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HSC 2012-2015 Chemistry Marathon (archive) (3 Viewers)

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HeroicPandas

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re: HSC Chemistry Marathon Archive

Haber process: N2(g)+3H2(g)->2NH3(g) delta(H)=-92 kJ/mol
Since nitrogen on the left hand side of the equation is in its elemental form its oxidation state is zero as is the oxidation state of hydrogen for the same reason. However, on the right hand side hydrogen has an oxidation state of +1 since it is part of a compound and nitrogen has an oxidation state of -3 in order to allow for a neutral NH3 molecule. Hence, redox has occurred and the Haber process is an electron-transfer reaction.
Oxidation half-equation: 3H2(g)->6(H+)+6e
Reduction half-equation: N2(g)+6e->2N^(3-)
Wasn't sure what to use as the states for the ions.
VERY NICE!

Be careful: Haber process is a reversible reaction and also include a catalyst (Fe3O4)
 

HeroicPandas

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Yeah thanks.

New question: Why is polystyrene rigid?
Flexibility of a polymer is determined by the side chains of its monomers. PS is made up of a monomer styrene whcich has a bulky benzene side chain. The bulky benezene side chain greatly reduces the flexibility, making PS inflexible and increasing rigidity. The total molecular weight of PS, results in stronger dispersion frces which accounts for its hardness rigidity and toughness

lol i need to revise module 1

Question: (module 1)

Things needed for a particular galvanic cell: magnesium solid, magnesium sulfate solution, carbon (graphite rods), iron(II) sulfate, iron(III) sulfate, filter paper, 2 beakers, voltmeter

a) Identify the anode and cathode (2 marks)
b) Explain what a salt bridge does. Identify a suitable substance for a salt bridge and justify your choice (3 Marks)
c) A student suggests to replace the graphite electrode rods for lead. Explain why this is unsuitable. (2 marks)
d) This galvanic cell was placed in someplace safe and secure for a very long time (3 months). There were MANY qualitative and quantitative changes. Predict 4 possible changes and explain them (8 marks)
 
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bleakarcher

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Flexibility of a polymer is determined by the side chains of its monomers. PS is made up of a monomer styrene whcich has a bulky benzene side chain. The bulky benezene side chain greatly reduces the flexibility, making PS inflexible and increasing rigidity. The total molecular weight of PS, results in stronger dispersion frces which accounts for its hardness rigidity and toughness

lol i need to revise module 1

Question: (module 1)

Things needed for a particular galvanic cell: magnesium solid, magnesium sulfate solution, carbon (graphite rods), iron(II) sulfate, iron(III) sulfate, filter paper, 2 beakers, voltmeter

a) Identify the anode and cathode (2 marks)
b) Explain what a salt bridge does. Identify a suitable substance for a salt bridge and justify your choice (3 Marks)
c) A student suggests to replace the graphite electrode rods for lead. Explain why this is unsuitable. (2 marks)
d) This galvanic cell was placed in someplace safe and secure for a very long time (3 months). There were MANY qualitative and quantitative changes. Predict 4 possible changes and explain them (8 marks)
a) The anode is the magnesium piece of metal and the cathode is the carbon (graphite) rod.
b) A salt bridge is required to complete the circuit in a galvanic cell and to maintain electrical neutrality in the electrolyte solution of each half cell allowing for a continuous flow of current through the galvanic circuit. Aqueous KNO3 would be a suitable solution in which to immerse the salt bridge (made from filter paper) as it will not form precipitates in the electrolyte solutions and therefore allow for electrical neutrality by transferring positive K+ ions to the FeSO4/Fe2(SO4)3 solution to negate the buildup of positive charge due to a decrease in the concentration of positive Fe^(2+)/Fe^(3+) ions and transferring NO3- ions to the MgSO4 solution to negate the buildup of positive charge due to an increase in the concentration of positive Mg^(2+) ions.
c) Not sure.
d) Oxidation half-equation: Mg(s)->Mg^(2+)(aq)+2e
Reduction half-equation(s): Fe(2+)(aq)+2e->Fe(s), Fe^(3+)(aq)+3e->Fe(s)
Observed changes after three months:
- Decolourisation of the FeSO4/Fe2(SO4)3 solution due to the discharge (reduction) of Fe(3+)/Fe(2+) ions from solution
- Decrease in the mass and size of the magnesium electrode (anode) due to the oxidation of magnesium as shown by the oxidation half-equation above.
- Deposit of a brown metal on the carbon rod due to the formation and subsequent corrosion of iron in the presence of water and oxygen.
- Sedimentation of iron hydroxide (insoluble) at the bottom of the container originating from the corrosion of the iron formed in the reduction half-equation.
Corrosion of iron
Oxidation half-equation: Fe(s)->Fe^(2+)(aq)+2e
Reduction half-equation: O2(g)+2H2O(l)+4e->4OH-(aq)
Formation of insoluble Fe(OH)2: Fe^(2+)(aq)+2OH-(aq)->Fe(OH)2(s)
 

