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Author: Subject: Decomposition of Potassium Permanganate with Alcohol
TheNaKLaB
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[*] posted on 19-11-2011 at 16:33
Decomposition of Potassium Permanganate with Alcohol


I was reading up about potassium permanganate and I found out that it decomposes in the presence of an alcohol. I conducted this experiment in a test tube and I found out that It did decompose the permanganate.
I used ethanol as the alcohol.

After about 10 minutes, I noticed that there was a brown precipitate at the bottom of the test tube, I'm guessing that this would be a manganese oxide of some sort.

I was wondering what the over all equation for this reaction would be?




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[*] posted on 19-11-2011 at 19:43


The brown precipitate you got is probably MnO2. If the solution was acidified prior to the reaction, one of the end products would have instead been Mn2+.

KMnO4 + C2H5O -->
MnO2+ C2H3COOK + H2O

I'm guessing that the eqn would look something like this
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[*] posted on 20-11-2011 at 06:25


Balancing redox reactions (in watery media):

Write out the unbalanced half-reactions first:

Reduction:
KMnO4 ===> K+ + MnO2

Oxidation:
C2H5OH === > CH3COOH

Balancing the reduction reaction:

Add water to the side that’s oxygen deficient:

KMnO4 === > K+ + MnO2 + 2 H2O

Now add H+ to the side that’s hydrogen deficient:

KMnO4 + 4 H+ === > K+ + MnO2 + 2 H2O

Add electrons to balance the charges:

KMnO4 + 4 H+ + 3e- ==== > K+ + MnO2 + 2 H2O … (R.1)

Same for the oxidation reaction:
C2H5OH + H2O === > CH3COOH

C2H5OH + H2O === > CH3COOH + 4 H+

C2H5OH + H2O === > CH3COOH + 4 H+ + 4 e- … (R.2)

To balance the overall redox reaction, balance the electrons of R.1 and R.2: multiply R.1 by 4 and R.2 by 3 and add up:

4 KMnO4 + 16 H+ + 12 e- === > 4 K+ + 4 MnO2 + 8 H2O
3 C2H5OH + 3 H2O === > 3 CH3COOH + 12 H+ + 12 e-

Eliminate excess H+ and H2O and e-:

4 KMnO4 + 3 C2H5OH + 4 H+ === > 4 K+ + 4 MnO2 + 3 CH3COOH + 5 H2O

Which you could re-write as:

4 KMnO4 + 3 C2H5OH + H+ === > K+ + 4 MnO2 + 3 CH3COOK + 5 H2O

And assuming one uses H2SO4 as the acid:

4 KMnO4 + 3 C2H5OH + ½ H2SO4 === > ½ K2SO4 + 4 MnO2 + 3 CH3COOK + 5 H2O

Homework: ;)

Balance the same type of redox reaction using potassium dichromate as oxidiser:

K2Cr2O7 === > 2 K+ + 2 Cr3+


[Edited on 20-11-2011 by blogfast25]




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[*] posted on 20-11-2011 at 08:57


Quote: Originally posted by theflickkk  
The brown precipitate you got is probably MnO2. If the solution was acidified prior to the reaction, one of the end products would have instead been Mn2+.

KMnO4 + C2H5O -->
MnO2+ C2H3COOK + H2O

I'm guessing that the eqn would look something like this


That's elementary school chemistry and should not be encouraged because it's basic and does not reflect the true nature of the reaction.

You don't get potassium acetate. Potassium cation does nothing and can be left out. These reactions (redox) are ought to be written down in ionic form, unless there's a precipitate such as manganese(IV) oxide.


This is a classic example of alcohol oxidation in organic chemistry. It actually proceeds over the aldehyde, but it goes quickly to acetic acid.




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[*] posted on 20-11-2011 at 11:33


Quote: Originally posted by Endimion17  
These reactions (redox) are ought to be written down in ionic form, unless there's a precipitate such as manganese(IV) oxide.




No, they don't for stoichiometric reasons. Or are you capable of weighing x mol of MnO4- anions? :(

Writing the equations with cations also stresses the reactions are electrically neutral.

There is nothing wrong with writing them that way AT ALL.




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[*] posted on 20-11-2011 at 13:04


But he wants to know what is happening in the reaction mixture, not amounts of reagents needed. Writing a complete stoichiometric equation is then useless because you won't get potassium acetate. There is no such species in the mixture.



