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Author Topic: VCE Chemistry Question Thread  (Read 2313577 times)  Share 

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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #1020 on: June 15, 2014, 09:58:24 pm »
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It's not that they're never reduced - you need to know some of the common oxidation pathways of alcohols as a part of the area of study 2 in unit 3. That's one of them. Hate to say it Bestie, but you need to revise/read over the study design.

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1021 on: June 16, 2014, 01:19:06 am »
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so alcohols are never reduced?

Well......I wouldn't say "never" reduced. Look up reactions out there that can convert alcohols to alkanes.
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #1022 on: June 16, 2014, 11:46:07 pm »
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when drawing the structure of an ester should the alcohol (alkyl) part go on the left or right and what about the alkyanoate part?
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #1023 on: June 17, 2014, 12:06:31 am »
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It honestly doesn't matter - if it helps you remember to do it one way, do it that way. But there is no correct way of drawing esters as to which side you should the alkyl and which you should put the alkyanoate (which did mean a bunch of strife for me last year, let me tell you).

hobbitle

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Re: VCE Chemistry Question Thread
« Reply #1024 on: June 17, 2014, 11:17:57 am »
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Hi guys.
In general, I'm quite confused by some aspects of electrolysis. 
In galvanic cells, using an ordered table of standard reduction potentials, we know that the strongest oxidant lies on the top left and the strongest reductant lies on the bottom right, so spontaneous reactions occur ‘along that diagonal’.  But in electrolysis, the reaction is not spontaneous, so is this principle switched?  Or do the reactions happen along the opposite diagonal (top right with bottom left)??
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hobbitle

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Re: VCE Chemistry Question Thread
« Reply #1025 on: June 17, 2014, 11:20:43 am »
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Attached are two slides from our lectures (Fundamentals of Chemistry).

I can do the balancing/cancelling and adding together part, but I don’t know how to ‘get’ the initial half-equations from the worded answer.  What happens to the K?  And the O?  How do we know which half-reactions to kick off with?
Basically there is a missing link from my brain between reading the question and saying "Oh ok, I need to use Cr2O72- --> Cr3+ to kick off this question".

Can anyone help me with the thought process here??
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1026 on: June 17, 2014, 11:46:26 am »
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Hi guys.
In general, I'm quite confused by some aspects of electrolysis. 
In galvanic cells, using an ordered table of standard reduction potentials, we know that the strongest oxidant lies on the top left and the strongest reductant lies on the bottom right, so spontaneous reactions occur ‘along that diagonal’.  But in electrolysis, the reaction is not spontaneous, so is this principle switched?  Or do the reactions happen along the opposite diagonal (top right with bottom left)??

Think of it this way. There is a quantity called the free energy that strives to reach a minimum at equilibrium. Free energy behaves a bit like potential energy in this regard. Now, a galvanic cell reaction is like having a ball roll down from the top of the hill; it's spontaneous as the energy decreases. An electrolytic cell reaction, however, is like having someone physically push the ball up the hill. Chemically, by applying a voltage, you're forcing the electrons to reduce a compound.

If you don't do the electrolysis properly and your products are in contact with each other, you're right; the top left bottom right principle will apply and the two products will react spontaneously. As an example, if you electrolyse molten sodium chloride, you'll get sodium metal and chlorine gas. The reason why they don't react with each other is they're physically separated from each other. If you let them make contact, they WILL react. Does this sort of help explain your queries?

Attached are two slides from our lectures (Fundamentals of Chemistry).

I can do the balancing/cancelling and adding together part, but I don’t know how to ‘get’ the initial half-equations from the worded answer.  What happens to the K?  And the O?  How do we know which half-reactions to kick off with?
Basically there is a missing link from my brain between reading the question and saying "Oh ok, I need to use Cr2O72- --> Cr3+ to kick off this question".

Can anyone help me with the thought process here??

OK. The potassium ions aren't affected at all. That's why you don't care about those.
I don't get your problem about the oxygen though.

