Sciencemadness Discussion Board - Sodium from sodium oxide

Sodium from sodium oxide

LMNts - 30-9-2018 at 14:05

Sodium can be produced by reacting sodium hydroxide and magnesium in a thermite like reaction;
2 NaOH + 2 Mg -> 2 Na + 2 MgO + H₂

Because this reaction is relatively inefficient I wondered if substituting the sodium hydroxide with sodium oxide would make the reaction more efficient;
Na₂O + Mg -> 2 Na + Mg

Could this reaction be possible or is sodium oxide less reactive towards magnesium than sodium hydroxide? All sodium producing reactions I could find that used magnesium as reducing agent use sodium hydroxide instead.

Currently I'm trying to make sodium oxide from sodium carbonate. This process is already somewhat challenging because small amounts of water in the carbonate seem to convert some of it to sodium hydroxide (Not sure?). When it is heated too fast it melts and never becomes solid again at high temperatures. Only when the temperature is slowly increased to 500 °C (100 °C per hour) the salt remains solid.

Ubya - 30-9-2018 at 14:18

i think sodium hydroxide is preferred just because sodium oxide is so hard to get pure, in contact with moist air it will turn in the hydroxide, so why not using the cheap hydroxide?

elementcollector1 - 30-9-2018 at 19:48

Sodium carbonate only decomposes at very high temperatures (851 degrees Celsius for the anhydrate, according to Wikipedia), presumably to sodium oxide and carbon dioxide in a process analogous to lime production. Any water present will immediately react with any produced sodium oxide to form sodium hydroxide, and this can happen within minutes in open air due to sodium oxide's extreme affinity for water vapor. Sodium hydroxide, once formed, does not appear to decompose back to sodium oxide upon heating (that I know of).

Sodium hydroxide is used as the sodium source because it's cheap and the closest thing to an oxide that's readily available. If efficiency is your goal, you'd be better off recreating the Castner cell using molten sodium hydroxide and a suitable electrical setup - lower heat requirements and greater reduction efficiency depending on your electrode and metal collection designs.

AJKOER - 4-10-2018 at 08:54

My opinion on an approach at more rational temperatures.

First, per Wikipedia on Sodium tert-butoxide (see https://en.wikipedia.org/wiki/Sodium_tert-butoxide ), to quote:

"Sodium tert-butoxide is the chemical compound with the formula (CH3)3CONa.[2] It is a strong base and a non-nucleophilic base. It is sometimes written in chemical literature as sodium t-butoxide. It is similar in reactivity to the more common potassium tert-butoxide.
The compound can be produced by treating tert-butyl alcohol with sodium hydride.[3] "

Then per Wikipedia on Potassium tert-butoxide (https://en.wikipedia.org/wiki/Potassium_tert-butoxide):

"Potassium tert-butoxide is the chemical compound with the formula K+(CH3)3CO−. This colourless solid is a strong base (pKa of conjugate acid around 17), which is useful in organic synthesis. It exists as a tetrameric cubane-type cluster. It is often seen written in chemical literature as potassium t-butoxide. The compound is often depicted as a salt, and it often behaves as such, but it is not ionized in solution."

Next, imbue the surface of Mg or Al with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH).

One may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-,H+) pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive leaching equation, to quote:

" PbS + 2 •H = Pb + H2S (5) " (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&... )

which I view functionally as follows:

Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)

Further, in the case of Sodium tert-butoxide, a possible corresponding reaction:

Na+(CH3)3CO− + (e-, H+) = Na + (CH3)3COH

So, one could attempt to adsorbed •H onto Mg or Al as a temporary reactive source, which could then be added to previously created Sodium tert-butoxide to hopefully liberate metal Na.

[Edited on 4-10-2018 by AJKOER]

unionised - 4-10-2018 at 11:17

Quote: Originally posted by elementcollector1

Sodium carbonate only decomposes at very high temperatures (851 degrees Celsius for the anhydrate, according to Wikipedia), .

That's a melting point, not a decomposition temperature.
The boiling point is about 1600.
Good luck finding a decomposition temp.

Quote: Originally posted by AJKOER

My opinion on an approach at more rational temperatures.
...

Na+(CH3)3CO− + (e-, H+) = Na + (CH3)3COH

[Edited on 4-10-2018 by AJKOER]

In the real world, sodium reacts quite quickly with tbutanol.

(also, the rest of your "Rational" plan apparently assumes that aluminium is more strongly reducing than sodium in "solution".

Guess again.

[Edited on 4-10-18 by unionised]

weilawei - 4-10-2018 at 13:44

Unionised, you raise a really good point.

It seems that molecules didn't exist before the temperature of the Universe dropped below 3000K, which was my rough recollection.

