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Author: Subject: Electrolysis of Aqueous Sodium Chloride
Joe
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[*] posted on 23-3-2005 at 15:30
Electrolysis of Aqueous Sodium Chloride


Hey,
A friend and I got into a debate over the products of electrolysis of a solution of NaCl. He said the salt was inert and only served to conduct the electricity and the products were only H2 and O2; whereas I believe that the sodium chloride reacts and creates sodium hydroxide and gives off Cl2 H2 and O2. Could anyone help me out here? Which one of us is right?
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hodges
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[*] posted on 23-3-2005 at 15:32


You will get NaOH and Cl2 if the salt is reasonably concentrated in the water.
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chemoleo
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[*] posted on 23-3-2005 at 15:42


There are about 100+ threads on the electrolysis of salt solutions. Have a look there, they contain more info that you could ever wish for.



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thalium
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[*] posted on 24-3-2005 at 10:30


they have pics too...:P



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Mahlzahn
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[*] posted on 28-4-2005 at 06:44


Are the electrolysis of soduim chloride good to prepare dilute HCl ?

2NaCl + 2H2O electrolysis (graphith electrodes)

-->

2Na + Cl2 + O2 + 2H2 (some HCl)

The chlorine gas is than bubbled into a new beaker(s) with H2O.

[Edited on 28-4-2005 by Mahlzahn]
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12AX7
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[*] posted on 28-4-2005 at 08:05


No, that'll make hypochlorous acid as well.
Cl2 + H+ + OH- = 2H+ + Cl- + OCl-
(Pretty sure that's the right polarity... if it isn't... WTF do I care :P )

Come to think of it, the HOCl can be destroyed by boiling, which converts OCl- to ClO3-, which is unstable in acid and releases as ClO2 gas (stay upwind).

But then, you also lose HCl, so you might as well just distill the whole damn thing.

You also need a salt bridge or semipermeable membrane, otherwise the NaOH produced at the cathode reacts with the OCl- produced (since the gas dissolves nicely right at the anode) and makes bleach and higher oxychlorides (I've got such a chlorate cell running right now).

Tim
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Mahlzahn
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[*] posted on 30-4-2005 at 06:17


Bubbling Cl2 through H2O producing
chlorin water and dilute HCl in the follow beakers. Chlorine some agressive ;).

2Cl2 + H2O -->
2HCl + O2
H2O*2Cl2
and
2HOCl
Strong enough for a synthesis. ;)

Intersted too , wich substance react with the NaCl and other substances in the
electrolytic beaker, which electrodes and the semipermeable membrane.

Electrolysis of for instance NH4Cl are of interest too:

H2O + 2NH4Cl -->
2NH3 + Cl2 + H2 (NH4OH*H2O,
ammonia solution)

[Edited on 30-4-2005 by Mahlzahn]
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12AX7
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[*] posted on 30-4-2005 at 07:40


Oof, careful with NH4Cl, if any chlorate forms, your ass is gone.

Tim
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[*] posted on 30-4-2005 at 23:48


it would actualy depend on what voltage was used.

by definition, your friend is right.

2H2O + 2e- ---> H2 + 2OH- Ered = -0.83
Na+ + e- ---> Na Ered = -2.71
Cl2 + 2e- ---> 2Cl- Ered = +1.36
O2 + 4H+ + 4e- ---> 2H2O Ered = +1.23

H2 and O2 should be produced.

experimentaly, Cl is oxidized and H is ruduced. But only due to the overpotential caused by the hydrogen bonding in water. The over-potential adds on about +2v to the +1.23 on the oxidation, so you end up with Cl, being the easier route. So your both kind of right.
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Mahlzahn
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[*] posted on 1-5-2005 at 06:36


You think your calculation accurate ?

qoute:
O2 + 4H+ + 4e- ---> 2H2O Ered = +1.23

It it of interest produce easy HCl

2H2O ---> O2 + 2H2
2H2O ---> O2 + 2H+ +2e +H2 ;)
2H2O ---> O2 + 4H+ + 4e [+1,23]
It`s only interest the easy route of Cl- bound to a H+ or somewhat (lectrodes ...)

The voltage (satturation of the solution) are only of interest for the speed of the reaction. (electric current flow)

(Car-Batterys 12 - 13,5, 22 - 24,6 volts)


[Edited on 1-5-2005 by Mahlzahn]
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[*] posted on 1-5-2005 at 06:56


Quote:
Originally posted by Mahlzahn
You think your calculation accurate ?

yes
Quote:

The voltage (satturation of the solution) are only of interest for the speed of the reaction. (electric current flow)

(Car-Batterys 12 - 13,5, 22 - 24,6 volts)
[Edited on 1-5-2005 by Mahlzahn]


Er... no, voltage doesnt control the speed of the reaction... the amount of electrons transfered controls the speed. aka the current. 2 different things. And voltage does matter. However in most typical cases it means either something happens or nothing happens.
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12AX7
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[*] posted on 1-5-2005 at 10:06


More correctly; current is the amount of reaction that proceeds, voltage is the rate at which current will flow <U>through a given resistance</U>.

Obviously, 5V on two nails isn't going to do nearly as much as 5V on industrial square footage electrodes. ;) Resistance is determined by the resistivity of the solution, electrode area, and geometries. Then I = V/R.

Tim
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Mahlzahn
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[*] posted on 4-5-2005 at 06:23


The voltage control the current flow and the current flow are dependently of the
eletricaly resistor of the solution (satturation).

R=U/I

Interested are the 2Cl- or the 2OH ;) wich goes in the electrolytic solution to the
electodes ;).

Or this organic incursions ?

CH + H2O2 --> COH + H2O
CH + HCl --> CCl + H +1e
CH2 + 2HCl --> CCl2 + 2H + 2e (H2)
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AllanD
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[*] posted on 17-7-2006 at 11:10


Both Sodium Hypochlorite and Sodium Chlorate
are produced commercially by electrolysis of solutions
of NaCl in H2O.

AllanD
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