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Author Topic: VCE Chemistry Question Thread  (Read 2313177 times)  Share 

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KiNSKi01

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Re: VCE Chemistry Question Thread
« Reply #8355 on: November 11, 2019, 11:48:21 pm »
+1
yeah i think that they would accept either

i was just doing a question with a calibration curve that could have been extrapolated, but VCAA used the value from the curve :// so i think it's safe to just go with whatever value they give you from the curve

also! i don't know if anyone will be able to answer this before the exam lol, but for Cl(g) + 2e- <–> 2Cl-(aq) and O2(g) + 4H+(aq) + 4e- <–> 2H2O(l) in the electrochemical series, if a solution of HCl is in question and its concentration is 1.0M, is it necessary to write reduction equations for both Cl and H2O? Our teacher taught us to write both equations if the concentration of HCl is between 0.5M and 2M, but i'm not sure if it's required in the exam?

thank you!

Mmmm good question, we've been told if concentration is greater than 2.0 M use the Cl reduction equation only - I reckon VCAA will make it fairly clear how they want it e.g a question might ask how may the reduction equation change if the concentration of HCl were to increase (or something like that)
I'm not sure if writing both Cl and H2O reduction equations is a good idea
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pugs

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Re: VCE Chemistry Question Thread
« Reply #8356 on: November 12, 2019, 08:04:04 am »
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Mmmm good question, we've been told if concentration is greater than 2.0 M use the Cl reduction equation only - I reckon VCAA will make it fairly clear how they want it e.g a question might ask how may the reduction equation change if the concentration of HCl were to increase (or something like that)
I'm not sure if writing both Cl and H2O reduction equations is a good idea
yeah i don't think i'm going to write both haha if it pops up
thank you for the reply!! best of luck for the exam!


2019 vce journal here

Bri MT

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Re: VCE Chemistry Question Thread
« Reply #8357 on: November 12, 2019, 08:06:46 am »
+3
Good luck everyone :)

Remember you can pause and take a breath if you feel yourself panicking - if you get initially lost on a question that's ok
« Last Edit: November 12, 2019, 08:08:51 am by Bri MT »

Uhhhh1204

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Re: VCE Chemistry Question Thread
« Reply #8358 on: November 12, 2019, 05:35:59 pm »
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Just wondering if it's fine if I forgot to write my VCAA number on the front of the chem exam today. I'll contact VCAA tomorrow but some reassurance that it'll be fine would be great :(

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8359 on: November 12, 2019, 06:17:32 pm »
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Just wondering if it's fine if I forgot to write my VCAA number on the front of the chem exam today. I'll contact VCAA tomorrow but some reassurance that it'll be fine would be great :(
There was a disussion about this last year here and it seems that you have nothing to worry about and that they can sort it out.
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jnlfs2010

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Re: VCE Chemistry Question Thread
« Reply #8360 on: December 09, 2019, 01:33:30 pm »
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What does it mean when the expected cell voltage is negative?

Thanks for any help if given!
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Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8361 on: December 09, 2019, 01:46:51 pm »
+2
What does it mean when the expected cell voltage is negative?

Thanks for any help if given!
Assuming you've done the adding correctly, a negative cell voltage means the reaction is no spontaneous and will require the input of a voltage of that magnitude in order to occur.
If the cell voltage was - 2v, the reaction would require a 2v input.
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jnlfs2010

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Re: VCE Chemistry Question Thread
« Reply #8362 on: December 09, 2019, 01:54:57 pm »
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Assuming you've done the adding correctly, a negative cell voltage means the reaction is no spontaneous and will require the input of a voltage of that magnitude in order to occur.
If the cell voltage was - 2v, the reaction would require a 2v input.

thanks!!
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redleafbun

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Re: VCE Chemistry Question Thread
« Reply #8363 on: December 13, 2019, 10:40:49 pm »
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How do you predict the reactions for electrolysis when water is involved?
For example:
Predict the products of the electrolysis of 1M NiSO4 (aq) using copper electrodes.
There are three equations involving water as the main reactant or product, which ones do I eliminate?

Bri MT

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Re: VCE Chemistry Question Thread
« Reply #8364 on: December 14, 2019, 09:32:47 am »
+3
How do you predict the reactions for electrolysis when water is involved?
For example:
Predict the products of the electrolysis of 1M NiSO4 (aq) using copper electrodes.
There are three equations involving water as the main reactant or product, which ones do I eliminate?

Hey,

Welcome to the forums :)

To predict the reactions that occur you need to identify all possible reactions and then look at the electrochemical series identify where the reactions are relative to each other. I liked to remember which ones occur by using a diagram with an arrow going up and then to the right, and another going down and to the left.

In this case our potential reactants are:
The nickel cations, solid copper & water.
Note that this gives us more than 3 possible reactions.

Now, we identify which the strongest oxidant and reductant - which predicts what reaction will occur   (like how you predict reactions in galvanic cells)

I've assumed here that you are confident in predicting reactions in galvanic cells - let me know if any part of this doesn't make sense or you're still unsure

redleafbun

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Re: VCE Chemistry Question Thread
« Reply #8365 on: December 14, 2019, 07:26:37 pm »
0
Hey,

Welcome to the forums :)

To predict the reactions that occur you need to identify all possible reactions and then look at the electrochemical series identify where the reactions are relative to each other. I liked to remember which ones occur by using a diagram with an arrow going up and then to the right, and another going down and to the left.

In this case our potential reactants are:
The nickel cations, solid copper & water.
Note that this gives us more than 3 possible reactions.