However the Sun's photosphere is given by Wikipedia as being 5772K, and molecules have been observed in it. (Hydrides and oxides primarily, it appears.) Also, there's one instance of CaOH being observed, though not in the Sun.

I think your comment wishing luck in finding a decomposition temperature is quite apt where the amateur chemist is concerned. That being said, I came across a paper on cavitation in liquids induced by ultrasound the other day, with insanely high recorded temperatures and pressures that make stars look just warm. Food for thought.

clearly_not_atara - 4-10-2018 at 14:35

Sodium oxide is more easily obtained by thermal decomposition of sodium nitrate, which decomposes at lower temperatures to release 1/2 O2 + 1/2 NO + 1/2 NO2.

Sodium peroxide can be produced under some conditions and can decompose to sodium oxide, but this reaction is usually a detonation IIRC.

This is not useful for making sodium, but it is an easy synthesis of sodium hydride via Na2O + H2 >> NaH + NaOH, IIRC.

[Edited on 4-10-2018 by clearly_not_atara]

symboom - 4-10-2018 at 17:47

Sodium hydride from sodium oxide and hydrogen doesn't seem possible to form at all. Just sodium hydroxide but hopefully I am wrong on this.

AJKOER - 4-10-2018 at 20:00

Quote: Originally posted by unionised

Quote: Originally posted by AJKOER

My opinion on an approach at more rational temperatures.
...

Na+(CH3)3CO− + (e-, H+) = Na + (CH3)3COH

[Edited on 4-10-2018 by AJKOER]

And perhaps at a temperature above the boiling point of t-butanol, some Na in a water free medium?

Also, neither Mg or Al are the reducers! It is hydrogen atom radical residing on the metal surface (albeit generated courtesy of the highly electropositive metals).

For more details, read the 'Make Potassium' thread which I view as a harder to execute one pot approach. The beginning part of that thread is actually quite interesting as the experimenters notice a profound influence of the surface properties of the Magnesium employed (for example, with magnesium turnings) as producing the best yield of potassium. This of course is completely understandable if one realizes one is working with a hydrogen atom embrittlement phenomenom, otherwise the observation is quite perplexing.

[Edited on 5-10-2018 by AJKOER]

unionised - 5-10-2018 at 02:55

Quote: Originally posted by AJKOER

Also, neither Mg or Al are the reducers! It is hydrogen atom radical residing on the metal surface!

[Edited on 5-10-2018 by AJKOER]

It would be good if you learned some chemistry.
That hydrogen is produced by reduction by the Mg or Al which will only happen if the metal is a better reductant than the hydrogen.

Here's a hint for you; only the so-called "Noble" metals are less good reducing agents than hydrogen- that's why they don't dissolve in non-oxidising acids.

So, what you are proposing is that we use hydrogen as a reducing agent to produce sodium.
But we know that hydrogen will oxidise sodium to form sodium hydride.

AJKOER - 5-10-2018 at 04:09

Quote: Originally posted by unionised

Quote: Originally posted by AJKOER

Also, neither Mg or Al are the reducers! It is hydrogen atom radical residing on the metal surface!

[Edited on 5-10-2018 by AJKOER]

-
No, not hydrogen, H2, but •H, which per Buxton (https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... ), is to quote:

"The hydrogen atom is the conjugate acid of e-(aq), and it is the major reducing species in acidic solution, Eq. (3).

e-(aq) + H3O+ → H (3) "

This source (https://www.researchgate.net/publication/272140510_Standard_... ) notes that the standard electrode potential of this radical reaction is -2.32 V.

I have displayed previously, an example of the apparent •H reaction with compounds, acting like a (e-, H+) pair. Here is yet another example per Eq (37) in the Buxton source, to quote:

" •H + OH- --> e-(aq) (37) "

which I would depict functionally as follows:

(e-, H+) + OH- --> e- + H2O = e-(aq)

which is also a restatement that the hydrogen atom is the conjugate acid (with conjugate base OH-) of the base e- in the acid water.

Now, there is a connection between the hydride ion, H-, and •H, in circumstances of an abundant of e-(aq):

e-(aq) + •H = H- (Source: see p. 16, Table 2 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )

[Edited on 5-10-2018 by AJKOER]

unionised - 5-10-2018 at 05:06

Quote: Originally posted by AJKOER

-
No, not hydrogen, H2, but •H, which per Buxton (https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... ), is to quote:

"The hydrogen atom is the conjugate acid of e-(aq), and it is the major reducing species in acidic solution, Eq. (3).

e-(aq) + H3O+ → H (3) "

This source (https://www.researchgate.net/publication/272140510_Standard_... ) notes that the standard electrode potential of this radical reaction is -2.32 V.

