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Electrolysis OXN/REDN reactions (1 Viewer)

menty

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Yes, iv done a search, and no i cant find it.

For electrolysis, say the electrolysis of concentrated NaCl, where the reduction is that of water, and oxidation is BOTH water and chloride ions, which one do we write for the oxidation reaction, both water and Chloride ions, or just the chloride ions?

How do we write the net ionic eqution if it is both equations for oxidation.
 

wanton-wonton

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menty said:
Yes, iv done a search, and no i cant find it.

For electrolysis, say the electrolysis of concentrated NaCl, where the reduction is that of water, and oxidation is BOTH water and chloride ions, which one do we write for the oxidation reaction, both water and Chloride ions, or just the chloride ions?

How do we write the net ionic eqution if it is both equations for oxidation.
I did this in Industrial Chem, it should be the same.

Anode reaction: 2Cl-(aq) --> Cl2(g) + 2e-
Cathode reaction: 2H2O(l) + 2e- --> H2(g) + 2OH-(aq)
Net ionic: 2Cl-(aq) + 2H2O(l) --> Cl2 + H2(g) + 2OH-(aq)


Oxidation is just chloride ions.
 

tennille

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Doesn't Cl only oxidise when it is highly concentrated? If you did this prac at school, water would oxidise. I doubt that Cl gas would have been produced since it is poisonous.
 

Templar

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You can produce a small amount of Cl gas if the voltage is high enough. It wasn't nice seeing yellowish green gas bubbling through water.
 

wanton-wonton

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Tennille said:
Doesn't Cl only oxidise when it is highly concentrated? If you did this prac at school, water would oxidise. I doubt that Cl gas would have been produced since it is poisonous.
The Cl- doesn't oxidise, it reduces since its a negative ion.
 

Abtari

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um, what are u talking about wanton-wonton

Cl- ions get oxidised i.e. they lose electrons to form atomic and molecular cl...

positive ions get reduced
negative ions get oxidised
 

rama_v

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Yes, in concentrated solutions (say >1M), chlorine ions get oxisided to form Chlorine Gas Cl2 (g)
 

Rax

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Ok we did this alright at school. When you calculate mixed solutions in Electrolysis you put all the possible reactions and use the standard reduction potentials table to find the actual overall reaction at the lowest voltage. So for NaCl Solution, the reactions involve water, Sodium or Chlorine.

Possible Reduction reactions
Na+ + e− -> Na(s) −2.71V
2H2O (l) + 2e− -> H2 (g) + 2OH (aq) − 0.83V

Possible Oxidation reactions
2Cl− (aq) -> Cl2 (g) + 2e− −1.36V
2H2O (l) -> O2 (g) + 4H+ + 4e− −1.23V

Thus the equasions with the lowest required voltages are
4H2O (l) + 4e− -> 2H2 (g) + 4OH (aq) − 0.83V
2H2O (l) -> O2 (g) + 4H+ + 4e− −1.23V

which balance to become the oxi-redox
6H20 -> 02(g) + 4H+ 2H2 (g) + 4OH (aq) -2.03V

This reaction will occur before the others and water will be reduced and oxidised at the cathode and anode. But as these are theoretical they will generaly work but not every time.
And if it is a concentrated solution or molten salts then the Cl and Na will be oxidised and reduced.
So when doing electrolysis remember the reacions with the lowest voltages needed will occur first. If you put more volts in the others will happen aswell. So you will have multiple gases forming at the anode and cathode. And this can be dangerous

Hopefully this will help
Seeya
 

Haku

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i think its actually +2.03V cause u gotta *-1 to the anode voltage, and the cathode voltage is +1.23 not minus.

so 0.80 + 1.23 = 2.03V positive.


its a good method to write out all the possible equations, and see which one is have more oxidation potential for anode and reduction potential for cathode. but you got to take note that with high voltages both reaction could occur, or when the concentration of Cl is >0.5mol. (i think thats what it is, pretty sure!)
 

Rax

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hmm I think it is -2.03 because if it was positive for one it would be Spontaneous and would not require any voltage thus it would be a galvanic cell. And eh I havent seen any water reacting on its own have you.
That and I already did the minus to the anode when I wrote it and a minus plus a minus is a minus (hope no one is confused there)
so its − 0.83V + −1.23V= −2.06...yeah I missed the other 0.03 but I am sure you all noticed that.
 

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