OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

For the question about acidic or alkaline at 60'C with a pH of 7, did no one use 10^-7 = [H+]? Because that statement has to true by definition doesn't it? So I then divided the Kw at 60'C, (which was 9.4 blah blah x 10^-14) by 10^-7, giving 9.4 blah blah x 10^-7 for [OH-] then I said that [H+] > [OH-] so the solution is acidic. Sorry to go against the flow, but my method makes complete sense to me and I think it's right, if anyone can show me why that's wrong, please go ahead (or if you think I'm talking sense please back me up lol)

Reply 3441

MrZebraCookie

2

Original post by ImNervous

Could you not react cr3+ with any thing below it when it came to showing it was reducing agent?

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lol yeah, i wrote like 6 equations for that one :/
I also reacted the Cr2O7 with the Cr3+ woops

Reply 3442

username1155040

5

Original post by _NMcC_

Yeah same, the OH- concentration was higher than the H+ concentration so it seemed logical to say that the the solution was Alkaline.

Isn't it neutral as pH is a measure of h+ conc which stays the same?

Reply 3443

postexamtalk

12

Original post by maisym00

For the equilibrium question I put: I- reacts with Ag to form agI (yellow ppt), therefore the position of equilibrium will shift to the left to restore I- conc. As a result, more aqueous iodine was formed so the position of equilibrium of 1 will also shift to the left, therefore more grey/black solid iodine is formed. For the electrochemistry question I needed up with two equations. The oxidising agent one where Cr+3 reacts with AL and the reducing agent one where Cr+3 reacts with feO4- or something. Anyone else?

I got the same except i left out the observations, did it explicitly ask you to mention what you observe in the question?

Reply 3444

MrZebraCookie

2

Original post by k.russell

For the question about acidic or alkaline at 60'C with a pH of 7, did no one use 10^-7 = [H+]? Because that statement has to true by definition doesn't it? So I then divided the Kw at 60'C, (which was 9.4 blah blah x 10^-14) by 10^-7, giving 9.4 blah blah x 10^-7 for [OH-] then I said that [H+] > [OH-] so the solution is acidic. Sorry to go against the flow, but my method makes complete sense to me and I think it's right, if anyone can show me why that's wrong, please go ahead (or if you think I'm talking sense please back me up lol)

10^-7<9.4*10^-7

Reply 3445

username2640523

8

Original post by ImNervous

Could you not react cr3+ with any thing below it when it came to showing it was reducing agent?

Posted from TSR Mobile

No because the Cr2+ would be oxidised and you were strictly looking at cr3+

There were two reactions

1 and 2 Where cr3+ is reduced to 2+
6 and 7(Bottom two) Where Cr3+ is oxidised to Cr6+

Posted from TSR Mobile

Reply 3446

four4

2

For the 6 marker on redox,
Would it be 2 marks for definitions
2 marks for equations and 2 marks for explaining?

For some reason i did overall equations with each redox system and mentioned 2 where Cr is being oxidised and reduced- didnt mention anything positive/negative eletrode potential of the cells.

Reply 3447

k.russell

13

Original post by MrZebraCookie

10^-7<9.4*10^-7

yep, just realised that myself lol thanks anyway
ffs oh well probs only -1

Reply 3448

ennha

Original post by HFancy1997

It was at 60 degrees C not 25, you dont use 25 degrees C Kw which is 10^-14, you use the one it gave in the question as it said 60 degrees.Think the answer was 10.76

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Think I missed the 60 degrees for this part! Where did it say this happened at 60 degrees? Also, how many marks was this question?

Reply 3449

tomlam

15

Original post by postexamtalk

I got the same except i left out the observations, did it explicitly ask you to mention what you observe in the question?

It was to do with the Iodide ions, either brown to colourless or colourless to brown.

Reply 3450

a209150

7

Posted from TSR Mobile

1466595087864.jpg34.5KB

Reply 3451

maisym00

8

Original post by MrZebraCookie

lol yeah, i wrote like 6 equations for that one :/
I also reacted the Cr2O7 with the Cr3+ woops

I don't think you could.. CR3+ needs to act as a reducing agent meaning it needs to be on the RHS on the equation, the key was to use the crO7/cr3+ equation and reverse it, meaning the only thing it could have reacted with was feO4

Reply 3452

cr7alwayz

8

Original post by HFancy1997

Thats correct method but x=4

Posted from TSR Mobile

I must have typed the wrong numbers into my calculator at some point but hopefully I got 3 or 4/5 marks.

Reply 3453

lai812matthew

10

what the f is a standard form??????????

Reply 3454

postexamtalk

12

Original post by Brandoncarlin

What I got for the 4-step mechanism:

Overall equation: H2O2 + 2I- + 2H+ forms I2 + 2H2O

H2O2 + I- forms IO- + H2O
IO- + H+ forms HIO
HIO + I- forms I2 + OH- (given in question)
OH- + H+ forms H2O

H2O2 + I- forms HIO + OH-
H+ + OH- forms H2O
HIO + I- forms I2 + OH- (given)
H+ + OH- forms H2O

Both valid?

(edited 7 years ago)

Reply 3455

k.russell

13

Original post by Brandoncarlin

What I got for the 4-step mechanism:

Overall equation: H2O2 + 2I- + 2H+ forms I2 + 2H2O

H2O2 + I- forms IO- + H2O
IO- + H+ forms HIO
HIO + I- forms I2 + OH- (given in question)
OH- + H+ forms H2O

yeah boi! Hope we're right as this does match the overall...

Reply 3456

screwedmofo

1

guys vote now on how you think the paper was.

http://www.thestudentroom.co.uk/showthread.php?t=4180760

Reply 3457

cr7alwayz

8

For using a calculation to explain why K2SO4 can dissolve at 25oC was the entropy change 0.02 or something which is positive so the reaction is feasible?