Chemistry: Post your doubts here!

[biba]

biba
sagar65265

Thank you so much for your help guys! It really helped a lot!
I really appreciate it !

no problem .. just pray for me

 

[Soldier313]

no problem .. just pray for me

i definitely will inshaAllah! May Allah grant you success in all that you do inshaAllah

 

[biba]

i definitely will inshaAllah! May Allah grant you success in all that you do inshaAllah

In sha allah! may Allah bless u with success too
Ameeen !

 

[Soldier313]

In sha allah! may Allah bless u with success too
Ameeen !

Aameen inshaAllah

 

[sagar65265]

Guys, I just have a question - In the electrolysis of aqueous solutions containing halide ions, how do we determine using standard electrode potential values (E nought values) what is produced at the anode? Assuming that conditions are all standard and concentration doesn't have a role to play, how do we determine what is produced at the anode?

Thanks in advance!

 

[littlecloud11]

Guys, I just have a question - In the electrolysis of aqueous solutions containing halide ions, how do we determine using standard electrode potential values (E nought values) what is produced at the anode? Assuming that conditions are all standard and concentration doesn't have a role to play, how do we determine what is produced at the anode?

Thanks in advance!

The anions in the solution are the halide ion and OH-
Identify the equations relating to these two ions in the data booklet. The one with the more negative E value is the one which is oxidized and liberated at the anode.

 

[Jaf]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_4.pdf
http://papers.xtremepapers.com/CIE/.../Chemistry (9701)/9701_s03_ms_1+2+3+4+5+6.pdf

Question 3)b).
Why doesn't SiO2 react with NaOH? The ER bluntly states that it just does not. The coursebook says that it DOES infact react with NaOH when the latter is hot and concentrated. So if the argument is that NaOH (aq) won't react with SiO2, then why is it reacting with SnO2. The coursebook says the NaOH needs to be hot and concentrated to react with SnO2 too.

 

[sagar65265]

The anions in the solution are the halide ion and OH-

Identify the equations relating to these two ions in the data booklet. The one with the more negative E value is the one which is oxidized and liberated at the anode.

Then how are we supposed the do Question 3 (c) for the following paper:

The first two solutions I can understand, Fluorine and Sulfate are not released, but why does the marking scheme say Bromine is released at the anode? is it because of the lack of standard conditions or is it some other factor we have to take into consideration? Because the electrode potential for Bromine is much higher than the electrode potential for OH- ions (the difference is more than that 0.30 V value) so Oxygen should either ways be released!

Thanks a lot in advance!
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_ms_42.pdf

 

[littlecloud11]

Then how are we supposed the do Question 3 (c) for the following paper:

The first two solutions I can understand, Fluorine and Sulfate are not released, but why does the marking scheme say Bromine is released at the anode? is it because of the lack of standard conditions or is it some other factor we have to take into consideration? Because the electrode potential for Bromine is much higher than the electrode potential for OH- ions (the difference is more than that 0.30 V value) so Oxygen should either ways be released!

Thanks a lot in advance!
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_ms_42.pdf

Yes, I noticed that too. The only way Br2 could have been produced was if the conditions were non-standard but they really should have mentioned this in the question. Br2 is not produced at the anode when the solution is aqueous and under standard conditions.

 

[sagar65265]

Yes, I noticed that too. The only way Br2 could have been produced was if the conditions were non-standard but they really should have mentioned this in the question. Br2 is not produced at the anode when the solution is aqueous and under standard conditions.

Neat, thanks a lot!

Good Luck for all your exams!

 

[littlecloud11]

Neat, thanks a lot!

Good Luck for all your exams!

You too!

 

[daniyal007]

In oct nov 2012 p42 Q4 , i have solved the whole question correct by drawing correct structural formulae..But in mark scheme skelatal formulae are written..will i be getting marks????

 

[strangerss]

How can prepare for the practical exam at home ?

 

[strangerss]

allasalamu allaykom ,Can some one please help in paper34 Oct/Nov 2012 question 1c? I have no clue how to calculate the concentrations

 

[Soldier313]

littlecloud11 sagar65265

so for the qn 3 c in mj/11 p42, i gather the answer in the marking scheme is wrong?

 

[littlecloud11]

littlecloud11 sagar65265

so for the qn 3 c in mj/11 p42, i gather the answer in the marking scheme is wrong?

As the question mentioned no conditions and we were asked to use the data booklet, O2 should be the correct answer. The mark scheme is wrong on this occasion.

 

[Soldier313]

As the question mentioned no conditions and we were asked to use the data booklet, O2 should be the correct answer. The mark scheme is wrong on this occasion.

Owkay cool, thank you

 

[KurayamiKimmi]

okay in d ( i ) we found the molecular formula as SiCl3... now in d ii the first Mr is 133 that is SiCl3 cuz (28+35+35+35=133)... now we have 247... let's see 247-133=114, we knw the formula of 133, find the formula of 114 which is SiCl2O (28+35+35+16=114) ,add both formulas of 133 and 114 i.e SiCl3 + SiCl2O = Si2Cl5O(Cl 3Si-O-SiCl 2 )
do the same for 263.. 263 - 247 = 16,meaning only O is added hence add Si2Cl5O + O= Si2Cl5O2 (.Cl 3Si-O-SiCl 2-O)
..

thanks !

 

[Jaf]

As the question mentioned no conditions and we were asked to use the data booklet, O2 should be the correct answer. The mark scheme is wrong on this occasion.

It's not wrong. The ER reiterates the answer and we know for a fact that whenever a halide ion is present in the solution of an electrolyte (of a salt), the halogen gas is liberated at the anode. It's a rule. Perhaps we can identify the problem if you tell us which standard potentials you're using to come to your answer (I'm specifically interested in the anode reactions).
Soldier313 sagar65265

 

[Soldier313]

It's not wrong. The ER reiterates the answer and we know for a fact that whenever a halide ion is present in the solution of an electrolyte (of a salt), the halogen gas is liberated at the anode. It's a rule. Perhaps we can identify the problem if you tell us which standard potentials you're using to come to your answer (I'm specifically interested in the anode reactions).
Soldier313 sagar65265

For the first, AgF :
F2 + 2e– ⇌ 2F– E value = +2.87V
O2+ 2H2O + 4e–⇌4OH– E value = +0.4 V

For the second, FeSO4 :
SO42–+ 4H++ 2e–⇌SO2+ 2H2O E value = 0,.17 V
O2+ 2H2O + 4e–⇌4OH– E value = +0.4 V

For the third, MgBr2 :
Br2 + 2e–⇌2Br– E Value = +1.07 V
O2+ 2H2O + 4e–⇌4OH– E Value = 0.4 V