I have displayed previously, an example of the apparent •H reaction with compounds, acting like a (e-, H+) pair. Here is yet another example per Eq (37) in the Buxton source, to quote:

" •H + OH- --> e-(aq) (37) "

which I would depict functionally as follows:

(e-, H+) + OH- --> e- + H2O = e-(aq)

Now, there is a connection between the hydride ion, H-, and •H, in circumstances of an abundant of e-(aq):

e-(aq) + •H = H- (Source: see p. 16, Table 2 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )

[Edited on 5-10-2018 by AJKOER]

It doesn't actually matter, does it?

You are making [magical species] by reducing something with metal.
There's no way that [magical species] is a better reductant than the metal, is there?

So, why did you just post a lot of dross?

You also seem not to have understood that, in the production of sodium hydride, there's every chance that hydrogen atoms are involved as the oxidant.

AJKOER - 5-10-2018 at 05:59

Unionised:

Forget my opinion, a pundit notes that the hydrogen atom is a major reducing species (not some magical species) with a significant standard electrode potential per yet other pundits (including giants in there field including Robert E. Huie and Pedatsur Neta).

I do agree with you that highly electropositive metals (including Mg and Al) are very instrumental in the process (per, I would argue, the liberation of electrons)!

I would note that the field of hydrometallurgy using the hydrogen atom in a reductive leaching application affirms the reality and science of the methodology.

I have now and previously provided sources and references including a role of surface chemistry (per experimental observations with Mg metal in the 'Make Potassium' thread) for all those interested in exploring the underlying science.

[Edited on 5-10-2018 by AJKOER]

AJKOER - 5-10-2018 at 13:13

For those interested, here is an extract from the 'Make Potassium' (http://www.sciencemadness.org/talk/viewthread.php?tid=14970&... ) to explore some postulated underlying reactions, which I will further comment on:

Quote: Originally posted by AJKOER

........
For those interested, one of my favorite thesis is from 2008, "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, link: https://www.google.com/url?sa=t&source=web&rct=j&...

I believe a similar electrochemical reaction scheme is occuring with Magnesium. Some extracts of interest for the brave, to quote:

"In addition to being a primary corrosion process, dissolution behavior of aluminum and its alloys in alkaline solutions is of considerable interest because it is the anode reaction in aluminum-air batteries.[4] ......The anodic half-reaction at the Al electrode is

Al + 4 OH − → Al (OH)4− + 3 e− (1.1)

which exhibits an electrode potential of -2.35 V in alkaline solutions(vs. NHE).

"2 Al + 6 H2O → 2 Al (OH)3 + 3 H2 (1.2)"

....."dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal, [9-14] ".....

"Another study of the dissolution of aluminum in aqueous solutions by Perrault revealed that the open circuit potential of aluminum in strongly alkaline solutions corresponds closely to the Nernst potential for oxidation of aluminum hydride to aluminate ions [25]

AlH3 + 7 OH− (aq) → Al (OH)4− + 3 H2O ( aq ) + 6 e− (1.3)

This suggests a role of surface aluminum hydride as a reaction intermediate in the dissolution process. Additional evidence for the presence of aluminum hydride was provided by Despic and co-workers.[26, 27] They found that aluminum hydride formation was one of the major processes apart from aluminum dissolution and hydrogen evolution, during the cathodic polarization of aluminum. Titanium corrosion in alkaline solutions is also thought to proceed through a hydride mediated mechanism.[28-30] "

"He found that the open-circuit potential in strongly alkaline media was determined by the equilibrium of the reaction

AlH3 + 7 OH− (aq) → Al(OH)4- + 3 H2O (aq) + 6 e− (3.7)

He obtained a standard chemical potential of 25 kcal/mol for AlH3 from his data, which was in reasonable agreement with prior thermochemical calculations done by Sinke et al who obtained a value of 11.1 kcal/mol for the chemical potential.[80] ...."

"The anodic reaction 3.7 is accompanied by the cathodic reduction of water to form hydrogen

H2O + e- → OH- + H (3.8)

and the reaction of hydrogen with aluminum to from hydride

Al + 3 H → AlH3 (3.9)"
..........................
[Edited on 24-5-2017 by AJKOER]

Interestingly, per a recent thread of mine on the attack of metals by radicals (see http://www.sciencemadness.org/talk/viewthread.php?tid=94166), here is an expanded comment based on the thesis reference to Eq (3.9) above where aluminum metal reacts directly with the hydrogen atom radical forming the important intermediate aluminum hydride:

Al --> Al(+++) + 3 e-
3 x [ •H + e- --> H- ] Source: See https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706...
--------------------------------
Net: Al + 3 •H → AlH3

which could be followed by (3.7) liberating six electrons while consuming seven OH- ions and one aluminum hydride (which consumed three electrons in its formation) and three •H (created from three electrons acting on three H+ from water per Eq 3.8):

AlH3 + 7 OH− (aq) → Al(OH)4(-) + 3 H2O (aq) + 6 e−

Recycling of the electrons implies a possible cyclic process consuming Al, H2O and OH-. However, the electrons could also interact with added organic compounds.

