FREE lectures this September + October | HSC Head Start + Exam Revision: book here | QCE Head Start: book here | VCE Exam Revision: book here

Welcome, Guest. Please login or register.

September 21, 2019, 07:10:18 am

Author Topic: VCE Chemistry Question Thread  (Read 933468 times)  Share 

0 Members and 4 Guests are viewing this topic.

fiona_atarnotes

  • Adventurer
  • *
  • Posts: 9
  • Respect: +2
Re: VCE Chemistry Question Thread
« Reply #7995 on: April 18, 2019, 06:09:41 pm »
+5
Are the cathode and anode always the electrodes or is the + or - terminals of the battery? If it is always the electrodes what if they are same such as carbon electrodes?

Hey!
The anodes and cathodes are the electrodes of the battery. For VCE, the cells (electrolytic and galvanic cells) generally have two electrodes where one is an anode and one is a cathode. Non-rechargeable batteries are galvanic cells. Here, the cathode is positively charged and the anode is negatively charged. For electrolytic cells, it's reversed so the cathode is negatively charged and the anode is positively charged. In the case where both electrodes are carbon, since carbon electrodes are inert, that just means the species that is oxidising/reducing is not carbon.

2017: Methods 50 + Premiere's | Physics 49 | Specialist 48 | Chemistry 46 | English 45
ATAR: 99.90
2018 - 2021: Research Science @ Monash

jasheel

  • Adventurer
  • *
  • Posts: 9
  • Respect: +4
Re: VCE Chemistry Question Thread
« Reply #7996 on: April 18, 2019, 06:16:40 pm »
+6
Are the cathode and anode always the electrodes or is the + or - terminals of the battery? If it is always the electrodes what if they are same such as carbon electrodes?

Hey, also adding to what Fiona said. If you have a battery attached to your electrolytic cell, the positive terminal of the battery will attach to the positive electrode (the anode), and the negative terminal will attach to the negative electrode (the cathode).

The way you can think about it is the negative side of the battery is 'forcing' negative electrons into the negative cathode (hence reduction). And therefore, the positive side is 'attracting' the negative electrons from the anode (hence oxidation).
ATAR: 99.70
2018 - 2023: Law (Honours) @ Monash.

peachxmh

  • Forum Regular
  • **
  • Posts: 84
  • \_(ツ)_/
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #7997 on: April 22, 2019, 10:46:06 pm »
0
Need help with a question! I cut out irrelevant bits of it as it was a multi-part question so let me know if anything doesn't make sense/there's information missing :p

In acidic solution, ions containing titanium can react according to the half-equation TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O (l) where E = -0.06 V. In the diagram (attached), A and B are inert graphite electrodes, and all ions have a concentration of 1M. If both A and B electrodes are replaced with iron, what equation will now occur at Electrode B?

I wrote 2TiO2+ (aq) + 4H+ (aq) + Fe (s) -> 2Ti3+ (aq) + 2H2O(l) + Fe2+(aq) but this is wrong - the answer is instead supposed to be Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g).

I don't understand this - why is 2H+(aq) + 2e- -> H2(g) a half-equation that is considered? Is it because the half-cell containing electrode B has H+ ions in solution and the question states it is an acidic solution? But the half-equation contains H2 which isn't part of the question or in the galvanic cell so shouldn't it be eliminated as an option?
2018: Biology [45]
2019: Chemistry [ ] English [ ] Methods [ ] Revolutions [ ] French [ ]

"Strive for progress, not perfection"

DBA-144

  • MOTM: APR 19
  • Trendsetter
  • **
  • Posts: 197
  • Respect: +31
Re: VCE Chemistry Question Thread
« Reply #7998 on: April 23, 2019, 07:27:22 am »
+3
I'm note sure about the below, but it might be because the iron is more reactive than the titanium? So the iron is reacting with the titanium rather than the iron. This would also mean that the metal would react with the acid by producing hydrogen gas. This is using the fact that metal + acid --> salt + hydrogen gas. The hydrogen ions come from the acidic solution and the hydrogen gas comes from the reaction between the metal and the acid.

