GrayGhost
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Manganese Acetate
Hello. Reading old book in the net over preparation manganese acetate (III) from Manganese acetate (II), acetic acid hot, and chlorine.
Someone have idea of mechanisms?????Thank
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blogfast25
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Go to library of this forum. Find 'Preparative Inorganic Chemistry' by Gregor Brauer. It has a procedure for Mn(III) acetate.
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GrayGhost
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Quote: Originally posted by blogfast25  | | Go to library of this forum. Find 'Preparative Inorganic Chemistry' by Gregor Brauer. It has a procedure for Mn(III) acetate. |
Yes Sir, I read but the procedure is not with Acetic and chlorine.
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blogfast25
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What was your original source? Describe the procedure PROPERLY. We are not mind readers or astrologists...
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GrayGhost
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The information is vague.
https://books.google.com.ar/books?id=Owuv-c9L_IMC&pg=PA6...
Only say manganese acetate is preparate by manganese permanganate or Chlorine in acetic acid hot.
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blogfast25
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Books like that are fairly lowly authorative.
I doubt if chlorine could oxidise Mn(II) to Mn(III) but I could be wrong on that.
Look up the Standard Reduction Potentials.
In water, Mn(III) will oxidise chloride to chlorine, first hand experience BTW. That strongly suggest the opposite
(what you want to do) is thermodynamically unfavourable.
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GrayGhost
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| Quote: Originally posted by blogfast25 | Books like that are fairly lowly authorative.
I doubt if chlorine could oxidise Mn(II) to Mn(III) but I could be wrong on that.
Look up the Standard Reduction Potentials.
In water, Mn(III) will oxidise chloride to chlorine, first hand experience BTW. That strongly suggest the opposite
(what you want to do) is thermodynamically unfavourable. |
That is the point, my dude. The book is wrong? Chlorine oxidise Mn (II) to Mn(III)? Apparently no.
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DraconicAcid
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Standard reduction potentials only apply in aqueous solution. They may differ in glacial acetic acid.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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blogfast25
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The only way I could possibly see that happen (Mn(II) to Mn(III) by Cl(0)) is if Mn(III) acetate was insoluble in the solvent. Removal by
precipitation could drive the equilibrium to the right (Delta G is changed).
[Edited on 28-5-2015 by blogfast25]
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GrayGhost
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Apparently the Mn(III) acetate should be soluble in acetic, but the chlorinated product of reaction if not is soluble , then equilibrium drive to the
right?
[Edited on 28-5-2015 by GrayGhost]
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blogfast25
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I didn't imply otherwise. You are correct.
The main difference must come from solvation enthalpies.
And if one reaction product is insoluble it changes the ΔG, so that could make the reaction thermodynamically favourable.
Classic example: oxidation of Fe(II) to Fe(III) is much 'easier' in neutral/alkaline than acid conditions because Fe(OH)<sub>3</sub> is so
insoluble.
[Edited on 28-5-2015 by blogfast25]
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blogfast25
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GrayGhost:
Can you describe your attempt in some detail?
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blogfast25
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| Quote: Originally posted by GrayGhost |
Apparently the Mn(III) acetate should be soluble in acetic, but the chlorinated product of reaction if not is soluble , then equilibrium drive to the
right?
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That would suggest the insoluble reaction product is something other than Mn(III) acetate.
[Edited on 28-5-2015 by blogfast25]
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DraconicAcid
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| Quote: Originally posted by blogfast25 |
Classic example: oxidation of Fe(II) to Fe(III) is much 'easier' in neutral/alkaline than acid conditions because Fe(OH)<sub>3</sub> is so
insoluble. |
An even better example- Cu(II) + Cu -> 2 Cu(I) is spontaneous in acetonitrile. The reverse is spontaneous in aqueous solution.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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GrayGhost
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I thank you your valuable answers guys.
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blogfast25
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| Quote: Originally posted by DraconicAcid |
An even better example- Cu(II) + Cu -> 2 Cu(I) is spontaneous in acetonitrile. The reverse is spontaneous in aqueous solution.
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Errmm. That works also in water, if there's chloride present:
Cu<sup>2+</sup>(aq) + Cu(s) + 2 Cl<sup>-</sup>(aq) === > 2 CuCl(s). Same reason though: Cu(I) being removed as solid
precipitate. A fairly standard way to prepare wet CuCl.
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