HSC Chemistry Question Thread

[Calley123]

Hey,
In the 2009 Exam, there was a question on galvanic cells which stated that " One half cell consists of an inert platinum electrode and a solution of Fe (2+) and Fe(3+). The other half cell consists of a lead electrode and a solution of Pb (2+).

The sample answer stated that oxidation occurred with Pb(s) to Pb(2+) and Reduction with Fe(3+) to Fe(2+)

1. Does the solution of Fe(3+) ions and Fe(2+) exist because the inert platinum just sits there?
2. I know that the more reactive metal is supposed to displace the ions of the less reductive ions but in this case how was I suppose to tell which would displace which?

Sorry for the spam before -it was an accident :/

tHnaks

First note that the answers BOSTES/NESA provide are not necessarily of band 6 level. They will just grab any random answer for their sample response.

I would say it is a great idea, because it makes the whole process a lot clearer. At the same time, where possible you could just replace it with some chemical equations.

THANK YOU!

It's not.

You don't have 1/2 Cl_(g). You have 1/2 Cl_2(g), which is half a mole of the chlorine element in the gaseous phase. Note that in the chlorine element molecule, you always have two chlorine atoms, not one.

And then when it gets reduced (i.e. gains electrons), you end up with Cl^-_(aq), which is one mole of the chloride ion. You only have one chloride ion appearing at a time.

Thanks

Merged posts - J41.

[Aaron12038488]

2003 hsc chem- 8. of MC, my answer was 6.18 close to 6.12. When i checked the solutions instead of multilplying the moles with 24.79 (RTP) they used 24.47L?

[Lear]

2003 hsc chem- 8. of MC, my answer was 6.18 close to 6.12. When i checked the solutions instead of multilplying the moles with 24.79 (RTP) they used 24.47L?

If i’m not mistaken these constants change over time due to new definitions. Since the exam in from 2003, the value could have been different.

[RuiAce]

Hey,
In the 2009 Exam, there was a question on galvanic cells which stated that " One half cell consists of an inert platinum electrode and a solution of Fe (2+) and Fe(3+). The other half cell consists of a lead electrode and a solution of Pb (2+).

The sample answer stated that oxidation occurred with Pb(s) to Pb(2+) and Reduction with Fe(3+) to Fe(2+)

1. Does the solution of Fe(3+) ions and Fe(2+) exist because the inert platinum just sits there?
2. I know that the more reactive metal is supposed to displace the ions of the less reductive ions but in this case how was I suppose to tell which would displace which?

Sorry for the spam before -it was an accident :/

tHnaks


THANK YOU!

Thanks

Merged posts - J41.

1. The platinum just sits there but I wouldn't say that the solution of Fe³⁺ and Fe²⁺ exist purely because of that. As dumb as it might seem, I'd say it exists because whoever made the cell poured some of both of the ions in there to begin with.

As for why it stays there (instead of reacting with the Pt for example), well yep that is because platinum is inert.

2. Use the data sheet. In the table of standard reduction potentials, whichever half-equation is higher up will need to be flipped (as it represents oxidation), whereas whichever is lower will represent reduction. the lead one lies above the iron one, so the former will be the one involved in the oxidation process.

2003 hsc chem- 8. of MC, my answer was 6.18 close to 6.12. When i checked the solutions instead of multilplying the moles with 24.79 (RTP) they used 24.47L?

If i’m not mistaken these constants change over time due to new definitions. Since the exam in from 2003, the value could have been different.

Yeah this should be the cause. A lot of minor details got changed in 2010 in the HSC sciences.
_______________________________________

Not too worried this time because both the above questions were fairly brief and clear. In the future, please provide the actual question, or at least link to the past paper itself for convenience.

[Calley123]

1. The platinum just sits there but I wouldn't say that the solution of Fe³⁺ and Fe²⁺ exist purely because of that. As dumb as it might seem, I'd say it exists because whoever made the cell poured some of both of the ions in there to begin with.

As for why it stays there (instead of reacting with the Pt for example), well yep that is because platinum is inert.

