HSC Chemistry Marathon

[RuiAce]

Before you can ask a question, you'll have to make an ATAR Notes account here. Please reserve this thread for questions intended to test others. If you need help, please post in our questions thread.

Hello chemists

The aim of this thread is to encourage students to test each other in their knowledge of the course. In this thread, you will post your own questions, be it your own made-up questions, past paper questions or anything appropriate, and have an opportunity to mark their response! The objective is so that you can figure out what you must and must not do in the exam, and help another student know where they're at.

The format is simple. When a student answers a question, they will type up their solution to the response, based off the mark allocated. You may choose to hand-write a response and post it up, but ensure that your handwriting is legible - I would only suggest this for calculations. Beneath their response, that student will post the next question for another student to answer. (It is recommended that they post in bold-font Next question to make it clear.)

When posting the question, they will also assign a mark allocation to their question. When the question is answered, they may choose to answer the next question, but they MUST MARK the response to their own. A brief explanation should be made whenever marks are lost explaining why they were.

Rules:
- One question at a time (multiple questions will be DELETED). Note that your question can have multiple parts.
(Exception - Nobody knows how to do it because they haven't been taught how to. In that case, you may request a new question be put up.)
- This is not for your questions you need help on. If you need help with a question, please use the thread linked at the top of this post.
- No breach of copyright (obtaining questions from exam companies like CSSA is not acceptable).
- Check back when someone answers your question so you can provide helpful feedback to them.

Good luck and enjoy!
_____________________________________________________________________

I will kick off with the first question from Production of Materials:

A galvanic cell is set up between two cells. The first cell involves magnesium metal dipped in magnesium nitrate solution. The second cell involves zinc metal dipped in zinc sulfate solution.
a) Identify the anode and the catholyte. (1)
b) The cell does not function without the presence of a salt bridge. Explain why the salt bridge is necessary. (2)
c) When the voltage of the cell was recorded, it was not equal to that of the theoretical value. Explain why this may have possibly been the case. (3)
d) Calculate the theoretical voltage this cell should produce. (2)

[QC]

A galvanic cell is set up between two cells. The first cell involves magnesium metal dipped in magnesium nitrate solution. The second cell involves zinc metal dipped in zinc sulfate solution.
a) Identify the anode and the catholyte. (1)
b) The cell does not function without the presence of a salt bridge. Explain why the salt bridge is necessary. (2)
c) When the voltage of the cell was recorded, it was not equal to that of the theoretical value. Explain why this may have possibly been the case. (3)
d) Calculate the theoretical voltage this cell should produce. (2)

a)Anode: Magnesium Metal
Catholyte: Zinc Sulfate 1M
b)A salt bridge is necessary in order to prevent the build-up of ions from either the cathode or anode. As the reduction and oxidation half cells react, they produce a negative charge and a positive charge respectively.
Anode: Mg(s)->Mg2+(aq)+2e-
Cathode: Zn2+(aq)+2e-->Zn(s)
Therefore, in the cathode, a negative charge builds up which must be neutralised by the use of an ionic solution such as KNO3 as the K+ ions transfer to the cathode and the NO3- to the anode where the Mg2+ is building up.
c)There are many errors that  occur in the experimental method that result in the incorrect theoretical value calculated. One of these is the Zn(s) and Mg(s) stripes not fully cleaned which can result in the oxidisation of the metal which means ZnO and MgO would react thus changing the expected/theoretical results. Another reason is that the anolyte and catholyte may not be 1M which results in a change in value from the theoretical as there may be more or less ions in solution. The wire/voltmeter may also produce resistance altering the results compared to the measured theoretical values that use more precise measuring instruments.
d)Cathode:Zn2+(s)++2e-->Zn(s) e0=-0.76V
Anode: Mg2+(aq)+2e-->Mg(s) e0=-2.36V, since it is undergoing oxidation, it is reversed i.e e0=2.36V
Therefore: e⁰cell=-0.76+2.36=1.6V
NEXT QUESTION (From 2016 HSC, I think I got the right answer but I can't seem to find it online): An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is 1.0 × 10⁵ L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation: C₈H₁₈ + 17/2 O_{2 (g)} + 8CO_(g) + 9H₂O_(l)
Exposure to carbon monoxide at levels greater than 0.100 g L^–1 of air can be dangerous to human health. 6.0 kg of octane was combusted by the car in this workshop. Using the equation provided, determine if the level of carbon monoxide produce in the workshop would be dangerous to human health. Support your answer with relevant calculations. (4 marks)

