Electrolysis of a NaCl solution

Electrolysis of a NaCl solution : what gases do you get?

Disclaimer : These are my video logs and are intended for others who are interested in chemistry/sciences. Do not attempt these experiments if you are not experienced. If you will, use safety measures and common sense and consider all possible problems that may arise before starting an experiment.

Category:

Science & Technology

Tags:

• Electrolysis
• NaCl
• solution
• sodium
• chloride
• hydrogen
• chlorine
• chemistry
• science
• vmelkon

[TheSejma]

Nice video, are you sure about the Na though?

I understand that:

Na + H20 -> NaOH + H2

But I read somewhere that the Na+ ion resists reduction and that the H+ ion gets reduced to form H and then H2.

No?

[vmelkon]

@TheSejma The most abundant positive ions are Na+. Also, the method used to prepare NaOH uses this exact concept along with a Hg cathode. I am in fact preparing KOH with Hg cathode. I'm sure it works with all alkali metals.

[TheLost4rmzone]

I thought the exact opposite of what you said. Thanks for letting the confusion out of me. Is sodium hypochlorate also formed during this reaction?

[vmelkon]

@TheLost4rmzone It would form sodium hypochlorite which is NaClO via the reaction

Cl2 + 2 NaOH → NaCl + NaClO + H2O

[bla287]

@vmelkon Well of course if you use an Hg anode, then of course results will be different because the Na(+)/NaHg (amalgam) pair has a higher potential, therefore some NaHg will be made which could be separated by evaporation (very toxic though).

But you don't need elemental Na to make NaOH, Look it's simple:

Water: H2O --> H(+) + OH(-)

Cathode: 2 H(+) + 2 e(-) --> H2

Anode: 2 Cl(-) --> Cl2 + 2 e(-)

==

Overall: 2 Cl(-) + 2 H2O --> H2 + Cl2 + 2 OH(-)

The Na(+) does nothing so you get NaOH

[vmelkon]

@bla287 I think it is much higher than 1%. This was my response to someone else "The most abundant positive ions are Na+. Also, the method used to prepare NaOH uses this exact concept along with a Hg cathode."

[bla287]

@bla287 (2/2)

Now look at the electrode potentials of H(+)/H2 and Na(+)/Na:

2 H(+) + 2 e(-) --> H2 | E0 = 0V (per definition)

Na(+) + e(-) --> Na | E0 = -2.71V

So even if the H(+) concentration is much lower than that of Na(+), the very huge differnce in potentials still makes the H(+) reduction the preffered reaction. Na(+) isn't reduced, maybe in very small amounts, far below 1% (I could calculate it exactly now but that'd take too long)

[bla287]

(1/2)

And Na is not being formed and then reacts with water, this is a misconception they tell you in the middle grades (~8th) at school to keep things simple. You must not forget that every aqueous solution contains H(+) ions because water undergoes autoprotolysis:

H2O --> H(+) + OH(-). The equilibrium is on the far left side, in a neutral solution you have 10^(-7) mol/l H(+) and the same amount of OH(-).

[bla287]

@bla287 (2/2)

- Lower the pH by adding some HCl, this prevents the Cl2 from dissolving and lowers the concentration of OH(-)

- Use graphite as anode, because it has a higher over-voltage for OH(-) to O2 oxidation than for Cl(-) to Cl2 oxidation

All these measures lead to a lower Cl(-)/Cl2 potential and a higher OH(-)/O2 potential therefore the former reaction will be preffered.

And besides, I'm pretty sure wax can also burn in chlorine, forming CCl4 and HCl instead of CO2 and H2O.

[bla287]

(1/2)The whole thing is actually quite complicated. Both O2 and Cl2 will be formed (since they have a very similar electrochemical potential), but you can force the relation of the two to be shifted in favour of chlorine by using the following tricks (having the Nernst equation in mind, google it if necessary):

-Have a fully saturated solution of NaCl (360 g/l), because the more Cl(-) ions the better