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Author Topic: VCE Chemistry Question Thread  (Read 2313035 times)  Share 

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anoushka_iyer

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Re: VCE Chemistry Question Thread
« Reply #8235 on: November 06, 2019, 02:05:35 pm »
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I have a question- When they say 'catalytic electrodes increase the efficiency of the reaction (in the context of a fuel cell)', what do they exactly mean by efficiency? Like is it energy efficiency or something else like the current output?

Thanks so much
« Last Edit: November 06, 2019, 03:41:49 pm by anoushka_iyer »
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Matthew_Whelan

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Re: VCE Chemistry Question Thread
« Reply #8236 on: November 06, 2019, 02:12:01 pm »
+3
I have a question- When they say 'catalytic electrodes increase the efficiency of the reaction', what do they exactly mean by efficiency? Like is it energy efficiency or something else?

Thanks so much

A catalyst lowers the activation energy for a reaction to occur which increases the proportion of successful collisions, therefore the process of two molecules colliding and reacting is more efficient.
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EmadMo

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Re: VCE Chemistry Question Thread
« Reply #8237 on: November 08, 2019, 12:29:58 pm »
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I was doing this company exam and I didn't quite get this question about electrochemistry. It says a( student set up the following electrolytic cells. Each electrode is made of graphite. Each battery provides 1 V, and each battery was attached to its respective cell for the same duration. Assume each cell is functioning under standard conditions.) In the solutions for the exam it says: Each battery provides 1 V, therefore under standard conditions no reaction occurs in the case of the Zn(NO3)2 cell as a voltage of at least 1.99 V is required according to the electrochemical series. How do you know that min voltage of at least 1.99 is required from the electrochemical series?

thanks

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8238 on: November 08, 2019, 01:20:07 pm »
+3
I was doing this company exam and I didn't quite get this question about electrochemistry. It says a( student set up the following electrolytic cells. Each electrode is made of graphite. Each battery provides 1 V, and each battery was attached to its respective cell for the same duration. Assume each cell is functioning under standard conditions.) In the solutions for the exam it says: Each battery provides 1 V, therefore under standard conditions no reaction occurs in the case of the Zn(NO3)2 cell as a voltage of at least 1.99 V is required according to the electrochemical series. How do you know that min voltage of at least 1.99 is required from the electrochemical series?

thanks
Just like how you can work out the voltage produced by a spontenous reaction occuring in a galvanic cell, you can work out the voltage needed to force the reaction to occur the opposite way.
using the example of a Zn2+(aq)/Zn(s) half cell and a Cu2+(aq)/Cu(s) half cell:
The relevant electochemical reactions are:

and

The discharging/spontaneous reaction will be

This will occur to produce a positive voltage of 0.34+0.76 =1.1V
Note that the E0 value for the Zn2+/Zn half cell shown above as -0.76 is made positive since the reaction occurs in the opposite direction to what is listed on the electrochemical series.

For this set up, the recharge/nonspontaneous reaction is the reverse of the discharge reaction:

And this will require a voltage to force the reaction to occur. The magnitude of the voltage input can be calculated as equal to the magnitude of the voltage output of the discharge reaction. This would mean that for this recharge reaction, 1.1V is required as an input. You could also think about this interms of calculating the hypothetical voltage produced by this recharging cell using V=-(0.34)+(-0.76)=-1.1V produced (meaning 1.1 V is consumed by the cell)

Applying this to your question, the relevant half cell reactions are:

and


^note that I am assuming the first reaction is relevant since I can't see the set up referred to in the question
The recharge reaction relevant here would thus be:


Just as we did above, we can calculate the neccasary input voltage for this to occur using the relevant E0 values. Thus V=-(1.23)+(-0.76) = -1.99V (indicating that -1.99 volts are produced --> meaning 1.99 volts must be inputted to force the reaction to occur).
Since the battery only supplied 1V, we can then say that no reaction will occur since the voltage input is not sufficient to force the recharge reaction.
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KiNSKi01

