Oxidation of ethanol

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123ash
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#1
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#1
Hello,

I am a little ocnfused, this is not homework Qs, but as Im writing up my practical, something has come into my mind. I have tried to use the internet, but teh answers are very confusing.

I'm sorry in advance if this is a silly Qs.

So when we oxidise ethanol into ethanal, we started off with sodium dichromate , to which we added sulphuric acid. We cooled teh solution in ice water until it was <10C. why did we cool it??

Then we added 0.5cm3 of ethanol and continued to monitor the mixture in the ice water, and tehre was an increase in tempereature. Why was this??

I probably should have asked my lecturer but I didnt really think of it then.

The below are homework Qs.

Explain how different quanitities of Na2Cr2O7.H2O and different reaction conditions allow different organic products formed. Density of ethanol = 0.70g cm-3.

I have no idea how to go about answering this Qs, even a lottel guidance to direct me in the right way will help lots.

Why is ethanoic acid higher strength acid than ethanal?

Is this something to do with the hydrogen bonding and/or boiling point?

Thank you
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KeepItUnreal
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#2
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#2
(Original post by 123ash)
Hello,

I am a little ocnfused, this is not homework Qs, but as Im writing up my practical, something has come into my mind. I have tried to use the internet, but teh answers are very confusing.

I'm sorry in advance if this is a silly Qs.

So when we oxidise ethanol into ethanal, we started off with sodium dichromate , to which we added sulphuric acid. We cooled teh solution in ice water until it was <10C. why did we cool it??

Then we added 0.5cm3 of ethanol and continued to monitor the mixture in the ice water, and tehre was an increase in tempereature. Why was this??

I probably should have asked my lecturer but I didnt really think of it then.

The below are homework Qs.

Explain how different quanitities of Na2Cr2O7.H2O and different reaction conditions allow different organic products formed. Density of ethanol = 0.70g cm-3.

I have no idea how to go about answering this Qs, even a lottel guidance to direct me in the right way will help lots.

Why is ethanoic acid higher strength acid than ethanal?

Is this something to do with the hydrogen bonding and/or boiling point?

Thank you
The temperature rise is probably because the reaction is exothermic so energy is released.
In terms of the question on how different quantities affect the organics products formed, if you have a greater concentration of the oxidising agent, ethanol undergoes full oxidation to form ethanoic acid instead of partial oxidation where ethanal is formed. For this one, you'd have to write equation for both reactions, where ethanal and ethanoic acid respectively and then show how different number of moles of the oxidising agents results in different organic products being formed.
Ethanoic acid is a higher strength acid because the extent of its disassociation in solution is greater than ethanal meaning it releases more H+ ions.
I don't know why they cool it but its probably to prevent any reactions occurring until ethanol is added.
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123ash
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#3
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#3
(Original post by KeepItUnreal)
The temperature rise is probably because the reaction is exothermic so energy is released.
In terms of the question on how different quantities affect the organics products formed, if you have a greater concentration of the oxidising agent, ethanol undergoes full oxidation to form ethanoic acid instead of partial oxidation where ethanal is formed. For this one, you'd have to write equation for both reactions, where ethanal and ethanoic acid respectively and then show how different number of moles of the oxidising agents results in different organic products being formed.
Ethanoic acid is a higher strength acid because the extent of its disassociation in solution is greater than ethanal meaning it releases more H+ ions.
I don't know why they cool it but its probably to prevent any reactions occurring until ethanol is added.
Hi,

Thank you for such a speedy reply. I guess suprisingly I was sort of on the right path.

when you say, write out the equations, do you mean:
Partial oxidation: Ethanol + [o] ---> Ethanal + Water
Complete oxidation: Ethanal + 2[o] ---> Ethanoic acid + Water

I dont understand exactly what you mean by the moles, do you mean the fact that ethanoic acid requires 2 [O]?
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Pigster
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#4
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#4
(Original post by 123ash)
Complete oxidation: Ethanal + 2[o] ---> Ethanoic acid + Water
That'll be ethanol rather than ethanal.

The trick is to oxidise the ethanol and separate out the ethanal before it has a chance to be further oxidised. Ethanal has a pretty low boiling point so will boil off even at low temps. If you heat it, the ethanal will likely be further oxidised before it can escape from the reaction mixture.
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123ash
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#5
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(Original post by Pigster)
That'll be ethanol rather than ethanal.

The trick is to oxidise the ethanol and separate out the ethanal before it has a chance to be further oxidised. Ethanal has a pretty low boiling point so will boil off even at low temps. If you heat it, the ethanal will likely be further oxidised before it can escape from the reaction mixture.
Thank you for pointing out the error in regards to ethanal and not ethanol. I did draw teh display formula of ethanol, but wrote ethanal on the paper. Thanks.

