Chemistry 3/4 2013 Thread

[Holmes]

Some students at my school are saying that the stuff we learnt for one of ammonia, nitric or sulphuric acid isn't applicable as examinable VCAA content, and it was only needed for the sac we did. I've seen how these types of questions are structured in the past chem papers (where you tick a box for your selected chemical), so can it still be examined in this year's paper?

[Jeggz]

Some students at my school are saying that the stuff we learnt for one of ammonia, nitric or sulphuric acid isn't applicable as examinable VCAA content, and it was only needed for the sac we did. I've seen how these types of questions are structured in the past chem papers (where you tick a box for your selected chemical), so can it still be examined in this year's paper?

In the study design it states that this isn't examinable anymore.

End-of-year examination
 
All outcomes in Units 3 and 4 will be examined. All key knowledge that underpins the outcomes in

Units 3 and 4 and the set of key skills listed on page 12 are examinable except:
• specific details related to the study of a selected chemical (one of ammonia, sulfuric acid or nitric acid).

The underlying principles related to factors that affect the rate of chemical reactions and the position
of equilibrium are examinable.

[jgoudie]

There is no hard and fast rule as to which reactant goes int he burette/pipette. 

In general you put the unknown in the conical flask but there is no reason that it can not go the other way around.  It all depends on the situation you are in, some times things like back titration required extra steps to convert your unknown into something so the conical flask is somewhere those extra reactions can happen.

I guess if you are looking at safety, it depends again on what you are handling.  I guess the stuff in the burette should be you least dangerous as it is harder to fill the burette and more likely to spill that stuff, but thats a bit of a stretch.

Overall, it is the circumstance which tells you what goes where, not a overall rule.

Hey guys,

why do we titrate a solution of known concentration (ie. in burette) against one of unknown concentration (and not the other way around)? Apparently it has something to do with safety?

[clıppy]

Just a quick question in regards to what is examinable. Are detailed studies still on the exam this year, or have they been removed?

[lzxnl]

Those are gone. They have to cut something from the exam as VCAA cannot seem to tolerate having a 3 hour exam for chemistry.

[barydos]

Aye yo,

Study design highlights key knowledge including:
'the construction and operation of fuel cells: advantages and disadvantages of fuel cells compared to conventional energy sources;'

Does this mean we have to know the equations and applications of the different fuel cells similar to table 27.1 of the Heinemann textbook (page 438)?

Sorry if it sounds stupid. I just want clarification.
Thanks again

Could anyone clarify this? I, too, would like to know.

[lzxnl]

Could anyone clarify this? I, too, would like to know.

So perhaps the differences between fuel cells and say, burning coal.

[thushan]

So perhaps the differences between fuel cells and say, burning coal.

Fuel cells are more expensive, and in running cars they have to be bloody hot to function (like 1000 centigrade i think?) and hydrogen transport is hazardous at best.

[Holmes]

VCAA 2006

Question 8
The passage of 0.019 faradays of electricity through a molten chromium compound yields 0.50 g of chromium metal.
The oxidation number of chromium in the compound is likely to be
A. +2
B. +3
C. +4
D. +6

Don't understand how to do this one? Thanks

[Alwin]

VCAA 2006

Question 8
The passage of 0.019 faradays of electricity through a molten chromium compound yields 0.50 g of chromium metal.
The oxidation number of chromium in the compound is likely to be
A. +2
B. +3
C. +4
D. +6

Don't understand how to do this one? Thanks

Hint 1: You can find the n(e^-)

Hint 2: You can find the n(Cr) solid

Hint 3: You can then find the ratio of n(Cr):n(e^-)

see what you get + hope it helps

Probably could have been muuch more helpful, so there's the ans if you're still stuck~

LAST RESORT, TRY IT YOURSELF FIRST!!

0.019 Faraday equals 0.019 mol of electrons --> n(e^-) = 0.019 mol
0.50 g of Cr equals 0.50/52 = 0.0096 mol --> n(Cr) = 0.0096 mol

n(Cr) : n(e^-) =  0.0096 : 0.019 = 1 : 2

Thus, for every one mol of Cr that reacted, two mol of electrons were required. This means that the oxidation number changed by 2.
Since the final oxidation number of Cr(s) is 0 and metals only exist as cations in solution/molten, originally the oxidation number of Cr was +2

Hence, I get [A]

[Holmes]

0.019 Faraday equals 0.019 mol of electrons --> n(e^-) = 0.019 mol
0.50 g of Cr equals 0.50/52 = 0.0096 mol --> n(Cr) = 0.0096 mol

n(Cr) : n(e^-) =  0.0096 : 0.019 = 1 : 2

Thus, for every one mol of Cr that reacted, two mol of electrons were required. This means that the oxidation number changed by 2.
Since the final oxidation number of Cr(s) is 0 and metals only exist as cations in solution/molten, originally the oxidation number of Cr was +2

I got all that info, I just had no idea how to use it or what I was doing. So, is this right?
A faraday is the charge on 1 mol of electrons. Therefore, 0.019 F is the charge on 0.019 mol of electrons. The reaction yields 0.0096 moles of solid chromium. So the ratio of n(e-) : n(Cr) is 2:1. The final oxidation number of the solid is 0. Chromium is a metal, so it exists as a cation. I don't understand the final leap in logic, as to why the 2:1 ratio indicates a 2+ ion. Is it because the equation would be Cr2+ + 2e- ---> Cr, hence a 2 + ion, and hence a 2+ oxidation number?

