + Hydroxyl. But where does Hydroxyl come from? From the elctrolyte (the cathode reaction) ... which is the original equation I asked about. but it comes from the elctrolyte, KOH (Potassium Hydroxide).

How does it work...? How does the KOH react with the air? If you put a "gas diffusion electrode" into KOH will it magically react and form Hydroxyl Ions?
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The formatting is bad again here let me retype it... (I'm taking out the charges)

Anode: Zn + 4OH- --> Zn(OH)4 + 2e (E0 = -1.25 V)
Fluid: Zn(OH)4 --> ZnO + H2O + 2OH-
Cathode: 1/2 O2 + H2O + 2e --> 2OH (E0 = 0.34 V pH=11)
Overall: 2Zn + O2 --> 2ZnO (E0 = 1.59 V)
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Sorry for the convoluted mess....
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Wow hard stuff maybe you should change majors........ Anyways, the KOH is already dissociated in solutions, existing as K+ and OH- ions.
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