AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

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i'm confused!

Forget about the emf values since you're given the equation of reaction. Work out from the equation the species that is reduced and the species that is oxidised.
In cell rep, species that is reduced is on the right in this case (I2/I-) and the species that is oxidised is on the left (in this case Br2/Br-) .

emf = Eright - Eleft. and this gives a negative value.

WHENEVER YOU ARE GIVEN THE EQUATION, YOU NEED TO WORK OUT THE OXIDISED AND REDUCED SPECIES.
But if you are not given the equation, do it the normal way (which I think you should understand).
If you do not understand still, I will break it down more yh.

Reply 401

SkyHunter

Original post by Crowned Temmy

I understandddd! :biggrin:

thanks a lot :smile:

)) x

Reply 402

NRican

Hey can anyone please explain to me how and when Pt is used when writing conventional cell representations? left, right both?

And when writing cell equations we don't need to include H+ and H20 in the overall eq. do we?

(edited 12 years ago)

Reply 403

Sparkly-Star

Can anyone please how to do Born Haber cycles for enthalpy of solution? I don't get it!! :cry2:

Thanks.

Reply 404

Sparkly-Star

Original post by NRican

Hey can anyone please explain to me how and when Pt is used when writing conventional cell representations? left, right both?

Both sides, I think it isn't used when it is done with hydrogen.

Reply 405

NRican

Original post by Sparkly-Star

Can anyone please how to do Born Haber cycles for enthalpy of solution? I don't get it!! :cry2:

Thanks.

I didn't bother I think in the exam calculation is enough? , just remember the equation E soln = E hydration + E lattice dissociation.

Reply 406

Sparkly-Star

Original post by NRican

I didn't bother I think in the exam calculation is enough? , just remember the equation E soln = E hydration + E lattice dissociation.

Oh ok thanks, I didn't know that was the equation!! :p:

I have seen past questions on it that's why so I don't really get those type. :p:

Reply 407

Sparkly-Star

Who's doing an all-nighter? (Or in my case... half-nighters lol)

Reply 408

al_habib

guys whats the O.S. of Cr in this compound Cr2O72- is it +6 or +12 and 2Cr3+ does 2 have any effect before Cr.thanx

Reply 409

Crowned Temmy

The spec states that we should know the half cell equations for fuel cells. Do we have to know the half equations for other types of electrochemical cell as well?

Reply 410

NRican

Original post by Sparkly-Star

Both sides, I think it isn't used when it is done with hydrogen.

But I've seen it done on one side only :s-smilie:

Reply 411

al_habib

Original post by Sparkly-Star

Who's doing an all-nighter? (Or in my case... half-nighters lol)

meeee

bt till 4 coz need to sleep for 3 hr after that then start cracking

Reply 412

Sparkly-Star

Original post by NRican

But I've seen it done on one side only :s-smilie:

Really? That's weird, I learnt it so that it is used on both sides.

Reply 413

jimmy303

Original post by NRican

But I've seen it done on one side only :s-smilie:

Sparkly-Star

It's used when there isn't a solid involved.

E.g. Zn2+(aq) + 2e- ---> Zn(s)

doesn't need platinum as it has a solid (Zn) which acts as the electrode so it becomes:

Zn(s)|Zn2+(aq) ||

However,

Fe3+(aq) + e- ---> Fe2+(aq)

there is no solid involved here so a platinum electrode is needed to transfer electrons, and the cell becomes:

Pt | Fe2+(aq), Fe3+(aq) ||

So if you see a half cell equation with no solid involved, chuck in platinum :smile:

It can be left or right depending on if the half cell without the solid is more negative or more positive, and can appear on both if neither half cell involves a solid as it does with the hydrogen fuel cell:

Pt|H2(g)|OH-(aq), H2O(l) || O2(g) | H2O(l), OH-(aq)|Pt

Hope this helped :redface:

(edited 12 years ago)

Reply 414

NRican

Original post by jimmy303

It can be left or right depending on if the half cell without the solid is more negative or more positive, and can appear on both if neither half cell involves a solid as it does with the hydrogen fuel cell:

Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt

Hope this helped :redface:

Thanks a lot :biggrin:

(edited 12 years ago)

Reply 415

Sparkly-Star

Original post by jimmy303

It can be left or right depending on if the half cell without the solid is more negative or more positive, and can appear on both if neither half cell involves a solid as it does with the hydrogen fuel cell:

Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt

Hope this helped :redface:

Thank you, that did really help! :grin:

I had no idea about all this! Shame on me. :facepalm:

Reply 416

choc1234

theres barely any synoptic in unit 5 chem ! its not like bio lol
i dont think there was any in the jan11 paper
i probs wont revise any synoptic

Reply 417

Willyg11

Original post by al_habib

guys whats the O.S. of Cr in this compound Cr2O72- is it +6 or +12 and 2Cr3+ does 2 have any effect before Cr.thanx

6+ for Cr2O72-

3+ for 2Cr3+ but counts as +6 due to the '2'

don't understand your last question

Reply 418

Willyg11

Original post by choc1234

theres barely any synoptic in unit 5 chem ! its not like bio lol
i dont think there was any in the jan11 paper
i probs wont revise any synoptic

There's the periodicity stuff that's a little synoptic :P

Reply 419

Miss Purple

Original post by Sparkly-Star

Can anyone please how to do Born Haber cycles for enthalpy of solution? I don't get it!! :cry2:

Thanks.

errrrr ok, ill try :smile:

?Hsol = ?Hlatt dissociation + ?Hhyd of gaseous ions

because lattice enthalpy and lattice dissociation both have the same enthalpy change symbols, the way to distinguish between them is the value.
because making bonds is exothemic, -ve value. (lattice enthalpy)
and because breaking bonds is endo thermic, +ve value. (lattice dissociation)

so now the born haber cycle (in terms of the diagram with the levels),

start with the solution of a ionic compound, eg. NaCl(s) ---> Na+(aq) + Cl-(aq)

so, ?Hlatt. diss. is +ve, so the products on the diagram increase to a higher energy level ......(Na+(g) + Cl-(g))

?Hhyd is -ve, so the products of this reaction is of a lower energy, but still higher than NaCl(s) .......(Na+(aq) + Cl-(aq))

Hess' law states ?H is the same independant of the route. so if we rewind back from (Na+(aq) + Cl-(aq)) to (Na+(g) + Cl-(g)) and then finally to NaCl(s), this is ?Hsol

the same can be applied to the cycle

write the solution reaction.
then from the ionic compound, write the lattice diss. and its products.
then from the products, the hydration of each of the ions.
bam. complete cycle

:confused:

i hope this isnt too much waffle. it'd be much easier if i had a diagram

EDIT: silly formatting, the question marks are meant to be triangles for delta