Sorry for dragging on the question, just want to understand it completely. Thanks for the help

[teletubbies_95]

Hey guys!
So I'm so confused about the electrolysis part of the hydrogen oxygen fuel cell( Pg 196 of the Heinemann Student Workbook) . Why do we have to do electrolysis and how do we find the electrolysis reactions?
I know this is a dumb question , but I can't see why we have to do it, other than finding the volume of gas produced in the reaction.

Thank you so much !

[Alwin]

I got all that info, I just had no idea how to use it or what I was doing. So, is this right?
A faraday is the charge on 1 mol of electrons. Therefore, 0.019 F is the charge on 0.019 mol of electrons. The reaction yields 0.0096 moles of solid chromium. So the ratio of n(e-) : n(Cr) is 2:1. The final oxidation number of the solid is 0. Chromium is a metal, so it exists as a cation. I don't understand the final leap in logic, as to why the 2:1 ratio indicates a 2+ ion. Is it because the equation would be Cr2+ + 2e- ---> Cr, hence a 2 + ion, and hence a 2+ oxidation number?

Sorry for dragging on the question, just want to understand it completely. Thanks for the help

Nah! its okay, good to get thorough understanding

Faraday's second law:
in order to produce one mole of a metal, one, two, three, or another whole number of moles of electrons must be consumed. This is a modernised statement of Faraday’s second law of electrolysis.
In equation form:
Mⁿ⁺(aq) + n e^- --> M(s)  for aqueous / molten to solid.
It could be a change in oxidation eg Fe3+ to Fe2+ but since we have a solid product in this reaction, it is not this particular example.

So yes, you're right it is Crⁿ⁺(aq) + 2 e^- --> Cr(s)  so n = 2

Hope it makes more sense! sounds like you've pretty much got it down pat ^^

Hey guys!
So I'm so confused about the electrolysis part of the hydrogen oxygen fuel cell( Pg 196 of the Heinemann Student Workbook) . Why do we have to do electrolysis and how do we find the electrolysis reactions?
I know this is a dumb question , but I can't see why we have to do it, other than finding the volume of gas produced in the reaction.

Thank you so much !

Sorry teletubbies, but I'm at a tad busy atm. Best I can do is this:

A previous post

I just have some electrolysis questions which I haven't quite understood yet:
(Image removed from quote.)
electrochemical series:

Spoiler

answers:

Spoiler

Thanks in advance!

7a. I'll go through the process with you step by step first so hopefully it makes sense, then you can go off and try the rest yourself! If you have any more problems, don't hesitate to post them ^^
Copper(II) Bromide (aq) --> Cu²⁺ and Br^- ions present in the solution. However, as we are dealing with electrolysis we cannot disregard water, H₂O.
So, on the electrochemical series we mark out Cu²⁺, Br^- and H₂O where they react as oxidants or reductants

STEP ONE

Now, we find the strongest oxidant and the strongest reductant as they will react together!

Mark it out on the electrochemical series:

STEP TWO

Now, we can write out the half equations:
Anode (+): 2Br^- (aq) --> Br₂ (aq) + 2e^-
Cathode (-): Cu²⁺ (aq) + 2e^- --> Cu (s)

We now have our products, Br₂ (aq) and Cu (s). You just need to go through the same process over and over ^^

At first you might need to print out some extra electrochemical series(s) to draw all over, but you should get better very quickly if you work at it!
Hope this made sense + good luck!!

EDIT: Always forget to put my pics in spoilers. #myamazingpaintskills LOL

In your case, water is both the strongest oxidant and reductant iirc that prac correctly. Adding KCl doesn't matter in this case ^^

Hope it helps you understand the process!

[lzxnl]

Hey guys!
So I'm so confused about the electrolysis part of the hydrogen oxygen fuel cell( Pg 196 of the Heinemann Student Workbook) . Why do we have to do electrolysis and how do we find the electrolysis reactions?
I know this is a dumb question , but I can't see why we have to do it, other than finding the volume of gas produced in the reaction.

Thank you so much !

Why you need electrolysis? Where are you going to get the hydrogen for the fuel cell? Although I don't think electrolysis is strictly part of the fuel cell.
How to find them...you have water, and you're going to somehow react that to form hydrogen and oxygen.
The electrochemical series has the equations for you; 2H2O => O2 + 4H+ + 4e- and 2H+ + 2e- => H2
Note how any H+ formed in the first reaction is consumed by the second.
As for why you need them, practice?

[teletubbies_95]

Thanks guys! I was confusing myself and overthinking the process too much