2. Use the data sheet. In the table of standard reduction potentials, whichever half-equation is higher up will need to be flipped (as it represents oxidation), whereas whichever is lower will represent reduction. the lead one lies above the iron one, so the former will be the one involved in the oxidation process.Yeah this should be the cause. A lot of minor details got changed in 2010 in the HSC sciences.
_______________________________________

Not too worried this time because both the above questions were fairly brief and clear. In the future, please provide the actual question, or at least link to the past paper itself for convenience.

It kinda makes sense now except how would I have know whether it was supposed to be Fe+2 (lies above lead) or Fe 3+  ( lies below lead)  that was involved in reduction process?

[RuiAce]

The equation \(Fe^{2+} + 2e^- \rightleftharpoons Fe_{(s)} \) was not of interest of us as we do not produce or react iron metal in our cell.

[Calley123]

The equation \(Fe^{2+} + 2e^- \rightleftharpoons Fe_{(s)} \) was not of interest of us as we do not produce or react iron metal in our cell.

I get it
However, does this mean that there is no mass gained on the Pt(s) cathode since there is no solid formed on it? I always thought there was mass gain on the cathode in a galvanic cell...

[RuiAce]

I get it
However, does this mean that there is no mass gained on the Pt(s) cathode since there is no solid formed on it? I always thought there was mass gain on the cathode in a galvanic cell...

Yes, there will be no mass gain.

It's fair enough to say that "usually" there is a gain in mass on the electrodes (and also loss in mass for the other electrode). But this isn't something that always occurs.

[Calley123]

Yes, there will be no mass gain.

It's fair enough to say that "usually" there is a gain in mass on the electrodes (and also loss in mass for the other electrode). But this isn't something that always occurs.

Thank you so much !!!!

[cnimm2000]

Hey guys,
For the nuclear chemistry section, i wanted to know how strontium-90 is produced
thanks in advance. 

[kauac]

Hey guys,
For the nuclear chemistry section, i wanted to know how strontium-90 is produced
thanks in advance.

Hi...
Not 100% sure on this, but I think strontium-90 is produced by bombarding a lighter strontium isotope with neutrons. (I worked this out as the neutron:proton ratio is very high, meaning that this is what made it unstable). The neutrons for this reaction are sourced from a nuclear fission reaction where uranium nuclei are split into neutrons + protons. Hope this is somewhat helpful!

[cnimm2000]

Hi...
Not 100% sure on this, but I think strontium-90 is produced by bombarding a lighter strontium isotope with neutrons. (I worked this out as the neutron:proton ratio is very high, meaning that this is what made it unstable). The neutrons for this reaction are sourced from a nuclear fission reaction where uranium nuclei are split into neutrons + protons. Hope this is somewhat helpful!

Thanks,
Also how do you know if the neutron: proton ratio is high?

[RuiAce]

Thanks,
Also how do you know if the neutron: proton ratio is high?

As a rule of thumb, the n:p ratio should be 1:1 for atoms with smaller atomic number, and as you go all the way to Uranium it should become 1.5:1.

There isn't really any rule of thumb as to when the transition from 1:1 to 1.5:1 starts to occur but you can consider the diagram on Wikipedia as a reference point.

[Mate2425]

Hi could i please have some help with a question in regard to the fermentation of glucose and monitoring mass changes dotpoint.

I was just wondering if you could please help with what the  best way to go about calculating the volume of CO2 released during the fermentation process of mass change 3.33g would be and also what the process of calculations needed to answer this question would look like.

Thank you,
Mate2425 😃

[kiwiberry]

Hi could i please have some help with a question in regard to the fermentation of glucose and monitoring mass changes dotpoint.

I was just wondering if you could please help with what the  best way to go about calculating the volume of CO2 released during the fermentation process of mass change 3.33g would be and also what the process of calculations needed to answer this question would look like.

Thank you,
Mate2425 😃

Hi!! So the equation for fermentation is
The mass change is due to the CO2 released, so 3.33g of CO2 was released. Now that we have the mass, we can find moles using n=m/M:
and then the volume using n=v/MV (assuming this is at 0^oC so MV=22.71, if it's at 25^oC then MV=24.79)
So the volume of CO2 released is 1.72L. Hope that helps!