HeroicPandas

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a) The anode is the magnesium piece of metal and the cathode is the carbon (graphite) rod.
b) A salt bridge is required to complete the circuit in a galvanic cell and to maintain electrical neutrality in the electrolyte solution of each half cell allowing for a continuous flow of current through the galvanic circuit. Aqueous KNO3 would be a suitable solution in which to immerse the salt bridge (made from filter paper) as it will not form precipitates in the electrolyte solutions and therefore allow for electrical neutrality by transferring positive K+ ions to the FeSO4/Fe2(SO4)3 solution to negate the buildup of positive charge due to a decrease in the concentration of positive Fe^(2+)/Fe^(3+) ions and transferring NO3- ions to the MgSO4 solution to negate the buildup of positive charge due to an increase in the concentration of positive Mg^(2+) ions.
c) Not sure.
d) Oxidation half-equation: Mg(s)->Mg^(2+)(aq)+2e
Reduction half-equation(s): Fe(2+)(aq)+2e->Fe(s), Fe^(3+)(aq)+3e->Fe(s)
Observed changes after three months:
- Decolourisation of the FeSO4/Fe2(SO4)3 solution due to the discharge (reduction) of Fe(3+)/Fe(2+) ions from solution
- Decrease in the mass and size of the magnesium electrode (anode) due to the oxidation of magnesium as shown by the oxidation half-equation above.
- Deposit of a brown metal on the carbon rod due to the formation and subsequent corrosion of iron in the presence of water and oxygen.
- Sedimentation of iron hydroxide (insoluble) at the bottom of the container originating from the corrosion of the iron formed in the reduction half-equation.
Corrosion of iron
Oxidation half-equation: Fe(s)->Fe^(2+)(aq)+2e
Reduction half-equation: O2(g)+2H2O(l)+4e->4OH-(aq)
Formation of insoluble Fe(OH)2: Fe^(2+)(aq)+2OH-(aq)->Fe(OH)2(s)
very nice, well done!
a) Correct
b) Correct
c) Lead mmm, there is something about lead.... here is a hint: unleaded fuel and what if i put carbon and lead in a solution of HCl
d) First 2 points are correct
Next 2 points...how do u produce Iron solid? it wasnt present at the beginning (reduction of Fe(2+)? where is the supply of electrons? oxidation of water?)

Some other predictions for part (d): (in the colour white)
No more voltage, concetration of the anolyte has increased, concentrating\\on of the the catholyte has decreased, salt bridge at very low (or no?) concetration of KNO3
 
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bleakarcher

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very nice, well done!
a) Correct
b) Correct
c) Lead mmm, there is something about lead.... here is a hint: unleaded fuel and what if i put carbon and lead in a solution of HCl
d) First 2 points are correct
Next 2 points...how do u produce Iron solid? it wasnt present at the beginning (reduction of Fe(2+)? where is the supply of electrons? oxidation of water?)