Now, we identify which the strongest oxidant and reductant - which predicts what reaction will occur   (like how you predict reactions in galvanic cells)

I've assumed here that you are confident in predicting reactions in galvanic cells - let me know if any part of this doesn't make sense or you're still unsure

Ahhh I see but I am still confused about why the two water equations are crossed out?
Thankss

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Re: VCE Chemistry Question Thread
« Reply #8366 on: December 14, 2019, 08:44:36 pm »
+1
Ahhh I see but I am still confused about why the two water equations are crossed out?
Thankss

When you look at the species in the equations crossed out, you will see that you lack the reactants present for it to be used; there is no oxygen gas or hydrogen peroxide as reactants given that the system is closed, thus you cannot use them!

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8367 on: December 14, 2019, 10:02:40 pm »
+2
Ahhh I see but I am still confused about why the two water equations are crossed out?
Thankss
To break down what Bri mentioned earlier about how the strongest reductand will react with the strongest oxidant present:
Oxidants are shown on the left hand side of the equations (they are reduced and gain electrons) and the strongest oxidant is the one closest to the top - In this case this is Nickle 2+ ions. Reductants are shown on the right (they are oxidised and gain electyrons) and the strongest reductant is the onle closest to the bottom - In this case solid copper. As such, we know that nickle ions will react with copper metal.
help visualising the increasing reductand and oxidant strengths
The oxidisation of water (the top equation crossed out) does not occur becuase theres is no hydrogen peroxide to be reduced (if hydrogen peroxide was present this reaction would occur) and becuase water is a weaker reductant than copper metal (if all the copper was used up then this reaction would  occur). The other crossed out equation doesn't occur since oxygen gas is not present (for this to be the case we would have to be told that ocygen gas is being bubled into the solution), and since hydroxide ions are not present (and even if they were they are a weaker reductant than the copper metal)

Hopefully this helps clarify some things, but let me know if this is still confusing. Understanding electrochemical reactions can be quite tricky!
« Last Edit: December 14, 2019, 10:06:55 pm by Erutepa »
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tay75

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Re: VCE Chemistry Question Thread
« Reply #8368 on: December 17, 2019, 10:15:33 pm »
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Hey everyone, couldn't find the answer I'm looking for online anywhere. I'm a little clueless when it comes to chemistry, sorry for the rookie question.

For conjugate redox pairs, I'm having some trouble identifying how much of a substance to include in my answer. All of these examples are from Chapter Review 4 of Heinemann, Q13.


1) Fe + I2 -> FeI2
The textbook answers say the redox pairs are Fe/Fe3+ and I2/I-. I'm guessing this is because it wouldn't be specific enough if you just gave the whole compound, FeI2, since the compound is involved in both oxidation and reduction.

2) Mg + FeCl2 -> MgCl2 + Fe
Conjugate redox pairs are Mg/Mg2+ and Fe2+ and Fe. Why isn't it listed as FeCl2 and MgCl2?

3) 10Br- + 2MnO4- + 16H+ -> 2Mn2+ + 8H2O + 5Br
Textbook answer says the reduction conjugate redox pair is MnO4-/Mn2+. So does that mean that for polyatomic ions, the whole thing is included in the answer?

4) Cu + 2NO3- + 4H+ -> Cu2+ + 2NO2 + 2H2O
For this one, the reduction conjugate redox pair is NO3- and NO2. Why is it NO2 rather than just N4+?

I guess I get equation 1, but to me it seems like 2 and 4 contradict themselves. Is there a particular trend based on what type of compound it is (ionic compound vs covalent compound) or is it just random and doesn't matter. Hopefully I was clear enough!
« Last Edit: December 17, 2019, 10:27:35 pm by tay75 »

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8369 on: December 18, 2019, 09:21:25 am »
+5
Hi tay 75!
Welcome to the forums!

I am fairly sure that for VCE we can assume redox reactions like these are taking place in water (unless otherwise stated).
Some Ionic compounds are water soluble (as the hydrogen bonds the ions form with water are stronger than the ionic bond between the ions). These water-soluble ionic compounds will disassociate (break apart) into their constituent ions when in water.
For question 1, FeI2 is a water soluble compound, meaning you can treat it as an Iron (II) cation (Fe2+) and Iodide anion (I-). As such you can think of the equation as being: Fe + I2 -> Fe2+ + 2I- since FeI2 essentially breaks up and exists as separate Fe2+ and I- ions. This means its more appropriate to sau the redox conjugate pairs are Fe/Fe2+ and I2/I-
The compounds in example 2 are also water-soluble, so the same logic applies.

However, some ionic compounds are water insoluble. These will not disassociate into constituent ions and thus when writing the conjugate redox pairs, it will be most appropriate to note the whole insoluble compound

the last two questions are dealing with covalently bonded compounds/ions. Polyatomic ions such as MnO4- contain covalent bonds (bonding the oxygen to the Manganese). These bonds are not broken apart in water and as such these covalent compounds/ polyatomic ions do not disassociate themselves, thus you must write the whole MnO4- ion in the redox conjugate pairs. The same is the case for the last question.
As a side note: If MnO4- was bonded to a potassium as KMnO4, this would be a water soluble ionic compounds that would disassociate in water to give the two constituent ions: K+ and MnO4-, however MnO4- does not itself break apart in water as its atoms are covalently bonded.

Hopefully this explanation helps you understand and hopefully I didn't get anything wrong (feel free to ask/correct me if you think I have made a mistake or if there's something that you still don't understand)


« Last Edit: December 18, 2019, 09:30:58 am by Erutepa »
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