Chemistry: Post your doubts here!

[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

 

[Jaf]

Ah, if that's what you're using then what you're doing is incorrect.

First of all, let me tell you WHY we choose the reaction (at the anode) with the lower electrode potential. If let's the say electrode potential at the cathode is +1.1V and 2 possible products at the anode have potentials +2.1V and +3.2V (for reduction halves; the signs are reversed when the product is being oxidised). So with the first product our electrode potential of the cell would be -1V and with the second product would be -2.1V. So we know that the first product will be produced as the more positive the electrode potential of the cell, the more favourable the reaction.

Keeping this in mind you'd obviously choose the product at the anode with the lower electrode potential so when you reverse the sign and find the electrode potential of the cell, it is greater.

For this reaction I'm afraid you're using the wrong equation:

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

The equation to be used here is:
O2 + 4H+ + 4e– ⇌ 2H2O E= +1.23V

Using this reaction, we see the electrode potential of Br2/Br- is lower and Br2 is produced at the anode. (products of electrolysis of AgF and FeSO4 remain the same).
I'll also tell you why it makes no sense if you use the OH-/O2 reaction here. Recall that the dissociation of water into H+ and OH- is very, very little. The Ka of water is 1x10^-14. Compare that with the Ka of weak acids which themselves dissociate very little (propanoic acid Ka = 1.3 x 10^-5) and you'll realize just how little it dissociates. Hence, we can not use the reaction with the OH- on the right side. We use the one with the H2O on the right side. I know we've been taught in O levels that if we wish to write the half-reaction at the anode where O2 is liberated during electrolysis, we use the OH-/O2 reaction but that was for simplicity and is completely wrong.

So where DO we use the OH-/O2 reaction? When the electrolyte has a considerable concentration of OH- ions like in NaOH(aq)/NH3(aq).

Can someone please address my question now? lol

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/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.

 

[Soldier313]

Jaf
Wow.....JazakAllah khair for that!
Thanx a lot.

 

[Student12]

ANALYTICAL chemistry is noooot nice with me!

 

[littlecloud11]

Ah, if that's what you're using then what you're doing is incorrect.

First of all, let me tell you WHY we choose the reaction (at the anode) with the lower electrode potential. If let's the say electrode potential at the cathode is +1.1V and 2 possible products at the anode have potentials +2.1V and +3.2V (for reduction halves; the signs are reversed when the product is being oxidised). So with the first product our electrode potential of the cell would be -1V and with the second product would be -2.1V. So we know that the first product will be produced as the more positive the electrode potential of the cell, the more favourable the reaction.

Keeping this in mind you'd obviously choose the product at the anode with the lower electrode potential so when you reverse the sign and find the electrode potential of the cell, it is greater.

For this reaction I'm afraid you're using the wrong equation:

The equation to be used here is:
O2 + 4H+ + 4e– ⇌ 2H2O E= +1.23V

Using this reaction, we see the electrode potential of Br2/Br- is lower and Br2 is produced at the anode. (products of electrolysis of AgF and FeSO4 remain the same).
I'll also tell you why it makes no sense if you use the OH-/O2 reaction here. Recall that the dissociation of water into H+ and OH- is very, very little. The Ka of water is 1x10^-14. Compare that with the Ka of weak acids which themselves dissociate very little (propanoic acid Ka = 1.3 x 10^-5) and you'll realize just how little it dissociates. Hence, we can not use the reaction with the OH- on the right side. We use the one with the H2O on the right side. I know we've been taught in O levels that if we wish to write the half-reaction at the anode where O2 is liberated during electrolysis, we use the OH-/O2 reaction but that was for simplicity and is completely wrong.

So where DO we use the OH-/O2 reaction? When the electrolyte has a considerable concentration of OH- ions like in NaOH(aq)/NH3(aq).

Can someone please address my question now? lol

Thank you so much for clearing that out! Really.

 

[Jinkglex]

Guys i have an urgent question, but its about calculators Can anyone please clarify to me for sure if fx-100MS is allowed or not in A level CIE's? Thanks!

 

[fuckingchemistry]

data booklet anyone

 

[syed1995]

Guys i have an urgent question, but its about calculators Can anyone please clarify to me for sure if fx-100MS is allowed or not in A level CIE's? Thanks!

It's allowed.

 

[sagar65265]

Ah, if that's what you're using then what you're doing is incorrect.

Can someone please address my question now? lol

Thanks a load for the earlier answer, it sure did clear up a lot!

I really wonder where SiO₂ does not react with NaOH - the textbook definitely mentions a reaction occurring, Wikipedia mentions a reaction occurring, and general concepts do confirm that! If, indeed, the acidity of the Group IV dioxides decreases down the group, why does SnO₂ react with a base while a supposedly more acidic oxide -SiO₂ - not react?

On the other hand, the examiners report did say "inert to both these reagents", so it might have been at a time when these reactions weren't really known? I have no idea, because they can't say that (aq) doesn't mean concentrated - SnO₂ requires the same reaction conditions! I'm pretty sure that must have been some sort of mistake or something. Can't be any other way to explain it, but if there is an explanation, it better be a good one!

Good Luck for all your exams!

 

[biba]

thanks !

np !! wel I think em pretty bad at explaining but *thumbs up* if u understood what I said

 

[Jinkglex]

It's allowed.

You just saved me from a world of worry my friend

 

[syed1995]

You just saved me from a world of worry my friend

No problem mate

 

[Farru]

Aoa.
Do we have to learn the structure of adenine, cytosine, thymine and guanine?? :/
Chem A2 booklet..
Rep asap!!!

 

[Student12]

Aoa.
Do we have to learn the structure of adenine, cytosine, thymine and guanine?? :/
Chem A2 booklet..
Rep asap!!!

I don't think so.. You just need to know the complementary base pairing and the no. Of hydrogen bonds.

 

[Student12]

Aoa.
Do we have to learn the structure of adenine, cytosine, thymine and guanine?? :/
Chem A2 booklet..
Rep asap!!!

And only about which r purine n pyramidine

 

[Farru]

okk.

I don't think so.. You just need to know the complementary base pairing and the no. Of hydrogen bonds.

and
A--- T (2 Hydrgn bondng)
C---G (3 H.b) Ryt??