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Re: VCE Chemistry Question Thread
« Reply #8239 on: November 08, 2019, 05:07:11 pm »
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Yo I am still confused how to deal with some half equations  :(

For example q10 a on the 2014 exam, why do you have to include the hydroxide ions? On both sides of the half equation they have a -1 charge so why do you need to include them?
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Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8240 on: November 08, 2019, 05:42:23 pm »
+5
Yo I am still confused how to deal with some half equations  :(

For example q10 a on the 2014 exam, why do you have to include the hydroxide ions? On both sides of the half equation they have a -1 charge so why do you need to include them?
This question just requires you to write out the revelant equations using the electrochemical series. From the information in the stem you know that its alkaline (therefore it can't have any H+ ions present, and instead will have OH-) and you know the two reactants (hydrogen gas and oxygen). We also know that the two half cell reactions must be spontaneous/produce a positive voltage.
As such we can identify the two relevant reactions from the electrochemical series as:

and


The reason why OH is present in both half cell reactions is becuase it is needed to balance the equations
If we were to construct these reactions by hand we would do the following:
1) write out the main elements involved:


2) Balance oxygens (since this is alkaline we can do this with OH- ions)


3) balance hydrogens with H+


4) since this is alkaline, remove H+ by adding OH- to both sides


which gives:


5) balance our charges/oxidation numbers by adding electrons where neccasary


6) add states


and hence you have your answer
« Last Edit: November 08, 2019, 05:48:16 pm by Erutepa »
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KiNSKi01

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Re: VCE Chemistry Question Thread
« Reply #8241 on: November 08, 2019, 05:54:09 pm »
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This question just requires you to write out the revelant equations using the electrochemical series. From the information in the stem you know that its alkaline (therefore it can't have any H+ ions present, and instead will have OH-) and you know the two reactants (hydrogen gas and oxygen). We also know that the two half cell reactions must be spontaneous/produce a positive voltage.
As such we can identify the two relevant reactions from the electrochemical series as:

and


The reason why OH is present in both half cell reactions is becuase it is needed to balance the equations
If we were to construct these reactions by hand we would do the following:
1) write out the main elements involved:


2) Balance oxygens (since this is alkaline we can do this with OH- ions)


3) balance hydrogens with H+


4) since this is alkaline, remove H+ by adding OH- to both sides


which gives:


5) balance our charges/oxidation numbers by adding electrons where neccasary


6) add states


and hence you have your answer

Hey thanks for the help but for the equation below I still dont get why you require the hydroxide ions for the anode half equation. Yeah i know that Zn(OH)2 is the overall product formed but I thought it was fine to ignore the hydroxide ions in the oxidation half equation because they are spectator ions - sorry if im not making myself clear


Zn(s) + 2OH–(aq) → Zn(OH)2(s) + 2e–
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xxxjss

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Re: VCE Chemistry Question Thread
« Reply #8242 on: November 08, 2019, 06:04:49 pm »
+2
Hey thanks for the help but for the equation below I still dont get why you require the hydroxide ions for the anode half equation. Yeah i know that Zn(OH)2 is the overall product formed but I thought it was fine to ignore the hydroxide ions in the oxidation half equation because they are spectator ions - sorry if im not making myself clear


Zn(s) + 2OH–(aq) → Zn(OH)2(s) + 2e–

I think the reason why OH is included as it would be incorrect to say that just Zn(s) forms as the product... it must be Zn(OH)(s), hence we require the hydroxide ions to balance the half equation

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8243 on: November 08, 2019, 06:05:02 pm »
+3
Hey thanks for the help but for the equation below I still dont get why you require the hydroxide ions for the anode half equation. Yeah i know that Zn(OH)2 is the overall product formed but I thought it was fine to ignore the hydroxide ions in the oxidation half equation because they are spectator ions - sorry if im not making myself clear