What did the previous replier mean in regards to moles?
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123ash
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#6
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Thank you both for your input, its really helped
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Pigster
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(Original post by 123ash)
What did the previous replier mean in regards to moles?
[O] is just shorthand for whatever oxidising agent you're adding, specifically the addition of one O atom to your compound (or 2H removed).

2[O] in this case means the addition of an O and the removal of 2H.

You'd need twice the number of mol of oxidising agent for the oxidation to EtOOH as you would to EtHO.
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123ash
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#8
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(Original post by Pigster)
[O] is just shorthand for whatever oxidising agent you're adding, specifically the addition of one O atom to your compound (or 2H removed).

2[O] in this case means the addition of an O and the removal of 2H.

You'd need twice the number of mol of oxidising agent for the oxidation to EtOOH as you would to EtHO.
Thank you so much again. Wish the tutur could explain like you. Ever thought of giving private tution? You're very good at explaining things.
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Pigster
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(Original post by 123ash)
Thank you so much again. Wish the tutur could explain like you. Ever thought of giving private tution? You're very good at explaining things.
After teaching A level chemistry for more than a decade, one would hope that I could explain this sort of thing.
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123ash
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(Original post by Pigster)
After teaching A level chemistry for more than a decade, one would hope that I could explain this sort of thing.
Well some are definitely better than others at explaining.

Can I ask why the ethanol density was given? if the answer is as simple as ethanol needing 2 moles of sodium dichromate to produce ethanoic acid. Whereas it requires 1 mole to produce ethanal?

I feel that I need to be doing something with the density?

Thank you.
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Pigster
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You're given a volume of pure EtOH, rather than the mass. How did you plkan to calculate the amount of EtOH?
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123ash
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#12
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(Original post by Pigster)
You're given a volume of pure EtOH, rather than the mass. How did you plkan to calculate the amount of EtOH?
Mass = Volume x Density

1.0cm3 x 0.76g = 0.76g which is the mass

Moles = Mass / Molar Mass

0.76/46 = 0.01652 moles of ethanol
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Pigster
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(Original post by 123ash)
Mass = Volume x Density
See... you found a use for the density.
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123ash
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#14
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(Original post by Pigster)
See... you found a use for the density.
Thank, but what do I do next, Im confused.

Do I now work out the moles of Na2Cr2O7.2H2O

3.986/297.9981 = 0.01338 moles of Na2Cr2O7.2H2O

which meas that Na2Cr2O7.2H2O is the limiting reagent, whoch I was already told.

But what do I do next, whats the next equation? Does that answer the Qs?

Sorry my understanding is very weak, this Qs has really got me thinking, i dont know what im doing.
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Pigster
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#15
(Original post by 123ash)
3.986/297.9981 = 0.01338 moles of Na2Cr2O7.2H2O
This is the first time you've mentioned 3.986. How are people supposed to help you if you are drip-feeding us with information?

In order to get the molar ratio between EtOH and Cr2O72-, you will either need to know how to do redox half and overall equations, or just google it.

Once you find the molar ratio (it is something horrible like 4:7), then should be a simple titration calc, followed by mass calc. But then again, there may be even more info you haven't told us.
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123ash
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(Original post by Pigster)
This is the first time you've mentioned 3.986. How are people supposed to help you if you are drip-feeding us with information?

In order to get the molar ratio between EtOH and Cr2O72-, you will either need to know how to do redox half and overall equations, or just google it.

Once you find the molar ratio (it is something horrible like 4:7), then should be a simple titration calc, followed by mass calc. But then again, there may be even more info you haven't told us.

I am sorry, It was not intensional. I dont really know what im doing myself.

Thank you for your help, really appreciated.
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Bree_Dav
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#17
(Original post by Pigster)
This is the first time you've mentioned 3.986. How are people supposed to help you if you are drip-feeding us with information?

In order to get the molar ratio between EtOH and Cr2O72-, you will either need to know how to do redox half and overall equations, or just google it.

Once you find the molar ratio (it is something horrible like 4:7), then should be a simple titration calc, followed by mass calc. But then again, there may be even more info you haven't told us.
Hiya,

Firstly thanks for all the help, this chain has helped me a lot and the explanations have helped a lot.

But I have a quick question.

It's with regards to the question "explain how the different quantities of Na2Cr2O7.2H2O and different reaction conditions allow different products to be formed."

I've used the density of ethanol - given in question - to calculate the moles of ethanol. I have also worked out the moles of Na2Cr2O7.2H2O based off the quantities and calculations from when I did the experiment.

With regards to the different quantities of Na2Cr2O7.2H2O I've spoken about limiting reactant.

However, I'm stuck at the point where we elaborate on different reaction conditions. What conditions do I need to talk about or mention?
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