Some other predictions for part (d): (in the colour white)
No more voltage, concetrating of the anolyte has increased, concentrating of the the catholyte has decreased, salt bridge at very low (or no?) concetration of KNO3
I couldn't think of any other points so I just put down ones according to my option topic Shipwrecks and Corrosion lol.
 

HeroicPandas

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I couldn't think of any other points so I just put down ones according to my option topic Shipwrecks and Corrosion lol.
haha, do u want more time to do (c)? (BTW anyone is free to do this)

Btw ur part (d), can u explain where did the iron solid originate from? (u need some iron for rusting or else, no rusting will occur)
 

bleakarcher

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haha, do u want more time to do (c)? (BTW anyone is free to do this)

Btw ur part (d), can u explain where did the iron solid originate from? (u need some iron for rusting or else, no rusting will occur)
I thought the iron came from the Fe^(2+)(aq)+2e->Fe(s).
 

bleakarcher

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where was the supply of electrons? from the oxidation of water?
From the formed iron Fe(s)->Fe^(2+)(aq)+2e.
I showed it in my previous answer here:
Corrosion of iron
Oxidation half-equation: Fe(s)->Fe^(2+)(aq)+2e
Reduction half-equation: O2(g)+2H2O(l)+4e->4OH-(aq)
 

HeroicPandas

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From the formed iron Fe(s)->Fe^(2+)(aq)+2e.
I showed it in my previous answer here:
Corrosion of iron
Oxidation half-equation: Fe(s)->Fe^(2+)(aq)+2e
Reduction half-equation: O2(g)+2H2O(l)+4e->4OH-(aq)
no....

Fe(s) --> Fe(2+) + 2e- only occurs when there is the presence of IRON SOLID

There was no iron solid present at the beginning, unless Fe(2+) is reduced, but where was the supply of electrons for Fe(2+) to be reduced? (it cant be from Fe(s) --> Fe(2+) +2e because there was no Iron SOLID to begin with)
 

bleakarcher

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no....

Fe(s) --> Fe(2+) + 2e- only occurs when there is the presence of IRON SOLID

There was no iron solid present at the beginning, unless Fe(2+) is reduced, but where was the supply of electrons for Fe(2+) to be reduced? (it cant be from Fe(s) --> Fe(2+) +2e because there was no Iron SOLID to begin with)
At the anode: Mg(s)->Mg^(2+)(aq)+2e
The electrons liberated through the oxidation of the magnesium electrode migrate to the cathode through the external wire where they reduce the Fe^(2+) ions in solution.
 

HeroicPandas

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Fe(2+) + 2e --> Fe(s) E= -0.44V (not spontaneous)

Versus

2Fe(3+) + 2e- --> 2Fe(2+) E = 0.77V (spontaneous)

These 2 reactions may occur, but since Fe(3+) was a greater tendency to be reduced, it will take in the electrons before Fe(2+) tries to take it (this is due to electron shells and becoming a stable atom)

Agree or disagree?

therefore, Fe(2+) + 2e --> Fe(S) does not occur
 

bleakarcher

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Fe(2+) + 2e --> Fe(s) E= -0.44V (not spontaneous)

Versus

2Fe(3+) + 2e- --> 2Fe(2+) E = 0.77V (spontaneous)

These 2 reactions may occur, but since Fe(3+) was a greater tendency to be reduced, it will take in the electrons before Fe(2+) tries to take it (this is due to electron shells and becoming a stable atom)

Agree or disagree?

therefore, Fe(2+) + 2e --> Fe(S) does not occur
Yeah, that makes sense. I just assumed the first half-equation was preferential. Thanks man.
 

bleakarcher

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very nice, well done!
a) Correct
b) Correct
c) Lead mmm, there is something about lead.... here is a hint: unleaded fuel and what if i put carbon and lead in a solution of HCl
d) First 2 points are correct
Next 2 points...how do u produce Iron solid? it wasnt present at the beginning (reduction of Fe(2+)? where is the supply of electrons? oxidation of water?)

