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November 01, 2025, 11:41:04 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448755 times)  Share 

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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1560 on: October 26, 2013, 04:05:12 pm »
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I think you'll find that generally, higher nucleon count is associated with elements with more protons and thus more electrons. The exceptions are few. It is true that later elements gain lots of neutrons, but by saying "higher molecular mass => more electrons", all I am doing is stating a correlation between molecular mass and electron count, not the exact relationship.
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Re: Chemistry 3/4 2013 Thread
« Reply #1561 on: October 26, 2013, 04:30:04 pm »
+2
What's the original question?

I think you misinterpreted the 75% efficient.
Let's say that you need 10000 moles of electrons. If it's 75% efficient, then 10000 moles of electrons IS 75% of the total amount of electricity. You divide by 0.75

yup thats it nliu. You are so wise haha.
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Holmes

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Re: Chemistry 3/4 2013 Thread
« Reply #1562 on: October 27, 2013, 11:05:57 am »
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When we're writing the oxidation of a primary alcohol to convert it into a carboxylic acid, the top of the reaction arrow will have a strong oxidant. At the bottom of the arrow is usually a H+. Do we need to write the H+ there? Because I'm not sure I understand the implications of having the oxidation take place in an acidic environment or under alkaline conditions.

Also, sorry for an annoying question, but are (non-/)renewable energy resources, such as coal and nuclear fission etc. still examinable as they were previously?
« Last Edit: October 27, 2013, 11:29:22 am by Holmes »

ahat

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Re: Chemistry 3/4 2013 Thread
« Reply #1563 on: October 27, 2013, 11:31:54 am »
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Hey,

I've always thought that in an electrolytic cell, it was the species with the lowest voltage requirement that would be preferentially discharged at an electrode.
b) I wrote the equation for Cu2+ as it is +0.34 V as the species that would react first.
c) I wrote Ag+ because it requires more voltage, +0.80V. The answers suggest that Ag+ will be the first to react because at +0.80V, it is the strongest oxidant.

Can someone please explain? To this point, I've always gone along the premise that it's the strongest oxidant reacting at the cathode in in a galvanic cell and the oxidant with lowest voltage requirement in an electrolytic cell. Thanks.
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brightsky

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Re: Chemistry 3/4 2013 Thread
« Reply #1564 on: October 27, 2013, 11:55:50 am »
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no, the strongest oxidant always gets oxidised preferentially. strong oxidant = good at receiving electrons. so when you turn on the electrolytic cell, both Ag+ and Cu2+ will migrate towards the cathode (since the driving cell makes the cathode negative). who will win? well Ag+ is hungrier than Cu2+ for electrons, so Ag+ will win.

you must realise that in both an electrolytic cell and a galvanic cell, the strongest oxidant present is reduced and the strongest reductant present is oxidised. this never changes. it's just that in a galvanic cell, you're looking for the smallest difference in Eo because electrons are being transferred from the top right hand corner to the bottom left hand corner. it is the CELL POTENTIAL that needs to be minimised, not the Eo of the oxidant.
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1565 on: October 27, 2013, 12:04:35 pm »
+2
When we're writing the oxidation of a primary alcohol to convert it into a carboxylic acid, the top of the reaction arrow will have a strong oxidant. At the bottom of the arrow is usually a H+. Do we need to write the H+ there? Because I'm not sure I understand the implications of having the oxidation take place in an acidic environment or under alkaline conditions.

Also, sorry for an annoying question, but are (non-/)renewable energy resources, such as coal and nuclear fission etc. still examinable as they were previously?

The implications are that either acid or base is necessary for the reaction to occur. One mechanism I know of for the oxidation of primary alcohols to carboxylic acids requires an acid catalyst, while Wiki says such oxidation can occur in 1 M KOH and KMnO4. For VCE purposes, put the acid hydrogen there. It is needed for the reaction.
Think enzymes for an extreme analogy.

I'm pretty sure energy resources are examinable.
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Patches

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Re: Chemistry 3/4 2013 Thread
« Reply #1566 on: October 27, 2013, 01:47:50 pm »
0
If I'm oxidising CH4 in the presence of O2- producing CO2, what's the balanced reaction?

I get CH4 + 2O2-> CO2 + 4H+ +8e-

While the solution is

CH4 + 4O2- > CO2 +2H2O + 8e-

The difference as far as I can tell is they've used H20 to balance the hydrogen in CH4, while I just used H+. Wouldn't that be correct, since H20 is used to balance oxygen which is already balanced with 2 on each side?

Thanks
« Last Edit: October 27, 2013, 01:49:26 pm by Patches »

lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1567 on: October 27, 2013, 02:07:37 pm »
+1
The oxide ion is a VERY VERY VERY VERY VERY powerful base. Even stronger than hydroxide ion. It is highly unlikely that H+ will be formed in the presence of oxide ion.