Zn(s) + 2OH–(aq) → Zn(OH)2(s) + 2e–
Sorry - you were being clear, I not only looked at the wrong question but also in the wrong exam  8)
Spectator ions are ions that exist in the same state on both sides of the equation.
For the half cell reaction you've stated, the hydroxide is in a solid state on one side of the equation, but in an aqeous state on the other, meaning it is not a spectator ion and thus you have to include it.
You are correct in ignoring spectator ions - its just that the OH- here is not a spectator ion.
Hopefully I actually answered your question this time haha
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KiNSKi01

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Re: VCE Chemistry Question Thread
« Reply #8244 on: November 08, 2019, 06:08:01 pm »
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Sorry - you were being clear, I not only looked at the wrong question but also in the wrong exam  8)
Spectator ions are ions that exist in the same state on both sides of the equation.
For the half cell reaction you've stated, the hydroxide is in a solid state on one side of the equation, but in an aqeous state on the other, meaning it is not a spectator ion and thus you have to include it.
You are correct in ignoring spectator ions - its just that the OH- here is not a spectator ion.
Hopefully I actually answered your question this time haha

Ah yes thank you very much  :)
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Re: VCE Chemistry Question Thread
« Reply #8245 on: November 08, 2019, 11:03:24 pm »
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Yo came across a NEAP question and I'm not sure if it's shitty or there's something I don't know

Basically does temperature affect both secondary and tertiary structure of enzymes whereas pH only affects tertiary

In the question, a result of no browning on an apple indicated that the enzyme polyphenoxidase had been denatured and you had to decide what level of enzyme structure was affected: Lemon juice spead on apple only affected tertiary structure whereas apple in boiling water affected both tertiary and secondary. So does temperature generally also affect secondary (I thought it would be dependent on the extent to which the enzymes are heated, therefore a question would be much more clear in indicating if secondary structure were also likely to have been affected)
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pugs

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Re: VCE Chemistry Question Thread
« Reply #8246 on: November 08, 2019, 11:07:14 pm »
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i'm just going through my chemistry notes and for some reason i don't understand this question - i think it's a very basic question so i'm sorry!!

In a low temperature environment, explain why it would be preferable to use pure petrodiesel or a biodiesel-petrodiesel blend instead of pure biodiesel
Pure petrodiesel or a blend would have a lower melting point so would remain a liquid at low temperatures. At low temperatures, a pure biodiesel would begin to crystalise and flow poorly along fuel lines. teacher answer btw

- I thought that biodiesel has a lower melting point than petrodiesel because of the presence of double bonds in biodiesel compared to none in petrodiesel?
- if you could talk about cloud point, wouldn't biodiesel still be preferable over petrodiesel as biodiesel has a higher cloud point?

could someone pls explain this question to me // thank you so muchh!


2019 vce journal here

KiNSKi01

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Re: VCE Chemistry Question Thread
« Reply #8247 on: November 08, 2019, 11:47:37 pm »
+3
i'm just going through my chemistry notes and for some reason i don't understand this question - i think it's a very basic question so i'm sorry!!

In a low temperature environment, explain why it would be preferable to use pure petrodiesel or a biodiesel-petrodiesel blend instead of pure biodiesel
Pure petrodiesel or a blend would have a lower melting point so would remain a liquid at low temperatures. At low temperatures, a pure biodiesel would begin to crystalise and flow poorly along fuel lines. teacher answer btw

- I thought that biodiesel has a lower melting point than petrodiesel because of the presence of double bonds in biodiesel compared to none in petrodiesel?
- if you could talk about cloud point, wouldn't biodiesel still be preferable over petrodiesel as biodiesel has a higher cloud point?

could someone pls explain this question to me // thank you so muchh!