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[*] posted on 20-11-2011 at 13:17


Both approaches are valid. You can use so-called half-reactions and combine these in a complete reaction, specified in terms of ions. Such reaction equations indeed tell more about what actually happens. But the complete neutral reactant equations, with ions like K(+) incorporated in the equation are useful as well. Although these are just spectator ions, they still have to go somewhere and in this particular reaction I indeed think that the only place where these ions can go is in potassium acetate and potassium (bi)sulfate. In a real mix, when it would be boiled down, you indeed would get a mix of potassium acetate, potassium (bi)sulfate and MnO2.



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[*] posted on 20-11-2011 at 13:35


Thank you guys or all your help! Just to clarify, there was no acid added to the mix. There was evolution of a gas (probably oxygen)

This information if on a wiki - Potassium Permanganate article:

Under acidic conditions, the alkene double bond is cleaved to give the appropriate carboxylic acid:
CH3(CH2)17CH=CH2 + 2 KMnO4 + 3 H2SO4 → CH3(CH2)17COOH + CO2 + 4 H2O + K2SO4 + 2 MnSO4

Potassium permanganate oxidizes aldehydes to carboxylic acids, such as the conversion of n-heptanal to heptanoic acid:
5 C6H13CHO + 2 KMnO4 + 3 H2SO4 → 5 C6H13COOH + 3 H2O + K2SO4 + 2 MnSO4

When solid KMnO4 is mixed with pure glycerol or other simple alcohols it will result in a violent combustion reaction.
3 C3H5(OH)3 + 14 KMnO4 → 14 MnO2 + 14 KOH + 9 CO2 + 5 H2O

[Edited on 19-09-2011 by TheNaKLaB]




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[*] posted on 20-11-2011 at 14:06


Quote: Originally posted by Endimion17  
But he wants to know what is happening in the reaction mixture, not amounts of reagents needed.


Which part of:

Quote: Originally posted by TheNaKLaB  
I was wondering what the over all equation for this reaction would be?


...didn't you undertstand?




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[*] posted on 21-11-2011 at 04:36


I seem to have slipped on the "over all". If there was "overall" I'd understand immediately. Therefore I've just read it too quickly and frankly, ignored it. Huh, it happens.

[Edited on 21-11-2011 by Endimion17]




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[*] posted on 21-11-2011 at 05:52


I'll let you off! ;)



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[*] posted on 21-11-2011 at 07:10


The reaction products will depend on the conditions.
Cold neutral permanganate will produce manganese dioxide and acetate, particularly if it is in excess. Acid permanganate will go all the way to manganese II. If the reaction mixture is above the boiling point of acetaldehyde then that will be favoured as it will evaporate before it can be further oxidised to acetic acid.
I feel that equations are a bit notional for a lot of organic reactions as few of them are quantitative and I feel that they mislead the less experienced chemist as to the reality of the situation.

[Edited on 21-11-2011 by ScienceSquirrel]
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[*] posted on 21-11-2011 at 08:56


Quote: Originally posted by ScienceSquirrel  
Cold neutral permanganate will produce manganese dioxide and acetate, particularly if it is in excess. Acid permanganate will go all the way to manganese II. [Edited on 21-11-2011 by ScienceSquirrel]


Well, that’s NOT my experimental experience. In acid conditions I did fully expect Mn [+II] to form but I got MnO2. The reaction ran quite hot but I didn’t get a whiff of acetaldehyde but very clearly of acetic acid (unmistakably so).

Maybe there’s a clue in the following. Oxalic acid is recommended as a primary standard for K permanganate titrant solutions (oxidising the oxalic acid to CO2 and reducing the permanganate to Mn [+II]). It’s recommended to heat the oxalic acid standard solution in the conical flask to about 80 C to speed up the oxidation but also to titrate fairly slowly to avoid formation of MnO2. I’ve done this several times w/o problems.

It could indicate that MnO2 is favoured in conditions of more concentrated permanganate. My experiments with KMnO4/alcohol oxidation used quite concentrated solutions…




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[*] posted on 21-11-2011 at 09:12


I suspect that temperature and the concentration of acid and permanganate play a big role.
Manganese dioxide and acetic acid are the preferred products under a lot of conditions.
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