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Author: Subject: Dissolution of Group 6 metals
j_sum1
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[*] posted on 23-3-2015 at 06:07


So...
6H2O2 + Mo --> O2 + H2MoO4 + 5H2O??

Why not
4H2O2 + Mo --> H2MoO4 + 3H2O

Is it not correct to consider the evolution of O2 an undesired side reaction catslysed by the Mo? And just because I have bubbles evolving, that is no indicator that I am actually dissolving my Mo.

Is the acetic acid doing nothing? I ask because I merely copied what was done upthread.

Nearly 48 hours and my wire is still there. This is depressingly slow.
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Mesa
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[*] posted on 23-3-2015 at 07:11


Molybdenum acetate only forms from molybdenum hexacarbonyl. Evolution of bubbles can only be one of two possibilities given whats in the reaction mixture.

edit:
Yes, the acetic acid is doing nothing. Acetic acid/hydrogen peroxide has been used industrially to oxidize group 6 metals, but the concentrations were essentially 1:1. The idea was that the peracid was a better oxidizer than the H2O2.

[Edited on 23-3-2015 by Mesa]
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Kitsune1
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[*] posted on 23-3-2015 at 07:51


Molybdenum Disulphide can be used as a catalyst in the decomposition of H2O (it has potential as a Pt replacement in Hydrogen Fuel Cells), so it is plausible that there is some decomposition of H2O2 however this reaction is so painfully slow and so little gas is actually evolved that it would be very hard to test.

My reason for the addition of Acetic Acid was purely to be in line with my other experiments which was to look at Acetate-metal complexes such as Copper (ii) Acetate or Iron (iii) Oxyacetate. I was interested to see whether Molybdenum (as well as Tungsten) could be dissolved into an organic acid in the same way that copper could through the simple action of a household oxidiser (H2O2) as a potential route to open up further studies into Mo compounds without reliance on purchasing compounds or using reagents such as HF. However what made things more interesting for me is that on the addition of Hydrogen Peroxide, Molybdic/Tungstic acid was created, not the respective acetate, thus rendering the addition of Acetic Acid unnecessary.

I don't know whether Acetic acid's action on Mo(CO)6 is the only route or whether it is the standard route used as I have not looked into Group 6 chemistry too much and the information out there seems very limited.
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bolbol
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[*] posted on 23-3-2015 at 22:49


Wouldn't Aqua regia (Nitric + Hydrchloric acid) get the job done?
I heard Ruthenium is the only metal that Aqua regia doesn't do anything against
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Kitsune1
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[*] posted on 24-3-2015 at 03:21


It does, especially in hot Aqua Regia, it will even react with hot Hydrochloric acid or Hot Sulphuric; Nitric acid can also dissolve it but it quickly passivates; the aim was to bring the metal into solution without the use of "dangerous" reagents, really testing how far I could go with organic acids. Ruthenium is indeed a strange metal, most room temperature or hot acids have no action on it but then it is vulnerable to to ClO- ion.
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Molecular Manipulations
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[*] posted on 24-3-2015 at 08:44


Quote: Originally posted by Kitsune1  
Molybdenum Disulphide can be used as a catalyst in the decomposition of H2O.

Ref? A catalyst causes a reaction to proceed faster or slower (usually faster) then it otherwise would. It can't cause a thermodynamically unfavorable reaction to proceed any more than it's own equilibrium would allow.
H2O(g) → H2 + 1/2 O2 ΔG°= 228.7 kJ/mol. For the opposite of this reaction (formation of water) the Keq = 1.22x10^83 That's an enormous constant, no visible amount of water can decompose without adding the 228.7 kJ/mol, catalyst or not.

[Edited on 24-3-2015 by Molecular Manipulations]




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[*] posted on 24-3-2015 at 12:53


Quote: Originally posted by Molecular Manipulations  
Quote: Originally posted by Kitsune1  
Molybdenum Disulphide can be used as a catalyst in the decomposition of H2O.

Ref? A catalyst causes a reaction to proceed faster or slower (usually faster) then it otherwise would. It can't cause a thermodynamically unfavorable reaction to proceed any more than it's own equilibrium would allow.
H2O(g) → H2 + 1/2 O2 ΔG°= 228.7 kJ/mol. For the opposite of this reaction (formation of water) the Keq = 1.22x10^83 That's an enormous constant, no visible amount of water can decompose without adding the 228.7 kJ/mol, catalyst or not.


The reference,
Gao, M.-R. et al. An efficient ​molybdenum disulfide/cobalt diselenide hybrid catalyst for electrochemical ​hydrogen generation. Nat. Commun. 6:5982 doi: 10.1038/ncomms6982 (2015).
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Molecular Manipulations
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[*] posted on 24-3-2015 at 12:57


And that's what I was expecting, electrochemical ​hydrogen generation. There's the energy source.

[Edited on 24-3-2015 by Molecular Manipulations]




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