TheNaKLaB
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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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theflickkk
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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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blogfast25
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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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Endimion17
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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
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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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blogfast25
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| 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.
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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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Endimion17
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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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woelen
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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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TheNaKLaB
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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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blogfast25
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Which part of:
...didn't you undertstand?
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Endimion17
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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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blogfast25
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I'll let you off!
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ScienceSquirrel
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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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blogfast25
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| 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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ScienceSquirrel
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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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woelen
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Thread Pruned
27-11-2011 at 23:58 |
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