mfilip62
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Nightmares with REDOX
One reaction is driving me nuts;
H2O2 + KMnO4 + KOH
What the hell do I get!!!
I failed on exam becouse of this.
By RedOx rules you get MnO4(2-) ion in strong alkaline solution,which I suppose this is,in weak alkaline and neutral media you get MnO2 and in acid
media Mn(2+) ion.
Only reaction i can get anything near equation is with MnO2...
Is there way to equate it with MnO4(2-)!?!?!?
Process and balanced molecular formula would be much appriciated!
Any if anybody knows exuation of etanol / K2Cr2O7
that will be appriciated too!
Thanks
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psychokinetic
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My redox isn't the greatest either, but isn't KOH a product of H2O2 + KMnO4? (Along with MnO2 and O2).
Therefore I presume you'd get MnO2, O2, and some more KOH.
“If Edison had a needle to find in a haystack, he would proceed at once with the diligence of the bee to examine straw after straw until he found
the object of his search.
I was a sorry witness of such doings, knowing that a little theory and calculation would have saved him ninety per cent of his labor.”
-Tesla
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woelen
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Try to find half reactions for the oxidizers and reductors. Oxygen atoms must be balanced with water and either acid or base to make the total balance
correct.
In KMnO4 it is the permanganate ion which is reduced:
MnO4(-) is converted to MnO2 and the solution becomes alkaline. Mn goes from oxidation state +7 to +4, so 3 electrons are consumed.
MnO4(-) + 3e --> MnO2 + 2[O(2-)]
I write the oxygens between [ ] because these do not form in this way. Compensate for this with water. Add water at the left and the right and use H2O
+ [O(2-)] --> 2OH(-)
(1): MnO4(-) + 3e + 2H2O --> MnO2 + 2H2O + 2[O(2-)] --> MnO2 + 4OH(-)
Hydrogen peroxide is reduced, one of the products being oxygen gas and hydroxide ion is consumed in this process.
H2O2 --> 2[H(+)] + O2 + 2e
Here I write H(+) between [ ] because this entity does not appear in this reaction. This can easily be compensator for by adding hydroxide ions at the
left and right:
(2): 2OH(-) + H2O2 --> 2OH(-) + 2[H(+)] + O2 + 2e --> 2H2O + O2 + 2e
Now you have two balanced half-reactions. Now you have to combine these to get the total reaction. Number of electrons produced must be same as number
of electrons consumed (left and right sides must have same number of electrons). You can achieve this by taking 2 times reaction (1) and 3 times
reaction (2) and adding them left and right of the arrow. After adding, you can simplify if both at the left and right hydroxide ion and/or water
appears. I leave this as an exercise for you.
-----------------------------------------------------------------------------------------------------
Now the second one. Dichromate is an oxidizer which works in acidic environment. Acid is consumed in this reaction. Cr goes from +6 to +3 and Cr(3+)
ions are formed. We have two Cr-atoms in dichromate, so 6 electrons are consumed per dichromate ion.
Cr2O7(2-) + 6e --> 2Cr(3+) + 7[O(2-)]
We know this reaction consumed acid, so we have to compensate for the O(2-) as follows: 2H(+) + [O(2-)] --> H2O. We have 7 of these beasts at the
right, so we need to add 14H(+) to compensate for these:
Cr2O7(2-) + 6e + 14H(+) --> 2Cr(3+) + 14H(+) + 7[O(2-)] --> 2Cr(3+) + 7H2O
Voila, this is the half-reaction for reduction of dichromate in acidic solution.
Ethanol is a somewhat more difficult thing. Ethanol can be oxidized to ethanal or further down to acetic acid. Let's assume that it is oxidized to
ethanal. We have to determine the halfreaction for that. For oxidation of organics in aqueous media a good method of deriving half-reactions is to
introduce the formal species [O] and use that as oxidizer.
[O] + CH3CH2OH --> CH3C(O)H + H2O
The species [O] can be though of as being made from water, releasing two electrons, which can be consumed by the oxidizer:
H2O --> 2H(+) + [O] + 2e
Note that this is all formal, it is just a tool for calculating things!
So, we have
H2O + CH3CH2OH --> 2H(+) + [O] + 2e + CH3CH2OH --> 2H(+) + 2e + CH3C(O)H + H2O
In this particular example we happen to have H2O at both sides (but this is not always the case), so we simplify
CH3CH2OH --> CH3C(O)H + 2H(+) + 2e
Now for writing the total equation we need 3 times the last one plus 1 time the eqation for dichromate such that both sides of the arrow have 6
electrons. Then simplify by taking away H(+) from left and right such that it only remains on one side of the arrow.
Now for you an exercise: Suppose the ethanol is oxidized to CH3COOH instead of CH3C(O)H, how many [O] are needed for oxidizing this and if you have
answered the question then derive the falf-reaction for oxidation of CH3CH2OH to CH3COOH.
I also would like to see you write out the complete reactions. Put some effort in it, do the math yourself, that is the only way to get used to this
kind of things.
Yet another exercise. Try to find the half reaction for reduction of colorless orthovanadate VO4(3-) to brown/red hypovanadate V4O9(2-) in strongly
alkaline solution. Vanadium acts as oxidizer. What is the change of oxidation state of vanadium? When you answered this question, then you can find
out how many electrons are consumed and then you can do the other math like I did for permanganate (remember, here we are talking about alkaline
environment and you have to work with hydroxide ions).
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mfilip62
pierced by a crossbow under a bridge while eating Billy goats
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I appreciate your help very much Woelen,thanks!