Also, the metal part of it is not that hard to figure out now either. It's a metal and is going to donate electrons, so the metal would just be donating electrons. 

If you don't mind, I have a question of my own- how can we be certain that the iron would form the 2+ ion? I suspect that this is due to the Hydrogen needing to accept 1 electron each, so Iron is donating 2 electrons total?

Just want to say that I am not certain about any of this, just a potential answer. However, I am not sure why else the Titanium would not be reacting like it was before. It's probably a lot more complicated than this but still I hope this helps, sorry if I am wrong.

persistent_insomniac

  • Forum Regular
  • **
  • Posts: 81
  • Respect: +1
  • School: -------
  • School Grad Year: 2019
Re: VCE Chemistry Question Thread
« Reply #7999 on: April 23, 2019, 07:28:29 pm »
0
Hi!
With electroplating cells does it matter what electrolyte you use? For e.g. A can wanting to be plated with tin - should we only ever use an Sn2+ electrolyte or can we use any solution below Sn2+ on the electrochemical series as Sn2+ will still be the strongest oxidant so regardless will still gain the electrons?
I hope I make sense....

fiona_atarnotes

  • Adventurer
  • *
  • Posts: 9
  • Respect: +2
Re: VCE Chemistry Question Thread
« Reply #8000 on: April 24, 2019, 07:11:53 pm »
+5
Hi!
With electroplating cells does it matter what electrolyte you use? For e.g. A can wanting to be plated with tin - should we only ever use an Sn2+ electrolyte or can we use any solution below Sn2+ on the electrochemical series as Sn2+ will still be the strongest oxidant so regardless will still gain the electrons?
I hope I make sense....
Hey! So generally, the answer is yes, you should only ever use an electrolyte with the metal ion you're plating. I think if you included other oxidants in the solution that are weaker than Sn2+ would be okay initially. However, it's just that with electroplating, you want the concentration of the metal ion you're plating onto the object to be unchanging. So if you included other oxidants in the solution that are weaker than Sn2+, as the reaction continues, the concentration of Sn2+ ions decreases whereas the concentration of the other weaker oxidants remain the same. Therefore, it would reach a point where the other oxidants reduce preferentially due to the decrease in concentration of the Sn2+ ions despite those other oxidants being weaker oxidising agents.

Hope that makes sense!

2017: Methods 50 + Premiere's | Physics 49 | Specialist 48 | Chemistry 46 | English 45
ATAR: 99.90
2018 - 2021: Research Science @ Monash

Rameen

  • Trailblazer
  • *
  • Posts: 25
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #8001 on: April 25, 2019, 11:59:32 am »
0
A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

sweetcheeks

  • Forum Obsessive
  • ***
  • Posts: 436
  • Respect: +38
  • School: ---
  • School Grad Year: 2016
Re: VCE Chemistry Question Thread
« Reply #8002 on: April 25, 2019, 12:26:28 pm »
0
A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

What answer are you getting? It would be helpful if you could post your working out. Are you using stoichiometry?

-_-zzz

  • Adventurer
  • *
  • Posts: 21
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #8003 on: April 25, 2019, 01:59:53 pm »
+2
A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

All you have to do first if find out how many moles of CaCl2 you have. So given that you have 0.030 moles of Cl-, then the moles of CaCl2 must be half of that because for every 1 mole of CaCl2, you have 2 moles of Cl-. Thus the moles of CaCl2 = 0.015 mol. You can then convert the moles to mass using n = m/M :)
2018: Accounting 49
2019: English Language | Specialist Maths | Methods | Chemistry | Biology

jasheel

  • Adventurer
  • *
  • Posts: 9
  • Respect: +4
Re: VCE Chemistry Question Thread
« Reply #8004 on: April 27, 2019, 04:02:01 pm »
+4
Need help with a question! I cut out irrelevant bits of it as it was a multi-part question so let me know if anything doesn't make sense/there's information missing :p

In acidic solution, ions containing titanium can react according to the half-equation TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O (l) where E = -0.06 V. In the diagram (attached), A and B are inert graphite electrodes, and all ions have a concentration of 1M. If both A and B electrodes are replaced with iron, what equation will now occur at Electrode B?