You need to know 'dichromate => chromium(III)' as dichromate is what you start off with and chromium(III) ions are the end product. You just have to know that unfortunately. Just like you just have to know sodium metal is oxidised to Na+.
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hobbitle

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Re: VCE Chemistry Question Thread
« Reply #1027 on: June 17, 2014, 12:01:45 pm »
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Think of it this way. There is a quantity called the free energy that strives to reach a minimum at equilibrium. Free energy behaves a bit like potential energy in this regard. Now, a galvanic cell reaction is like having a ball roll down from the top of the hill; it's spontaneous as the energy decreases. An electrolytic cell reaction, however, is like having someone physically push the ball up the hill. Chemically, by applying a voltage, you're forcing the electrons to reduce a compound.

Yep this makes sense (we haven't covered free energy but yeah)!

Quote
If you don't do the electrolysis properly and your products are in contact with each other, you're right; the top left bottom right principle will apply and the two products will react spontaneously. As an example, if you electrolyse molten sodium chloride, you'll get sodium metal and chlorine gas. The reason why they don't react with each other is they're physically separated from each other. If you let them make contact, they WILL react. Does this sort of help explain your queries?

Hmm yes I get it but that's not exactly what I was trying to ask (it's difficult to articulate what I'm finding confusing, sorry!).  I think there's just something I'm not thinking about in the right way.  Let me have a think about it and see if I can find some specific questions.

Quote
You need to know 'dichromate => chromium(III)' as dichromate is what you start off with and chromium(III) ions are the end product. You just have to know that unfortunately. Just like you just have to know sodium metal is oxidised to Na+.

Thanks, that explains that one!  :) Haha.  One of the weirdest things about chem is knowing what to 'rote learn' and what should be intuitive.
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1028 on: June 17, 2014, 01:45:33 pm »
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Chem is, after all, based on experiment
By definition you have to rote learn some parts of it
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #1029 on: June 17, 2014, 07:47:28 pm »
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when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?

how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid
« Last Edit: June 17, 2014, 08:00:48 pm by Jason12 »
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Re: VCE Chemistry Question Thread
« Reply #1030 on: June 17, 2014, 08:02:00 pm »
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when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?
UV light provides energy to break the Cl-Cl bond to create free radicals (unpaired electrons). These radicals are very reactive and able to attack the C-H bond.
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Re: VCE Chemistry Question Thread
« Reply #1031 on: June 17, 2014, 08:07:42 pm »
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when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?

how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid
If you are provided with the formula, try to work out from there. For linoleic fatty acids which is C17H31COOH. you can try to use DOB formula (which I'm too lazy to use :( ) or just replace the COOH with a Hydrogen, hence, you have C17H32, which is polyunsaturated (2 double bonds).
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #1032 on: June 17, 2014, 10:08:59 pm »
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how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid

A saturated fatty acid has the formula

every double bond is 2H less than the saturated fatty acid

So for example, if we have
the saturated version of that would be   <--- 35 came from doubling 17, and adding 1.
Since its 31H rather than 35H, there are 2 double bonds.

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Re: VCE Chemistry Question Thread
« Reply #1033 on: June 18, 2014, 11:58:38 am »
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Okay lznxl and others... Here's a question (attached) on electrolysis that kind of outlines where I'm finding things confusing. Part b) is where I'm stuck. 

I've also attached the answer for part b.

I thought reduction was happening at A and C, and oxidation happening at B and D but the answer indicates otherwise.

Just focussing on A and B, I wrote that the species in the beaker were Al3+, SO4, and H2O.
Then I wrote what I thought was the four relevant half reactions and their Eo voltage value:

O2 + 4H+ + 4e- <--> 2H2O (Eo = +1.23)
SO4 + 4H+ + 2e- <--> H2SO4 + H2O (Eo = +0.17)
2H2O + 2e- <--> H2 + 2OH- (Eo = -0.83)
Al3+ + 3e- <--> Al (Eo = -1.66)

But I don't know if these are the right half reactions... And how they use them to get the answer... Which two of the four are the ones that occur and why???

Question:

Answer:
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hobbitle

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Re: VCE Chemistry Question Thread
« Reply #1034 on: June 18, 2014, 03:44:42 pm »
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The attached picture has been named in our notes as 3-ethyl-2methylhexane. Makes sense but couldn't it also be 3-isopropylhexane? There are two ways to get the six carbon backbone...
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