[RuiAce]

a)Anode: Magnesium Metal
Catholyte: Zinc Sulfate 1M
b)A salt bridge is necessary in order to prevent the build-up of ions from either the cathode or anode. As the reduction and oxidation half cells react, they produce a negative charge and a positive charge respectively.
Anode: Mg(s)->Mg2+(aq)+2e-
Cathode: Zn2+(aq)+2e-->Zn(s)
Therefore, in the cathode, a negative charge builds up which must be neutralised by the use of an ionic solution such as KNO3 as the K+ ions transfer to the cathode and the NO3- to the anode where the Mg2+ is building up.
c)There are many errors that  occur in the experimental method that result in the incorrect theoretical value calculated. One of these is the Zn(s) and Mg(s) stripes not fully cleaned which can result in the oxidisation of the metal which means ZnO and MgO would react thus changing the expected/theoretical results. Another reason is that the anolyte and catholyte may not be 1M which results in a change in value from the theoretical as there may be more or less ions in solution. The wire/voltmeter may also produce resistance altering the results compared to the measured theoretical values that use more precise measuring instruments.
d)Cathode:Zn2+(s)++2e-->Zn(s) e0=-0.76V
Anode: Mg2+(aq)+2e-->Mg(s) e0=-2.36V, since it is undergoing oxidation, it is reversed i.e e0=2.36V
Therefore: e⁰cell=-0.76+2.36=1.6V
NEXT QUESTION (From 2016 HSC, I think I got the right answer but I can't seem to find it online): An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is 1.0 × 105 L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation: C8H18(l) + 17 O (g) 8CO(g) l 2 2 + 9H2O( ) Exposure to carbon monoxide at levels greater than 0.100 g L–1 of air can be dangerous to human health. 6.0 kg of octane was combusted by the car in this workshop. Using the equation provided, determine if the level of carbon monoxide produce in the workshop would be dangerous to human health. Support your answer with relevant calculations. (4 marks)

Easy 1/1 for a)

b): Some teachers will give a 2/2 here, but I am going to purposefully mark harsh and give 1/2. It also serves an additional role of simply completing the circuit - without a circuit, you cannot have current, thus the cell would not work. No real explanation of the first mark was really necessary here; the second mark was a second point.

c): 3 valid points have been placed, but depending on what the marker feels like, this might still be a 2/3. I will give it a 3/3.
The undisputable ones are the oxidation on the metal (impurities), and the 1M concentration of the electrolytes. The voltmeter arises to question - in general, you assume that the materials are always functioning properly.

It might help to add (not replace the voltmeter with this) that we must maintain ALL standard conditions. The 1M solution is one part of it, but the other part is 25 deg C, 100 kPa pressure.

d) This looks correct going by my vague memory of the data sheet. 2/2.


Also, with that question, might want to add a bit of formatting (and preferably a new line) to the equation you inserted.

[Kekemato_BAP]

a)Anode: Magnesium Metal
Catholyte: Zinc Sulfate 1M
b)A salt bridge is necessary in order to prevent the build-up of ions from either the cathode or anode. As the reduction and oxidation half cells react, they produce a negative charge and a positive charge respectively.
Anode: Mg(s)->Mg2+(aq)+2e-
Cathode: Zn2+(aq)+2e-->Zn(s)
Therefore, in the cathode, a negative charge builds up which must be neutralised by the use of an ionic solution such as KNO3 as the K+ ions transfer to the cathode and the NO3- to the anode where the Mg2+ is building up.
c)There are many errors that  occur in the experimental method that result in the incorrect theoretical value calculated. One of these is the Zn(s) and Mg(s) stripes not fully cleaned which can result in the oxidisation of the metal which means ZnO and MgO would react thus changing the expected/theoretical results. Another reason is that the anolyte and catholyte may not be 1M which results in a change in value from the theoretical as there may be more or less ions in solution. The wire/voltmeter may also produce resistance altering the results compared to the measured theoretical values that use more precise measuring instruments.
d)Cathode:Zn2+(s)++2e-->Zn(s) e0=-0.76V
Anode: Mg2+(aq)+2e-->Mg(s) e0=-2.36V, since it is undergoing oxidation, it is reversed i.e e0=2.36V
Therefore: e⁰cell=-0.76+2.36=1.6V
NEXT QUESTION (From 2016 HSC, I think I got the right answer but I can't seem to find it online): An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is 1.0 × 10⁵ L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation: C₈H₁₈ + 17/2 O_{2 (g)} + 8CO_(g) + 9H₂O_(l)
Exposure to carbon monoxide at levels greater than 0.100 g L^–1 of air can be dangerous to human health. 6.0 kg of octane was combusted by the car in this workshop. Using the equation provided, determine if the level of carbon monoxide produce in the workshop would be dangerous to human health. Support your answer with relevant calculations. (4 marks)

This post is a bit old but I'll have a crack at it.

Volume of workshop= 100 000 L
m(octane)= 6000g
n(octane)= m/MM= 6000/114.23= 53.526mol
1:8 mole ratio,
n(CO)= 420.205mol
m(CO)= n*g/mol= 420.205*28.01= 11770g
c(CO)= m/L= 11770/100000= 0.1177g/L
Safe c(CO)= 0.100g/L
Therefore, it is dangerous.
Also because the reaction used up most of the room's oxygen too.