Some other predictions for part (d): (in the colour white)
No more voltage, concetration of the anolyte has increased, concentrating\\on of the the catholyte has decreased, salt bridge at very low (or no?) concetration of KNO3
lol I just saw the white texts. Question though man, are those actual observations, besides the first one to do with voltage? Also, why exactly is there no voltage (I have an idea just want to know what you think)?
 

HeroicPandas

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Yeah, that makes sense. I just assumed the first half-equation was preferential. Thanks man.
alright cool! :)

Answer for part (c) if u wanna know: (in the colour white)
-Lead a toxic heavy metal
-Lead is not inert in this solution:
Pb(2+) + 2e --> Pb(s) E = -0.13V
2Fe(3+) + 2e --> 2Fe(2+) E = 0.77V

.: Pb(s) --> Pb(2+) + 2e E = 0.13V
2Fe(3+) + 2e --> 2Fe(2+) E = 0.77V

Positive EMF, means a spontaneous reaction and so this reaction WILL occur at the cathode compartment, therefore Pb will be an unsuitable electrode. ALSO electrons liberated by the magnesium will be absorbed by the lead cathode which will trigger another reaction, we only want to look at 1 reaction
 
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HeroicPandas

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lol I just saw the white texts. Question though man, are those actual observations, besides the first one to do with voltage? Also, why exactly is there no voltage (I have an idea just want to know what you think)?
The question states:
"There were MANY qualitative and quantitative changes. "

The student will measure everything (concentration of salt bridge, anolyte, catholyte) - these are quantitative changes (a quantitative observation)

About voltage, the salt bridge works hard during the 3months, but soon, so much of the anode (magnesium) has been oxidised into Mg(2+) and the salt bridge's negative ions (NO3-) are out (magnesium ions may choose to go in the salt bridge if they like).

For cathode compartment, same thing

If the salt bridge fails, the current cannot be maintained, thus the galvanic cell will fail (no more voltage)

if i have made a mistake, please tell me
 
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bleakarcher

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The question states:
"There were MANY qualitative and quantitative changes. "

The student will measure everything (concentration of salt bridge, anolyte, catholyte) - these are quantitative changes

About voltage, the salt bridge works hard during the 3months, but soon, so much of the anode (magnesium) has been oxidised into Mg(2+) and the salt bridge's negative ions (NO3-) are out (magnesium ions may choose to go in the salt bridge if they like).

For cathode compartment, same thing

If the salt bridge fails, the current cannot be maintained, thus the galvanic cell will fail (no more voltage)

if i have made a mistake, please tell me
Yeah, I need to read the question more carefully. Also, I get what you're saying man and I understand there would be no current but why would there be no voltage? Wouldn't there still be a EMF between the two cells? Or actually, is there no voltage because the instantaneously induced EMF is immediately negated by the electric field produced as a result of the electrolyte solutions no longer being electrical neutral (since there is no working salt bridge)?
 

HeroicPandas

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Yeah, I need to read the question more carefully. Also, I get what you're saying man and I understand there would be no current but why would there be no voltage? Wouldn't there still be a EMF between the two cells? Or actually, is there no voltage because the instantaneously induced EMF is immediately negated by the electric field produced as a result of the electrolyte solutions no longer being electrical neutral (since there is no working salt bridge)?
I think my udnerstanding of EMF has holes in it, and now i've lost my concentration and im too confused lol

'electric field' what? lol - i dont know the actual definition and what causes it (does it mean electricity between 2 ions?)

Does cations or anions conduct electricity? (or only 1 does?)
 
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bleakarcher

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I think my udnerstanding of EMF has holes in it, and right i lost my concentration and im too confused lol

'electric field' what? lol

Does cations or anions conduct electricity? (or only 1 does?)
Sorry man. Say you have a galvanic cell with no salt bridge. Wouldn't you still have an instantaneous EMF (essentially a voltage) between the two cells until almost immediately this EMF is negated by an electric field as positively charged ions begin to dominate in the anolyte and negatively charged ions begin to dominate in the catholyte? I doubt this is much clearer.

In this example, electricity is conducted through the external wire. We're obviously drifting off the chemistry lol, but I just want to know.
 
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