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Hi

Question, why in this lattice enthalpy cycle do you not multiply the atomisation of fluorine by 0.5 if it's going from 1/2F2(g) ---> F(g)

http://ocr.org.uk/Images/131289-question-paper-unit-f325-equilibria-energetics-and-elements.pdf
Slightly confused about the mark scheme for question 1 d in the June 2012 paper.

I wrote F- ions are more attracted to Mg2+ ions than Na+ ions hence has a greater lattice enthalpy. And Na+ ions are more attracted to F- than Cl-, and hence NaF has a greater lattice enthalpy than NaCl.

Which would explain the pattern in lattice enthalpies.

However, the mark scheme just says F- has greater attraction for Na+ (but surely it'll have greater attraction for Mg2+).
Also Mg2+ has greater attraction for F- ions, however this doesn't explain why MgF2 is the most exothermic :s

Note; I got the two earlier marking points correct so I've already talked about charge density etc.
Reply 2422
can someone help:

when you have the cells questions and they ask for strongest oxidising agent or reducing agent... weakest oxidising agent and weakest reducing agent-- how do you decide ?

thank you :smile:.xx
Original post by LegendX
No, it's askin for the pH of the buffer solution, ur method is invalid and is disregards the the conjugate base formed due to the salt.


Which paper again?
Do we need to remember these half equations:
Mn2+
Hydrogen fuel half cell
thiosulfate
or any other ones

Someone please answer.
Original post by FailedMyExams
Hi

Question, why in this lattice enthalpy cycle do you not multiply the atomisation of fluorine by 0.5 if it's going from 1/2F2(g) ---> F(g)

http://ocr.org.uk/Images/131289-question-paper-unit-f325-equilibria-energetics-and-elements.pdf


because the figure is to do with what you're making. you're only making one F atom so you leave it. if say, you were making 2Cl, you would x2.
Original post by X44
Yes my point legendx can you answer please?


Posted from TSR Mobile


Gosh, i hate explaining over a forum but here goes:

You're reacting NaOH ( a base ) with an acid to form a salt, so you must do the moles of the acid - the moles of the naoh to see how many moles of acid are left over.
The mol of NaOH reacted = moles of the salt formed and you should know from your chemistry that the salt is going to decompose to form conjugate base and sodium ( this is where the large conjugate base reservoir comes from ) however the final step is realised that 50cm3 of volume came from each mixture so the total volume of the solution is 100cm3, convert that into dm3 and then work out the concentrations and then basically plug it into the formula :smile:
Original post by rukan
can someone help:

when you have the cells questions and they ask for strongest oxidising agent or reducing agent... weakest oxidising agent and weakest reducing agent-- how do you decide ?

thank you :smile:.xx


Strongest oxidising agent is the most positive electrode potential as the most positive will be the best at oxidising and also easily reduced. Similarly, strongest reducing agent is the one that is the most negative and is best oxidised.
my worst thing is the long-winded calculations, particularly to do with redox titrations and neutralisations :// is there anywhere online with some of these to practise (not past paper ones?)
Original post by LegendX
Hi otrivine, it was just question 8 in jan 2013 which we were talking about


are you doing the exam tomorrow?
Original post by rukan
can someone help:

when you have the cells questions and they ask for strongest oxidising agent or reducing agent... weakest oxidising agent and weakest reducing agent-- how do you decide ?

thank you :smile:.xx


Oxidising agent is the substance reduced and takes electrons from the substance being oxidised, hence a good oxidising agent takes electrons.

A reducing agent is the substance being oxidised and provides electrons for the substance being reduced, hence BAD reducing agent would suck at giving electrons.

Therefore, the best oxidising agent in a list would be the same as the worst reducing agent.

Vice versa for the weakest oxidising agent = best reducing agent.
Original post by samzurai
Equilibrium moves to the left in a buffer soloution in addition of acid to reduce the h+ ions so position always moves to the left so minimise h ions produced by Acid


Thanks, I think what was getting confused about is the fact that when you add H+, the E shifts left to get rid of extra H+, but then I was also thinking that surely the system would 'want' to shift back right because the conc. of conjugate base has decreased.

I guess, though, that the H+ added is far more significant because you also have the conjugate base produced by the salt's dissociation, and the acid wouldn't of dissociated that much anyway.

:smile:
Reply 2432
Original post by _HabibaH_
Strongest oxidising agent is the most positive electrode potential as the most positive will be the best at oxidising and also easily reduced. Similarly, strongest reducing agent is the one that is the most negative and is best oxidised.


thank you :smile: so what about weakest reducing agent and weakest reducing agent...?
Quick question:

When you are told to identify acid 1, base 1, acid 2 and base 2, how do you know from the LHS of the equilibrium, which to choose as the acid and which as the base?
Reply 2434
Original post by eggfriedrice
Slightly confused about the mark scheme for question 1 d in the June 2012 paper.

I wrote F- ions are more attracted to Mg2+ ions than Na+ ions hence has a greater lattice enthalpy. And Na+ ions are more attracted to F- than Cl-, and hence NaF has a greater lattice enthalpy than NaCl.

Which would explain the pattern in lattice enthalpies.

However, the mark scheme just says F- has greater attraction for Na+ (but surely it'll have greater attraction for Mg2+).
Also Mg2+ has greater attraction for F- ions, however this doesn't explain why MgF2 is the most exothermic :s

Note; I got the two earlier marking points correct so I've already talked about charge density etc.

I found comparing the cations with the cations only and the Anions only as a good rule of thumb, with the charge most overruling and the f- ions is smaller than cl
Original post by otrivine
are you doing the exam tomorrow?


Indeed.
Original post by _HabibaH_
Which paper again?


Not too sure :smile:, i just saw the question that the person had posted
Original post by rukan
thank you :smile: so what about weakest reducing agent and weakest reducing agent...?


The weakest reducing would be the strongest oxidising, presumably. And the weakest oxidising would be the strongest reducing presumably. Just ask someone else to be sure though.
Original post by georgiaaaxo
my worst thing is the long-winded calculations, particularly to do with redox titrations and neutralisations :// is there anywhere online with some of these to practise (not past paper ones?)



http://www.docbrown.info/page07/SSquestions/redox_vol_calcs1.htm

I find these are really good practice!
Reply 2439
Original post by nightsky/
Thanks, I think what was getting confused about is the fact that when you add H+, the E shifts left to get rid of extra H+, but then I was also thinking that surely the system would 'want' to shift back right because the conc. of conjugate base has decreased.

I guess, though, that the H+ added is far more significant because you also have the conjugate base produced by the salt's dissociation, and the acid wouldn't of dissociated that much anyway.

:smile:


In theory that makes sense but as u said that's why they mix salts and weak acids together to overcome it

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