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# Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread watch

1. (Original post by keres)
Attachment 554239

jan 2015 IAL paper - ew
Find the oxidation state change from NO3- to NH3 it goes from +5 ----> 3
I then balanced it by adding 8 electrons to the left so +5 + 8e- ----> -3

do the same to Al you get 0 -----> -3
Balance to give 0 + 3e- -----> -3

Combine equations, you should find that you have to multiply NO3 by 3 and the Al by 8 to balance the electrons.

Does that make any sense??
2. (Original post by DoctorMire)
It cant be A becaue for Ag silver to be a reducing agent you would have to flip the second half eqution and therefore the sign would change from +0.8 to -0.8 and since -0.8 is less than 0, it is a weaker reducing agent.
Remeber the values they give you are the reduction potentials . In that equation silver was being oxidised .
Do you get it now?
I thought the more negative the value the stronger the reducing agent and -0.8 is more negative than 0?
3. (Original post by noctisff)
I thought the more negative the value the stronger the reducing agent and -0.8 is more negative than 0?
The values they have given u are reduction potentials
The Emf values in the data booklet and the ones in that table are all reduction potentials- so the value stands for the abilty to reduce
the higher the value the greater the reducing power
therefore since reduction potential for Ag wasnt directly given you flip the equation and the sign and then notice that 0 is higher than -0.8 therefore hydrogen is a greater reducting agent than Ag from the data provided
make sense?
4. (Original post by DoctorMire)
The values they have given u are reduction potentials
The Emf values in the data booklet and the ones in that table are all reduction potentials- so the value stands for the abilty to reduce
the higher the value the greater the reducing power
therefore since reduction potential for Ag wasnt directly given you flip the equation and the sign and then notice that 0 is higher than -0.8 therefore hydrogen is a greater reducting agent than Ag from the data provided
make sense?
I see now, thanks man
5. (Original post by noctisff)
I see now, thanks man
no problem , do you by anychance know what specific titrations we need to know for tomorrow and any questions on back titrations and copper and chromium complexes?
6. (Original post by DoctorMire)
The values they have given u are reduction potentials
The Emf values in the data booklet and the ones in that table are all reduction potentials- so the value stands for the abilty to reduce
the higher the value the greater the reducing power
therefore since reduction potential for Ag wasnt directly given you flip the equation and the sign and then notice that 0 is higher than -0.8 therefore hydrogen is a greater reducting agent than Ag from the data provided
make sense?
Wait but for the Zn2+/Zn equation in the data booklet, the E value = -0.76V. I thought the more negative the value, the stronger the reducing power of the metal?
7. (Original post by DoctorMire)
no problem , do you by anychance know what specific titrations we need to know for tomorrow and any questions on back titrations and copper and chromium complexes?
The titrations we need to know are the sodium thiosulfate one with iodine, and the potassium manganate one, not sure about back titrations
8. Can someone explain how you do this please? Thanks in advance

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9. (Original post by TeaAndTextbooks)

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Kind of hard to tell what step 1 is without being given it...xD
10. (Original post by Z.e.e)
Find the oxidation state change from NO3- to NH3 it goes from +5 ----> 3
I then balanced it by adding 8 electrons to the left so +5 + 8e- ----> -3

do the same to Al you get 0 -----> -3
Balance to give 0 + 3e- -----> -3

Combine equations, you should find that you have to multiply NO3 by 3 and the Al by 8 to balance the electrons.

Does that make any sense??
yes that does make sense thank you! but i thought the Al goes from 0 ---> +5 ?
11. (Original post by keres)
yes that does make sense thank you! but i thought the Al goes from 0 ---> +5 ?
Al(OH)4-

Mathematically speaking, try this

x + (-4) = -1 where x is the oxidation state of Al and -4 is the oxidation state of OH

x is 3
12. (Original post by keres)
yes that does make sense thank you! but i thought the Al goes from 0 ---> +5 ?
Nope, the oxidation state of Al in Al = 0 and in Al(OH)4- = +3.

think of it as x - 4 = -1 ---> x = +3.

the OH ion has a charge of -1, since there are 4 of them, you have -4 in total.
13. guys when is the oxidation state of H -1 again? and does the fact that the oxidation state of O being -1 when bonded to fluorine have something to do with the fluorine being more electronegative?
14. Yeah you're right lol I've attached step 1

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15. (Original post by ihaspotato)
I have a Bromoalkane and a Benzene I want to make them react in the presence of a halogen carrier...

What is the halogen carrier?

AlCl3 or AlBr3?
Bromoalkane, so AlBr3.
Electrophile formation : R-Br + AlBr3 = R+ + AlBr4-

If you were using a chloroalkane then you would use R-Cl and AlCl3

At the end of the reaction you would have an extra H+, which would react with the AlBr4- to give back the catalyst AlBr3 and HBr.
16. C0balt Ayman! samb1234 I really need some clearance on this whole positive/negative electrode issue...

So let's say you have a Zn2+/Zn electrode as one half cell and Cu2+/Cu as the other. Zinc is the stronger reducing agent, so it is more easily oxidised. This means that the equilbrium from zinc: Zn2+ + 2e- <--> Zn will lie further to the left. There is therefore a build up of electrons on the zinc electrode, making it the negative electrode.

However, I thought that oxidation always occurs at the anode (the positive electrode)??

EDIT: Okay so I just found out that in these cells, anode is negative but in electrolytic cells it is positive!!!! MAKES SO MUCH MORE SENSE!

Also, what would happen if the voltmeter did not have infinite resistance?
17. Anyone done the Jan 2014 Ial paper get how to do this question

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18. (Original post by tayloryeah)

why does benzene not decolourise bromine water, but phenol does?

the reaction of the formation of a salt between methylamine and sulphuric acid?

Why does benzene burn with a 'smokey' flame ?

finally, could anyone write the formation of the electrophille for the fuming sulphuric acid reaction ? ( i know it is not in our specification to know the formation of this particular electrophille, but from having done unit 4 chem and bio, its clear that EDEXCEL dont really know there spec well)
- Benzene does not decolourise bromine water because it doesn't undergo an addition reaction(because this would involve breaking up the delocalised system). However phenol reacts by addition reaction, the reason is because of the OH group which increases the electron density and makes it more susceptible to attack from electrophiles.

<--- Click to enlarge

- Benzene burns with a smoky flame because it's carbon to hydrogen ratio is very low. Basically, lower the ratio, sootier the flame.

- I'm sorry I don't know because it's not in the spec
19. (Original post by Don Pedro K.)
guys when is the oxidation state of H -1 again? and does the fact that the oxidation state of O being -1 when bonded to fluorine have something to do with the fluorine being more electronegative?
-H is always +1 except in metal hydrides e.g. LiH, where it's -1.
-O is always -2, except when bonded to to F because F is much more electronegative

there's a handy table here: http://www.chemguide.co.uk/inorganic...idnstates.html

20. Any ideas how to figure these out?
thanks!
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