AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

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Reply 880

jamest92

Original post by Sparkly-Star

So Pt is used in conventional cell diagrams if there is no solid? Someone said this yesterday. :p:

Just wanna confirm. Thanks. :smile:

Yep, used as the electrode

Reply 881

jwest

7f) and 9c ii) on january 20100 any ideas??? any help would be appreciated

Reply 882

lemontang1

Im sorry if this question has already been asked....but why is first electron affinity exothermic and second electron affinity endothermic??? Thanks

Reply 883

Stirlo

Original post by Sparkly-Star

So Pt is used in conventional cell diagrams if there is no solid? Someone said this yesterday. :p:

Just wanna confirm. Thanks. :smile:

Yes because it isn't possible to get a non-solid electrode

Reply 884

fivepoundoff

When drawing the structures for complexes like in Jan 2011 question 4, do you need to show co-ordinate bonds, or are you just ok with a normal dash?

Reply 885

mkb230

Good luck everyone ! My last exam, only need a C in this one and feeling confident so no more revision for me :smile:

Reply 886

colabottles

Original post by mkb230

Good luck everyone ! My last exam, only need a C in this one and feeling confident so no more revision for me :smile:

I only need a C too but I can't possibly go in without knowing absolutely everything which is daft but hey :s-smilie:

Reply 887

Anon1993

Hey, Im looking at Jan 2010 q3b - and I dont understand why they've used that particular O2 half equation?

Will rep :smile:

Reply 888

fivepoundoff

Original post by lemontang1

Im sorry if this question has already been asked....but why is first electron affinity exothermic and second electron affinity endothermic??? Thanks

Negative ion after the first E.A. therefore they repel - something like this anyway (for endothermic). For exothermic (first E.A.) its because adding the electron makes it more stable

Reply 889

Anon1993

Original post by fivepoundoff

When drawing the structures for complexes like in Jan 2011 question 4, do you need to show co-ordinate bonds, or are you just ok with a normal dash?

You have to draw an arrowed bond with the arrow pointing from the atom donating the lone pair to the atom accepting :smile:

Hope this helps, and good luck!

Reply 890

jamest92

Original post by jwest

7f) and 9c ii) on january 20100 any ideas??? any help would be appreciated

7 f) [Co(Cl4)]2-

This is because Cl- are bigger than H20 molecules so you can only fit 4 around

9 c) It would start off slow because both the ions repel each other. As some Mn2+ is produced the reaction gets faster because this catalyses the reaction which produces more Mn2+ and further speeds up the reaction (may want to check this )

Reply 891

gildartz

Original post by king101

Can you please explain this dissociation/Enthalpy of... I hate it :P Only part im a little confused on

Ok, it'll be better if we use chlorine as an example:
The enthalpy of atomisation of chlorine results in 1 mole of Cl. So if you have NaCl, you keep this value the same. If you have MgCl2 on the other hand, you double it.

The enthalpy of dissociation, as well as the mean bond enthalpy, of chlorine results in 2Cl since you're breaking one mole of Cl-Cl bonds. If you have NaCl, you halve this figure, if you have MgCl2, you keep it the same.

Original post by strawberry_cake

I meant for the jan ones :smile:

Don't think there are any Jan ones for those years

Reply 892

F1Addict

Original post by lemontang1

Im sorry if this question has already been asked....but why is first electron affinity exothermic and second electron affinity endothermic??? Thanks

1st is exo because you're adding 1 electron to a gaseous atom, and the electron is attracted to the positive nucleus, so energy is released.
2nd is endo because you're adding 1 electron to a negative ion. Both are negatively charged; like charges repel, so you've got to put energy in to make the negative ion accept the electron.

Why does Chapter 15 have to have so many equations.. theres too many! :frown:

Reply 893

IFondledAGibbon

Original post by jwest

7f) and 9c ii) on january 20100 any ideas??? any help would be appreciated

7f) It's [CoCl4]^2-

Because [Co(H2O)6]^2+ + 4Cl- --> [CoCl4]^2- + 6H2O

And the explanation for the co-ordination number: Cl- is too large of an ion to fit 6 around the transition metal. The repulsive forces make it impossible.

9c ii) You'd draw the curve starting slowly because the ions repel each other. Then rapidly decreasing because Mn2+ ions autocatalize the reaction. The more Mn2+ are formed they catalyse their own production. It would level out at the end because all of the MnO4- would be used up as it goes to completion.

Edit: Too slow lol

Reply 894

angel1992

Original post by lemontang1

Im sorry if this question has already been asked....but why is first electron affinity exothermic and second electron affinity endothermic??? Thanks

first electron affinity is exothermic because there is no repulsion in this reaction and so the say its a cl atom it can more easily gain an electron without any problem, while say cl- adding an electron to this is second elctrona ffinity and has a nagative charge so there is repulsion which means that it is endothermic as energy has to be put in to keep this electron and negatively charged ion together

Reply 895

fivepoundoff

Original post by Anon1993

Hey, Im looking at Jan 2010 q3b - and I dont understand why they've used that particular O2 half equation?

Will rep :smile:

If they used the first O2 equation then the 4H+ would react with the 4OH- on the bottom equation, therefore it wouldn't be in alkaline conditions (I think). But using the second one you'll end up with 4OH- cancelling but its still in alkaline conditions.

...thats my guess

Reply 896

WJD

I only need a D to get my predicted grade overall yet I'm still bricking it.

Reply 897

jwest

Original post by IFondledAGibbon

ok, thanks for the help

Reply 898

angel1992

Original post by WJD

I only need a D to get my predicted grade overall yet I'm still bricking it.

how do you know this, did you do chem4 in jan and you got youe prac mark?

i sat chem 4 a week ago, and did the chem empa so jus gonna get as high as i can. got a fairly decent as grade so hoping for an a overall

Reply 899

Anon1993

Original post by fivepoundoff

If they used the first O2 equationn then the 4H+ would react with the 4OH- on the bottom equation, therefore it wouldn't be in alkaline conditions (I think). But using the second one you'll end up with 4OH- cancelling but its still in alkaline conditions.

...thats my guess

Thanks. . i kind of get it, but in the next part, 3c, i dont understand why they have an OH- in the hydrogen half cell bit?