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

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I can't explain it in detail now but the trick with redox is to start with last step and work your way up for step 1 using stoichiometry to connect one step to the next via a common compound.

Without looking at the question I'm pretty sure you'll have to do is calc moles of titrating solution. Usually chromate, manganate, or thiosulfate

Then do the ratio method to calculate moles on the other product in the last equation. E.g. 2:1 ratio thiosulfate:iodine. So half mols from the paragraph above. Now you can move up a step as you've calculated the moles of, say, the iodine, to continue the example.

If there is a volume change (commonly 25cm sample from 250cm total) then you times the moles by 10 to account for this.

It gets pretty self-explanatory from there. Once you can do it fine, F325 final questions (usually redox calc) you'll be able to finish in under a minute.

Original post by greenorange
Does anyone know a method to form balanced redox equations and balanced half-equations, like Question 8.a on this paper.

Thanks
Original post by greenorange
Does anyone know a method to form balanced redox equations and balanced half-equations, like Question 8.a on this paper.

Thanks


Iron starts as Iron (III) Oxide Therefore its formula is likely Fe2O3 and when reacted with a solution containing hydroxide ions forms FeO4 2- Where the oxidation number of the iron is +6.
So a change from +3 to +6 means it is being oxidised.

To start with you have to balance atoms

Fe2O3 + Cl2 + OH- ---> FeO4 2- + 2Cl-
You need 2 iron atoms on the right hand side making it now
Fe2O3 +OH- + Cl2 --> 2FeO4 2- + 2Cl-
Now there are too many oxygens on the right hand side so you have to increase the number of OH-
Fe2O3 + 5OH- +Cl2 --> 2FeO4 2- +2Cl-
As there are more hydrogens on the left than right you add water molecules to the right hand side. However each water molecule is H2O so you have to double the amount of OH-
Fe2O3 +10OH- +Cl2 --> 2FeO4 2- +2Cl- + 5H2O

Now all the atoms are balanced you have to balance charges.
The charge of the left hand side is 10- and of the right hand side is 6-
If you increase the number of Cl- to 6Cl- you will have balanced charges. Then change the number of Cl2 so it balances. Overall you will get
Fe2O3 + 10OH- + 3Cl2 --> 2FeO4 2- + 6Cl- +5H2O
Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?
Original post by ShannonD_1697
Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?


Typically the lattice enthalpy and enthalpy of solution will be on the same side and the enthalpies of hydration on another.
The sum of the left = the right
Original post by ShannonD_1697
Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?


You have to draw a Born-Haber cycle with gaseous ions at the top, being hydrated, one at a time (X+(g) --> X+(aq)), after hydration of cation and anion you have the dissolved ions.
Draw from the gaseous ions down to the ionic solid (this is lattice enthalpy) and draw a final arrow from the ionic solid to the aqueous ions. That's how to draw an energy cycle, you can work out missing values if you have all the energy changes bar 1
Formation = Sum of everything else
Therefore
Formation = Lattice + X
Therefore
Lattice = Formation - X

X is the sum of all the known values (excluding formation obviously)

Lattice is always negative (hence down arrow). Don't forget to times things like atomisation (and affinity and ionisation if no second affinity/ionisation value provided) by 2 if necessary.

E.g. MgCl2. You'd times atomisation Cl by 2. Along with the affinity if you're only given first affinity (because they're needs to be two for each ion)

-

If you're wondering why it's formation = sum of rest, look at the cycle and you'll see that formation arrow goes on way whilst the other arrows all go in another direction. So different routes for the same value
Original post by Rust Cohle
Formation = Sum of everything else
Therefore
Formation = Lattice + X
Therefore
Lattice = Formation - X

X is the sum of all the known values (excluding formation obviously)

Lattice is always negative (hence down arrow). Don't forget to times things like atomisation (and affinity and ionisation if no second affinity/ionisation value provided) by 2 if necessary.

E.g. MgCl2. You'd times atomisation Cl by 2. Along with the affinity if you're only given first affinity (because they're needs to be two for each ion)

-

If you're wondering why it's formation = sum of rest, look at the cycle and you'll see that formation arrow goes on way whilst the other arrows all go in another direction. So different routes for the same value



Thank-you this makes sense it's always something i forget to do I realise afterwards
Original post by k.russell
You have to draw a Born-Haber cycle with gaseous ions at the top, being hydrated, one at a time (X+(g) --> X+(aq)), after hydration of cation and anion you have the dissolved ions.
Draw from the gaseous ions down to the ionic solid (this is lattice enthalpy) and draw a final arrow from the ionic solid to the aqueous ions. That's how to draw an energy cycle, you can work out missing values if you have all the energy changes bar 1


Thank-you :smile:
Original post by VMD100
Typically the lattice enthalpy and enthalpy of solution will be on the same side and the enthalpies of hydration on another.
The sum of the left = the right


Thank-you :smile:
https://goo.gl/mGFVGf

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how to do f325 faster?
Reply 2831
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alow
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Original post by lai812matthew
how to do f325 faster?


Practice.
Original post by RayMasterio
Okay, suppose it just asked for the emf of Copper. By this logic, if the equilibrium shifted left then the overall cell emf becomes positive? because it's losing electrons?


Edit: Or is it because it moves to the left, the cu2+ is gaining more electrons and so the emf of the cell is negative?


The eq shifted left, decreasing conc of Cu2+ AND decreasing the electrode potential (also, the E is not standard as conc is no longer standard, below 1moldm-3)

Hope that made sense :smile:
guys can anyone help me with June 2015 question 2d part i, q 4c part ii, and part d iii
Original post by zirak46
The eq shifted left, decreasing conc of Cu2+ AND decreasing the electrode potential (also, the E is not standard as conc is no longer standard, below 1moldm-3)

Hope that made sense :smile:


I cant rep you but it makes sense. Thanks mate.
Can someone explain rate determining steps to me?
Reply 2836
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four4
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Original post by Dinasaurus
Can someone explain rate determining steps to me?


Check out "machemguy" on YouTube. He explains it pretty well.
Original post by RayMasterio
I cant rep you but it makes sense. Thanks mate.


No worries, lets go 619 that exam
Original post by zirak46
No worries, lets go 619 that exam


Lol

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Reply 2839
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four4
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When making up this years grade boundaries for f324, does OCR have to take last years grade boundaries into consideration? Can take they jump straight 8 marks and go from 42 from last year to 50 this year for an A?

Just a little curious since thats not happened in the past, the biggest difference between 2 years has been 5 marks...

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