I wrote 2TiO2+ (aq) + 4H+ (aq) + Fe (s) -> 2Ti3+ (aq) + 2H2O(l) + Fe2+(aq) but this is wrong - the answer is instead supposed to be Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g).

I don't understand this - why is 2H+(aq) + 2e- -> H2(g) a half-equation that is considered? Is it because the half-cell containing electrode B has H+ ions in solution and the question states it is an acidic solution? But the half-equation contains H2 which isn't part of the question or in the galvanic cell so shouldn't it be eliminated as an option?

In a galvanic cell the strongest oxidant will always react with the strongest reductant.

In this case they gave you the E value of TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O(l), which is -0.06 V. If you place this in the electrochemical series, you'll find that 2H+(aq) is a stronger oxidant than TiO2+. Because of this 2H+(aq) will be reduced in preference (hence: 2H+(aq) + 2e- -> H2(g)).

The reaction is considered because it's part of the half-cell containing electrode B. It doesn't matter that H2 isn't part of the galvanic cell because its a product (if H2 was a reactant than it would be a different matter). It only matters that H2+ is being reduced. The H2 is just going to be released into the air as a gas!
ATAR: 99.70
2018 - 2023: Law (Honours) @ Monash.

jasheel

  • Adventurer
  • *
  • Posts: 9
  • Respect: +4
Re: VCE Chemistry Question Thread
« Reply #8005 on: April 27, 2019, 04:07:45 pm »
+2
If you don't mind, I have a question of my own- how can we be certain that the iron would form the 2+ ion? I suspect that this is due to the Hydrogen needing to accept 1 electron each, so Iron is donating 2 electrons total?
Iron (Fe(s)) on the electrochemical series will lose two electrons, it can't ever lose one electron only! (Fe2+ can lose another electron to form Fe3+ later on however). 
ATAR: 99.70
2018 - 2023: Law (Honours) @ Monash.

f0od

  • Forum Regular
  • **
  • Posts: 61
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #8006 on: April 29, 2019, 09:08:14 pm »
0
In our recent reduction potentials prac, our predicted cell voltage (0.57) was just over the actual cell voltage (0.56). I was wondering what might be the reason for this (slight, but still a) difference aside from it not being in SLC conditions?

Thanks! :)
class of 2019

-_-zzz

  • Adventurer
  • *
  • Posts: 21
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #8007 on: May 03, 2019, 05:15:40 pm »
0
In our recent reduction potentials prac, our predicted cell voltage (0.57) was just over the actual cell voltage (0.56). I was wondering what might be the reason for this (slight, but still a) difference aside from it not being in SLC conditions?

Thanks! :)

Even if the investigation was conducted under SLC you would still probably get a different value due to the degree of precision exhibited by the voltmeter. When the ECS was created I'd assume that super precise instruments were used.
2018: Accounting 49
2019: English Language | Specialist Maths | Methods | Chemistry | Biology

samyaks123

  • Adventurer
  • *
  • Posts: 5
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #8008 on: May 09, 2019, 06:08:53 am »
0
Just have a question about Q 8b on the vcaa 2018 chemistry exam, since there are four solar panels which each produce a current of 5.20A, would you multiply that by 4 when working out volume of gas for that question

jasheel

  • Adventurer
  • *
  • Posts: 9
  • Respect: +4
Re: VCE Chemistry Question Thread
« Reply #8009 on: May 09, 2019, 05:50:08 pm »
+3
Just have a question about Q 8b on the vcaa 2018 chemistry exam, since there are four solar panels which each produce a current of 5.20A, would you multiply that by 4 when working out volume of gas for that question

In the question it seems to be making it really clear the four solar panels are each operating separately and for 8 hours each to produce a current of 5.20 A each. In that case I think in that case you would multiply the mol of electrons by four before you calculated the mol of H2. Unfortunately the examiners report isn't out yet which is annoying!
ATAR: 99.70
2018 - 2023: Law (Honours) @ Monash.