To answer your question about cloud point - If you think about what a higher cloud point means then you would realise that because biodiesel has a higher cloud point than petrodiesel it will be less effective at cold temps. Think about this: at a cold temperature which one is more likely to have already reached its cloud point (biodiesel) - because you would have to go to even lower temp to reach cloud point of petrodiesel (slightly counter intuitive due to lower temp and higher cloud point)
« Last Edit: November 08, 2019, 11:50:03 pm by KiNSKi01 »
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Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8248 on: November 09, 2019, 08:50:18 am »
+4
Yo came across a NEAP question and I'm not sure if it's shitty or there's something I don't know

Basically does temperature affect both secondary and tertiary structure of enzymes whereas pH only affects tertiary

In the question, a result of no browning on an apple indicated that the enzyme polyphenoxidase had been denatured and you had to decide what level of enzyme structure was affected: Lemon juice spead on apple only affected tertiary structure whereas apple in boiling water affected both tertiary and secondary. So does temperature generally also affect secondary (I thought it would be dependent on the extent to which the enzymes are heated, therefore a question would be much more clear in indicating if secondary structure were also likely to have been affected)
In the past (2017 nht) they did accept denaturation due to heat as occuring due to the breaking of hydrogen bonds in the tertiary strucuture (no mention of the secondary strucutre). However previous multiple choice asnwers have wanted you to know that denaturation can involve breakage of the hydrogen bonds in the secondary strucutre. I personally do mention the breakage of hydrogen bonding in the secondary, tertiary (and quaternary if the protein is quaternary) but it seems VCAA will be happy with you just saying tertiary.
Also, I don't think VCAA wants you to know about the relative strength of hydrogen bonding within the secondary and tertiary strucutres as I haven't seen it come up. As such, I think it would be safe to say that heating that denatures a proteins tertiary strucure will also denature a proteins secondary strucure (atleast at a VCE level). I might be wrong though, so please correct me if this is utter nonsense.

i'm just going through my chemistry notes and for some reason i don't understand this question - i think it's a very basic question so i'm sorry!!

In a low temperature environment, explain why it would be preferable to use pure petrodiesel or a biodiesel-petrodiesel blend instead of pure biodiesel
Pure petrodiesel or a blend would have a lower melting point so would remain a liquid at low temperatures. At low temperatures, a pure biodiesel would begin to crystalise and flow poorly along fuel lines. teacher answer btw

- I thought that biodiesel has a lower melting point than petrodiesel because of the presence of double bonds in biodiesel compared to none in petrodiesel?
- if you could talk about cloud point, wouldn't biodiesel still be preferable over petrodiesel as biodiesel has a higher cloud point?

could someone pls explain this question to me // thank you so muchh!
Just to add to what KiNSKi01 has said,
Cloud point is essentialy the temperature at which a fuel begins to solidify. We want our fuels to be liquid so that they can flow through piplines and through engines and what not. Thus if biodiesel has a higher cloud point, it will start to solidify at higher temperatures, meaning it becomes inneffective at a higher temperature than pertrodiesel with a lower cloud point.
For example: if the temperature was aproximately 8 degrees, biodisel may start to solidify (due to its higher cloud point) whereas petrodisel might not solidify (due to its lower cloud point) which makes petrodisel a better fuel option in this circumstance. Note that youdon't need to know the precise temperatures, just that biodisel has a higher cloud point
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Re: VCE Chemistry Question Thread
« Reply #8249 on: November 09, 2019, 09:51:51 am »
+2
Yo came across a NEAP question and I'm not sure if it's shitty or there's something I don't know

Basically does temperature affect both secondary and tertiary structure of enzymes whereas pH only affects tertiary

In the question, a result of no browning on an apple indicated that the enzyme polyphenoxidase had been denatured and you had to decide what level of enzyme structure was affected: Lemon juice spead on apple only affected tertiary structure whereas apple in boiling water affected both tertiary and secondary. So does temperature generally also affect secondary (I thought it would be dependent on the extent to which the enzymes are heated, therefore a question would be much more clear in indicating if secondary structure were also likely to have been affected)

I suspect that with pH they were trying to test that you understood zwitterions in an indirect way

Imo asking whether temperature could affect secondary structure would be a lot more reasonable than expecting you to know if it did