[Edited on 5-10-2018 by AJKOER]

unionised - 5-10-2018 at 16:38

Quote: Originally posted by AJKOER

Unionised:

Forget my opinion, a pundit notes that the hydrogen atom is a major reducing species (not some magical species) with a significant standard electrode potential per yet other pundits (including giants in there field including Robert E. Huie and Pedatsur Neta).

I do agree with you that highly electropositive metals (including Mg and Al) are very instrumental in the process (per, I would argue, the liberation of electrons)!

I would note that the field of hydrometallurgy using the hydrogen atom in a reductive leaching application affirms the reality and science of the methodology.

I have now and previously provided sources and references including a role of surface chemistry (per experimental observations with Mg metal in the 'Make Potassium' thread) for all those interested in exploring the underlying science.

[Edited on 5-10-2018 by AJKOER]

It doesn't actually matter, does it?

You are making [magical species] by reducing something with metal.
There's no way that [magical species] is a better reductant than the metal, is there?

Unless you plan to tell me that you can move electrons to somewhere they would rather not be, by moving them from somewhere where they are happy.
This is not an "appeal to authority" issue.

AJKOER - 5-10-2018 at 19:24

Comparison of Standard Electrode Potential

Mg2+(aq) + 2e- -> Mg(s) -2.38 V (see http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.h...)

e-(aq) + H3O+ --> •H -2.32 V (see https://www.researchgate.net/publication/272140510_Standard_... )

e- + nH2O --> e·-(aq) -2.89 V (same source)

However, reaction will not be taking place at standard conditions, so comparison is informative only.

unionised - 6-10-2018 at 08:22

Quote: Originally posted by AJKOER

Comparison of Standard Electrode Potential

Mg2+(aq) + 2e- -> Mg(s) -2.38 V (see http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.h...)

e-(aq) + H3O+ --> •H -2.32 V (see https://www.researchgate.net/publication/272140510_Standard_... )

e- + nH2O --> e·-(aq) -2.89 V (same source)

However, reaction will not be taking place at standard conditions, so comparison is informative only.

No.
However, the reaction will not be taking place at standard conditions, so comparison is not informative.

You can't get round the conservation of energy.
The better reducing agent reacts with an oxidant to make a ledd good reductant.

AJKOER - 7-10-2018 at 04:13

Unionised:

My meaning in the expression "so comparison is informative only" is like in FYI which means not informative (English is apparently not a science like math).
----------------------------------

For those who feel strongly about the "Standard Electrode Potential", a possible word of caution. The Nernst Equation has a component, Q defined as the thermodynamic reaction quotient (see, for example, http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/electrod...). Now, consider a hypothetical situation of a simple galvanic cell involving say iron (and iron ions) and silver (and it's ions). If one adds to that system, say copper ions, then as a consequence of a side reaction (namely a REDOX couple induced equilibrium between the iron and copper ions, Fe(ll) + Cu(ll) = Fe(lll) + Cu(l) also assuming the presence of a complexing agent like chloride and a favorable low pH), I would argue there could be a change in the concentration of Fe ions due to the further presence of Cu ions, impacting cell voltage. In other words, there is also an implicit assumption of a homogeneous and not heterogeneous system.

[Edited on 7-10-2018 by AJKOER]

unionised - 7-10-2018 at 04:19

So, you have just explained, in some detail, why the standard electrode potentials are irrelevant.

Why did you post them?

Anyway, you still can't make a stronger reductant (like Na) from a weaker one (like Mg) unless there's something else happening to drive the reaction.

It also seems you don't understand what FYI means. The whole point of something sent "FYI" is that it is purely informative (as opposed to something which needs action).

symboom - 7-10-2018 at 05:01

Anyway, you still can't make a stronger reductant (like Na) from a weaker one (like Mg) unless there's something else happening to drive the reaction.

Like nurdrages video on sodium metal from sodium hydroxide with magnesium
Could you describe how that was even possible

unionised - 7-10-2018 at 05:43

Quote: Originally posted by symboom

Could you describe how that was even possible

Sorry, I thought I had. Did you miss this bit?

Quote: Originally posted by unionised

unless there's something else happening to drive the reaction.

AJKOER - 7-10-2018 at 07:06

That something else like being per my comment above, "And perhaps at a temperature above the boiling point of t-butanol, some Na in a water free medium?"

where the t-butanol is a gas and the Sodium a solid resting in an inert medium separated from water vapor as well.

[Edited on 7-10-2018 by AJKOER]