CHEM thread.

[Hutchoo]

Can someone help me understand these MCQ =)?

[Hutchoo]

And 2009 VCAA Questions 5 and 11 MCQ.

Thanks in advance.

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2009chem2-w.pdf

[sin0001]

Alright, question 11 (mcq):
n(CH4)= 1 mol & n(O2)= 0.5 mol. Therefore CH4 is limiting reagent, so use the 'n(CH4)' to calculate energy RELEASED, so we can see that 1 mol of CH4 releases 889 kJ, so since we have 1 mol of CH4, 889 kJ is released (D)
Question 6:
[H+] = 10^(-4.9) = 1.26 x 10^-5, [CN-]=[H+]=1.26 x 10^-5
Therefore, Ka=([H+][CN-])/[HCN] = 6.3 x 10^-10
So (1.26 x 10^-5)^2 / [HCN] = 6.3 x 10^-10   
Therefore, [HCN] = 0.25 M (Hence C)
Question 18:
No zinc is deposited at anode (eliminate D)
Overall equation is Cu2+  +  Zn ---> Zn2+  + Cu, so you can see that mass of copper deposited (produced) will be same as mass of Zinc lost at anode (Hence A)
Question 20:
No. of Faradays is basically n(e-) required
So n(Cl2) = 2.24 L / 22.4 L per mol = 0.1 mol
Remember that n(Cl2) produced = 1/2 x n(e-) used. So n(e-)/no. of faradays = 0.2 mol (Hence C)

VCAA MCQ 5:
Initial value of concentration fraction is 800, so Q < K, therefore equilibrium needs to shift backwards, using up Hydrogen Fluoride (HF). So we can see that [HF]eq. < [HF]initial (0.400M). So only answer that suits this is D.

[daniel034]

Q11 on your picture => 16g of O2 which is the limiting reagent n(O2)= 16/32 = 0.5mol as it takes 2 mols of O2 for every 1 mol of methane only 0.25 mol of CH4 will react therefore 889*0.25 kJ will be released = B

[daniel034]

2009 Q11 Shown by polarity of electrodes
Cd is oxidised by Co2+ therefore Cd is below Co and Co is oxidised by Mn3+ so the order from strongest to weakest oxidant goes Mn3+ > Co2+ > Cd2+

Therefore Mn2+ will not react spontaneously with any of them because it can't form a negative gradient on the electrochemical series so the answer is D

Hope that helps

[Hutchoo]

Hmm, I don't get Q7 2008 exam 2.

I know it's a simple question, but I thought that because Nickle has a higher voltage (from data book) than the Cd, that it would be the stronger oxidant, meaning the ni would be the anode.

I drew the arrow to the right, but it's actually to the left. D=

Someone please explain!


http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2008chem2-w.pdf

[charmanderp]

Hmm, I don't get Q7 2008 exam 2.

I know it's a simple question, but I thought that because Nickle has a higher voltage (from data book) than the Cd, that it would be the stronger oxidant, meaning the ni would be the anode.

I drew the arrow to the right, but it's actually to the left. D=

Someone please explain!


http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2008chem2-w.pdf

Anode is oxidation. The anode is where the strongest reductant is oxidised. In this case cadmium is the stronger reductant so it reacts at the anode.

[Hutchoo]

Ah, I understand that, thanks.

Can you explain question 8ai/ii?

[Somye]

Refer to your databook to find strongest oxidant/reductant with species given in the fuel cell

In this case, H2 is the strongest reductant, and O2 is the strongest oxidant (in a acid electrolyte)

so H2 is oxidised (at anode) to H+ and O2 reduced to H20 (at cathode)-

[Hutchoo]

Refer to your databook to find strongest oxidant/reductant with species given in the fuel cell

In this case, H2 is the strongest reductant, and O2 is the strongest oxidant (in a acid electrolyte)

so H2 is oxidised (at anode) to H+ and O2 reduced to H20 (at cathode)-

OMG LOL. I feel like the biggest dumbass. THANKS DUDE!~

[charmanderp]

Just do it systematically. Somye and I have both laid out our orders of thinking and they're pretty much foolproof. Just go through that every time you do a redox question and you can't go wrong.

[Hutchoo]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem2-w.pdf

VCAA 2011 - Question 10 MC.

I read through VCAA's solution, but I still don't get it.

Thanks in advance.

[charmanderp]

First we look at the case on the left hand side. E1 is connected to the positive electrode, the cathode, indicating that it is being reduced. E3 is connected to the negative electrode, the anode, indicating that it is being oxidised.

So what we know is that E1 is oxidising E3, hence E1 is the stronger oxidant and thus has the higher standard electrode potential.

Now we move across to the case on the right hand side, where we have an electrochemical cell made up of an E1 half cell and an E2 half cell.

E1 is connected to the negative electrode, the anode, indicating that is is being oxidised. E2 is connected to the positive electrode, the cathode, indicating that is is being reduced.

So what we know is that E2 is oxidising E1, hence E2 is the stronger oxidant and thus has the higher standard electrode potential.

So from case 1: E1 has a higher standard electrode potential than E3.
From case 2: E2 has a higher standard electrode potential than E1.

Let's construct a small electrochemical series:
E2
E1
E3

Hence the values of the electrode potentials in order from highest to lowest is: E2, E1, E3, which corresponds with option C.

[Hutchoo]

That's so damn clear, I finally understand what I was doing wrong!
Thanks!

[charmanderp]

Like I said earlier, split all redox questions up into this exact system of thinking and you cannot go wrong.