Balancing these equations isn't just a procedure we go through to make everything fit. It is also meant to reflect the chemical reality.
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ahat

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Re: Chemistry 3/4 2013 Thread
« Reply #1568 on: October 27, 2013, 02:56:38 pm »
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no, the strongest oxidant always gets oxidised preferentially. strong oxidant = good at receiving electrons. so when you turn on the electrolytic cell, both Ag+ and Cu2+ will migrate towards the cathode (since the driving cell makes the cathode negative). who will win? well Ag+ is hungrier than Cu2+ for electrons, so Ag+ will win.

you must realise that in both an electrolytic cell and a galvanic cell, the strongest oxidant present is reduced and the strongest reductant present is oxidised. this never changes. it's just that in a galvanic cell, you're looking for the smallest difference in Eo because electrons are being transferred from the top right hand corner to the bottom left hand corner. it is the CELL POTENTIAL that needs to be minimised, not the Eo of the oxidant.

Hey, so according to your info brightsky, what would be the products in a "dilute aqueous solution of NaCl is electrolysed"?

I found each of the possible reactions:

ANODE: where Eo = -1.23V
where Eo = -1.36V
Therefore, because Chlorine ions are the strongest reductant (most 'negative' Eo), they should preferentially discharge right? The solutions says it's water though.

CATHODE: where Eo = -0.83V
where Eo = -2.71V
Water will be preferentially discharged as it is the strongest oxidant (most 'positive' Eo)

Help appreciated. I thought I'd mastered this but now I'm more confused than before D:
And which way do electrons flow in a secondary cells? Is it the opposite of a primary cell, ie. from cathode to anode? Thanks.

Okay, I think I've understood my error I've prematurely flipped the equation before judging which would react. So, if I hadn't of flipped the anode half-reactions, the water equation would have had the lowest Eo value (most 'negative'). Thanks for the help so far though :)
Could you still please clarify the electron flow question though? Thanks.
« Last Edit: October 27, 2013, 03:29:12 pm by ahat »
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1569 on: October 27, 2013, 03:37:21 pm »
+1
Ah, problem. Don't reverse the reactions around. Keep them in the reduction format. Only that way can you compare reduction and oxidation strength in the way teachers have taught you. E0 values are, by definition, reduction potentials. If you wanted to compare oxidation potentials, then higher oxidation potentials suggest higher reducing strength. Think about it. You'd then have lithium all the way at the top at around +3.

Hey, so according to your info brightsky, what would be the products in a "dilute aqueous solution of NaCl is electrolysed"?

I found each of the possible reactions:

ANODE: where Eo = -1.23V
where Eo = -1.36V
Therefore, because Chlorine ions are the strongest reductant (most 'negative' Eo), they should preferentially discharge right? The solutions says it's water though.

CATHODE: where Eo = -0.83V
where Eo = -2.71V
Water will be preferentially discharged as it is the strongest oxidant (most 'positive' Eo)

Help appreciated. I thought I'd mastered this but now I'm more confused than before D:
And which way do electrons flow in a secondary cells? Is it the opposite of a primary cell, ie. from cathode to anode? Thanks.

Okay, I think I've understood my error I've prematurely flipped the equation before judging which would react. So, if I hadn't of flipped the anode half-reactions, the water equation would have had the lowest Eo value (most 'negative'). Thanks for the help so far though :)
Could you still please clarify the electron flow question though? Thanks.

Read my electrochemistry thread in the chemistry resources thing. Cathode and anode refer specifically to where the electrons come from. The anode is ALWAYS where the electrons leave.
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Re: Chemistry 3/4 2013 Thread
« Reply #1570 on: October 27, 2013, 03:57:01 pm »
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Thankyou :)
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barydos

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Re: Chemistry 3/4 2013 Thread
« Reply #1571 on: October 28, 2013, 11:45:35 am »
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This was in the VCAA 2010 Exam 1 where you had to write an equation for the ionisation of ethanol in mass spec:
Spoiler

I'm just wondering what the importance of the square brackets is, would we lose marks if we did not include the square brackets?
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1572 on: October 28, 2013, 12:02:38 pm »
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I honestly don't see the significance of them and my teacher has never used them.
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Re: Chemistry 3/4 2013 Thread
« Reply #1573 on: October 28, 2013, 12:25:32 pm »
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alright cheers man

Another question: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2010chem2-w.pdf
Question 7d, why isn't the electrons going anti clockwise during discharge?
I thought the electrons go toward the cell where reduction is happening?
« Last Edit: October 28, 2013, 12:48:23 pm by Anonymiza »
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #1574 on: October 28, 2013, 02:08:31 pm »
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alright cheers man

Another question: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2010chem2-w.pdf
Question 7d, why isn't the electrons going anti clockwise during discharge?
I thought the electrons go toward the cell where reduction is happening?

During discharge, electrons go towards the positive half-cell, which is the LiCoO2. The half equations given are when the cell is being recharged.
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