Oxidation of ethanol
17
Write a half-equation for the overall oxidation of ethanol into ethanoic acid.
This was the answer:
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
But, i dont know how they got it. Can anyone explain?
This was the answer:
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
But, i dont know how they got it. Can anyone explain?
Reply 1
19
What's the product when you oxidise an alcohol?
Original post by 123chem
Write a half-equation for the overall oxidation of ethanol into ethanoic acid.
This was the answer:
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
But, i dont know how they got it. Can anyone explain?
This was the answer:
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
But, i dont know how they got it. Can anyone explain?
The oxygen of water (boldt print) reacts with ethanol.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
Reply 3
123chem
OP
17
Original post by Kallisto
The oxygen of water (boldt print) reacts with ethanol.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
That does help quite a lot! I've always thought of it in terms of oxidation number, and i couldnt work out these oxidation numbers for them!
Reply 4
123chem
OP
17
Original post by alow
What's the product when you oxidise an alcohol?
aldehyde or an acid?
Original post by 123chem
That does help quite a lot! I've always thought of it in terms of oxidation number, and i couldnt work out these oxidation numbers for them!
Sometimes it is better to have a closer look at the atoms and functional groups to understand it.
to be honest, oxidation numbers are too complicated. Original post by 123chem
aldehyde or an acid?
Aldehyde! whenever alcohol dehydrated, it leads to aldehyde. Here is a mnenomic for you: -dehyde in the name stands for dehydrated.
Reply 6
123chem
OP
17
Original post by Kallisto
Sometimes it is better to have a closer look at the atoms and functional groups to understand it. to be honest, oxidation numbers are too complicated.
Aldehyde! whenever alcohol dehydrated, it leads to aldehyde. Here is a mnenomic for you: -dehyde in the name stands for dehydrated.
to be honest, oxidation numbers are too complicated. Aldehyde! whenever alcohol dehydrated, it leads to aldehyde. Here is a mnenomic for you: -dehyde in the name stands for dehydrated.
But if its under reflux, its an acid, right?
Reply 7
123chem
OP
17
Original post by 123chem
But if its under reflux, its an acid, right?
And how did you know there was water molecule on left side?
Original post by 123chem
But if its under reflux, its an acid, right?
An acid comes into being when an aldehyde oxidises. Example:
CH3CHO (Ethanal) + O -> CH3COOH (ethanoic acid)
So, there are three steps of oxidation in total:
alcohol -> aldehyde -> acid
Reply 9
17
Original post by 123chem
And how did you know there was water molecule on left side?
So start off with
This is not balanced, there are too many oxygens on the right hand side. So we add water to the left hand side to fix this
Now we need to balance the hydrogens on the right hand side so we add the 4 missing hydrogens in the form of H+
The charge needs to be balanced too so add 4 electrons
Reply 10
Original post by Kallisto
The oxygen of water (boldt print) reacts with ethanol.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
At the same time, the hydrogen atoms of water and the hydrogen atoms of the methylene group (CH2) in ethanol are ionized (in bold print above). That is why 4 H+ ions and 4e- come into being.
CH3CH2OH + H2O = CH3COOH + 4H+ + 4e–
i know this is 8 years ago but you have really helped understand this
Original post by roshanxd
i know this is 8 years ago but you have really helped understand this
Erm... thanks! I can't even remember what an answer I have posted.
Good to know that it helped you eight years later.Reply 12
1
I've seen that the 4H+ and 4e- cancel out but if you look at the oxidation states of the carbon in ethanol you get -2 and if you looked at the oxidation state of carbon in ethanoic acid you get 0, which would mean there should be a +2e
Reply 13
Original post by RKJ1
I've seen that the 4H+ and 4e- cancel out but if you look at the oxidation states of the carbon in ethanol you get -2 and if you looked at the oxidation state of carbon in ethanoic acid you get 0, which would mean there should be a +2e
In CH3CH2OH, the methyl carbon is -3 (the adjacent carbon has no effect and the 3 hydrogens each decrease its oxidation state by 1), the methylene carbon is -1 (the adjacent carbon has no effect, the oxygen increases its oxidation state by 1 and each hydrogen decreases its oxidation state by 1). The oxygen is of course -2 and each hydrogen is +1. You can do a quick sanity check to find that the sum of all the oxidation states is 0, e.g (-3) + 3(+1) + (-1) + 2(+1) + (-2) + (+1) = 0.
In CH3COOH, the methyl carbon is also -3 (same rationale as last time), but the methylene carbon is now +3 (the carbon has no effect, the carbonyl oxygen increases the oxidation state by 2 and the other increases it by 1). Both oxygens are -2 again and each hydrogen is +1. As a sanity check, (-3) + 3(+1) + (+3) + 2(-2) + (+1) = 0 and so all the oxidation states deduced work.
So the carbon of interest changes its oxidation state from -1 to +3, which is a loss of 4 electrons. So it is a + 4e^- and not a + 2e^-.
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