Only possible solution I got was:
2 KOH + 2KMnO4 + 3H2O2 -->
2 MnO2 + 2 H2O + 3O2+ 4 KOH
Problem is that 2 OH(-) comes from the H2O2,which makes this equation false,because that is one whole other half-reaction!
Another issue here is MnO2.
Professor told us that MnO2 (and half-reaction you wrote)
happens in neutral to mildly alkaline solution.
An I am sure that KOH is VERY strong lie.
PLEASE can you make WHOLE BALANCED equation!!!???
I need it tomorrow on exam!
Same thing for ethanol oxidized to CH3COOH in reaction with
Dichromate.
I know half-reactions for chromate and dichromate,but organic stuff is what bothers me!
I will try to solve all examples later on my own,but I need the urgently right NOW so I can memorize them for tomorrow.
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woelen
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Your equation is correct, but it could be simplified further. You can subtract 2KOH from both sides. But this equation is exactly what I would expect.
The hydroxide is not coming from the H2O2, but from H2O. What you see here is only the NET equation. In reality the process goes through all kinds of
intermediate steps in which water is involved.
The half reaction I wrote also happens at high pH. At high pH another half reaction is possible in which dark green soluble MnO4(2-) is formed, but in
the presence of H2O2 the MnO4(2-) will not form, only MnO2 is formed.
-------------------------------------------------------------
I do not spoonfeed complete reactions, because memorizing a hard to remember balanced equation makes no sense at all, tomorrow they will ask something
which just is somewhat different (e.g. propanol instead of ethanol, or methanol oxidized to CO2 instead of ethanol). So, you really have to teach
yourself how to completely balance reactions, when the half reactions are given.
I will help you with some half reactions:
ethanol to acetic acid: CH3CH2OH + H2O --> CH3COOH + 4H(+) + 4e
ethanol to ethanal: CH3CH2OH --> CH3C(O)H + 2H(+) + 2e
ethanal to acetic acid: CH3C(O)H + H2O --> CH3COOH + 2H(+) + 2e
For propanol/propanal just keep the same equations, but replace the left CH3-group with a CH3CH2-group.
The reactions, shown above, are in acidic media with excess acid (e.g. from sulphuric acid). Dichromate oxidations always are in excess acidic
solution, such as dilute sulphuric acid. Without the dilute sulphuric acid the dichromate oxidation hardly proceeds, if at all.
I hope that this helps you somewhat with your exam! Success!
[Edited on 23-2-10 by woelen]
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mfilip62
pierced by a crossbow under a bridge while eating Billy goats
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OK,is this correct;
for ethanal
4 H2SO4 + 3 CH3CH2OH + K2Cr2O7 = 3 CH3CHO + 2 KCr(SO4)2 + 7H2O +3H2SO4
or more widely
Cr2O7(2-) + 3 C2H5OH + 14H(+) = 2 Cr(3+) + 7H2O + 3 CH3COH
and for acetic cid
6H2O + 6 C2H5OH + 4 Cr2O7(2-) + 32H(+) = 6 CH3COOH + 8 Cr(3+) + 28 H2O
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woelen
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The redox part of all your equations is exactly as it should be. The only problem is the compensation for the H(+) and H2O.
In the first equation you have the 3H2SO4 at the right. Omit these and it is OK, it should be
4 H2SO4 + 3 CH3CH2OH + K2Cr2O7 = 3 CH3CHO + 2 KCr(SO4)2 + 7H2O
If you have this, then deriving the second equation is very simpe. Just write the acid and salts in the form of ions and remove any spectator ions
from the equation (sulfate, potassium) then you get the equation you wrote, except with 8H(+) instead of 14H(+). This 8 H(+) is so, because the half
reaction for ethanol to ethanal yields 2 H(+) at the right for a single molecule. You have 3 of them, so you will have production of 6H(+) at the
right. You have consumption of 14H(+) at the left and the net effect of this is consumption of 8H(+) at the left (14 - 6).
The last equation is entirely correct. You only have to simplify it somewhat. At the left and at the right you have H2O. Remove 6 H2O from both sides.
Another simplification is possible. All coefficients have a common factor 2. Divide the entire equation by 2. When you do both of these steps then you
get the following:
removal of 6H2O from left and right: 6 C2H5OH + 4 Cr2O7(2-) + 32H(+) = 6 CH3COOH + 8 Cr(3+) + 22 H2O
division by 2 of both sides: 3 C2H5OH + 2 Cr2O7(2-) + 16H(+) = 3 CH3COOH + 4 Cr(3+) + 11 H2O
Based on your answers I conclude that you understand things quite well. Just do the final math a little bit more precise and try to make the equations
as simple as possible (if a compound or ion appears on both sides, then you can simplify, if the coefficients have a common factor larger than 1, you
can also simplify).
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JohnWW
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That reaction of ethanol with dichromate is the working principle of "breathalyzers", as used by the Traffic Pigs to detect drunken drivers. A certain
amount of ethanol vapor in the average lung capacity of an human is required to reduce the orange crystals of dichromate in a packed tube to green
Cr(III) acetate down to a certain depth, beyond which depth any color change indicates a "failed" breath test. (And then the unlucky driver is
required to accompany the Pigs to their station for more accurate evidential breath and blood tests, being charged if those are also over the alcohol
limit.).
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mfilip62
pierced by a crossbow under a bridge while eating Billy goats
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And what you get by mixing
PbO2 + CH3COOH + KI + Na(CH3COO)2
I need aquation asap,no teaching,just "spoon-feeding"
this stupid equation is just part of one very big math problem,and I am out of time.
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