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Topic: Electrolysis process  (Read 2530 times)

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Offline Tiani2709

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Electrolysis process
« on: May 26, 2018, 04:57:32 PM »
In the electrolysis process of H2O to H2 and O2 with the electrolite NaOH, what the NaOH converts to?

Offline billnotgatez

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Re: Electrolysis process
« Reply #1 on: May 26, 2018, 05:16:00 PM »
You have to show your attempts or thoughts at solving the question to receive help.
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http://www.chemicalforums.com/index.php?topic=65859.0

Offline Tiani2709

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Re: Electrolysis process
« Reply #2 on: May 26, 2018, 05:33:24 PM »
You have to show your attempts or thoughts at solving the question to receive help.
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Oh, thank you for the link, I didn't notice.

I tried to equation balance and got the point where in one tube I got H2 and in the other one O2 (got this by noticing how a normal screw started oxydising in one of the tubes). I tried to explain this with:

H2O -> 2H2 + O2


Then I thought about the NaOH splitting into Na and OH and doing this:

4OH -> O2 + 2H2O

But that's all I got, I don't see where the Na goes to. As far as I remember Na + Water causes and explosion, so it can't be alone.

By the way, is the threat locked and should I post a new one with the information I got? Or it's fine posting it as a reply. Thank yoi.


Offline Borek

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Re: Electrolysis process
« Reply #3 on: May 26, 2018, 05:59:49 PM »
Oh, thank you for the link, I didn't notice.

Funny thing, you were asked to read the rules while registering, you even clicked to confirm you did read them.

Quote
H2O -> 2H2 + O2

And that's all that is happening. Do you know the concept of spectators and the net ionic reaction?

Quote
But that's all I got, I don't see where the Na goes to. As far as I remember Na + Water causes and explosion, so it can't be alone.

You are ignoring charges of ions. Na and Na+ are two very different things, you don't have Na here.

Quote
By the way, is the threat locked and should I post a new one with the information I got? Or it's fine posting it as a reply. Thank yoi.

No idea what you mean, it is perfectly OK to continue discussion in this thread.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Tiani2709

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Re: Electrolysis process
« Reply #4 on: May 26, 2018, 10:47:58 PM »
Oh, thank you for the link, I didn't notice.

Funny thing, you were asked to read the rules while registering, you even clicked to confirm you did read them.

Quote
H2O -> 2H2 + O2

And that's all that is happening. Do you know the concept of spectators and the net ionic reaction?

Quote
But that's all I got, I don't see where the Na goes to. As far as I remember Na + Water causes and explosion, so it can't be alone.

You are ignoring charges of ions. Na and Na+ are two very different things, you don't have Na here.

Quote
By the way, is the threat locked and should I post a new one with the information I got? Or it's fine posting it as a reply. Thank yoi.

No idea what you mean, it is perfectly OK to continue discussion in this thread.

Thank you for your response, I started reading about the spectators and got an idea about what does that means.

So basically NaOH -> Na + OH, then the OH -> O2 and H2O which breaks into 2H2 + O2, and the water which is already in the solution breaks into 2H2 + O2. I'm right? And the Na stays in the water like a Ion Na.

Offline Borek

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Re: Electrolysis process
« Reply #5 on: May 27, 2018, 03:39:35 AM »
So basically NaOH -> Na + OH, then the OH -> O2 and H2O

You are again ignoring charges, enough to click on the buttons above the edit field to enter them.

Quote
which breaks into 2H2 + O2, and the water which is already in the solution breaks into 2H2 + O2. I'm right? And the Na stays in the water like a Ion Na.

A bit chaotic, but in general yes, you are about right. Na+ is just a spectator, OH- gets consumed by one of these processes, but is then produced in the other one, so the net effect is its concentration stays constant. As I said earlier - all that is happening is water electrolysis.
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Offline Tiani2709

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Re: Electrolysis process
« Reply #6 on: May 27, 2018, 11:36:18 AM »
Thank you for your reply, I think now I understand all the process. Now I can continue with the little experiment I'm doing.

Again, thank you for your help.

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