VCAA Chemistry Examination - AN Official Solutions (Draft 1)

[Emily C]

Your MC answers are scaring me

Multiple choice answers are freaking me out too!

[Edward21]

Interested to hear your reasoning for question 13 Thush.

Also for 30, I'd have thought the answer would be D, not C. Given that all three solutions are aqueous, there shouldn't be any metal deposited onto the electrode in the solution containing K+ because H2O would be reduced in preference to K+ for an infinitely long amount of time, producing no metal (K+(aq) is never reduced to K(s))

OMG YES. I was so suspicious of the last MC just because they're usually tricky, and I picked up on the K+ all those practice exams didn't go to waste! 

[charmanderp]

Hey, for qn 30,
is it me or is the answer C?

Since the same current is provided, that means n(e-) is the same throughout.

Let's say the charge equivalence of 1mole of n(e-) is supplied,
then 39.1g of K, 0.5(55.9)g of Fe and 107.9g of Ag?

0g of K is produced. This is because the electrolyte containing K+ is aqueous, meaning it contains water. Water is a stronger oxidant than potassium ions are, and since there is an 'infinite' amount of water in aqueous solutions, all of the electrons supplied end up reducing water until they've been exhausted. Thus no K+ ions are reduced to potassium metal.

[DefyingGravity]

Shouldn't there be 55.9/2g of Fe? Because Fe is 2+ therefore n(e-)=2? Kind of thing... And there is no K+ reduced, therefore the answer is potassium ---> iron ----> silver? Which is D.

[charmanderp]

Shouldn't there be 55.9/2g of Fe? Because Fe is 2+ therefore n(e-)=2? Kind of thing... And there is no K+ reduced, therefore the answer is potassium ---> iron ----> silver? Which is D.

Exactly! Although you can't calculate the exact mass of Fe, because you don't know what current was used exactly. You only know  that the current used for each solution was the same.

According to the respective electron:metal ratios for the reduction of Fe2+ and Ag+ them, you're going to have half the amount of Fe2+ that you do Ag+ holding all else constant. So increasing amount in mole of the three metals is potassium, iron, silver. Since silver is heavier than iron and there is no potassium, in increasing mass it's also potassium, iron, silver.

[sandbum1]

what if you wrote C2O42- (aq) + Ca2+ (aq) --> CaC2O4 (s) ...

[saraoj]

For question 4, doesn't a decreased volume cause a shift to the left and thus decreased hydrogen yield? (Question 4a ii)
For the next part (iii), would an increased temperature not decrease the reaction rate for Reaction 2 because it is exothermic and thus the back reaction (endothermic) is favoured?
Thanks!

anyone?

[drake]

saraoj, an increase in temperature always increases the rate, regardless if endothermic or exothermic!

[Edward21]

saraoj, an increase in temperature always increases the rate, regardless if endothermic or exothermic!

Too true I started off with a yield explanation, but then shifted towards increase in reaction rate for both once I read the question again!!

[saraoj]

Too true I started off with a yield explanation, but then shifted towards increase in reaction rate for both once I read the question again!!

oh my gosh so many silly mistakes, thanks heaps!

[lzxnl]

saraoj, an increase in temperature always increases the rate, regardless if endothermic or exothermic!

For individual reactions, yes, but for anyone who has done the ATAR Notes chemistry exam 4...they'll know what I mean...

[beastly101]

For individual reactions, yes, but for anyone who has done the ATAR Notes chemistry exam 4...they'll know what I mean...

What do you mean? lol

[lzxnl]

For that one, an increase in temperature reduced the rate of the overall reaction.
It was a two-step reaction, and the first step was exothermic and very fast, so it reached equilibrium very. Increasing the temperature wouldn't noticeably increase the rate of that reaction, but it reduced the yield. The second step was rather slow, and the rate increase from the temperature arise apparently did not compensate for the lower yield from reaction 1; one of the products in reaction 1 was a reactant in reaction 2. So here, a temperature increase REDUCED the overall rate.

But that was just one of Thushan's troll questions

[beastly101]

For that one, an increase in temperature reduced the rate of the overall reaction.
It was a two-step reaction, and the first step was exothermic and very fast, so it reached equilibrium very. Increasing the temperature wouldn't noticeably increase the rate of that reaction, but it reduced the yield. The second step was rather slow, and the rate increase from the temperature arise apparently did not compensate for the lower yield from reaction 1; one of the products in reaction 1 was a reactant in reaction 2. So here, a temperature increase REDUCED the overall rate.

But that was just one of Thushan's troll questions

I had reduced for the exothermic one as well.

[thushan]

I'm a trollface.