Chemistry: Post your doubts here!

[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??

 

[Farru]

And only about which r purine n pyramidine

we have to learn the struc. of purine and pyramidine? right??

 

[freakybandi]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_4.pdf
qs 2 anybody?

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_4.pdf
qs 2 anybody?

hey there..

a) iodine is purple so the colourless solution will turn purple..

b) only reactants of the slow reaction gets written down in our rate equation so for reaction one we got one mole of hydrogen peroxide and one mole of iodide ion but no moles of hydrogen ions so it will not be written in the rate equation so we have a = 1 b = 1 and c = o .... now for the second equation, IO- is made from H2O2 and iodide ion so u can pretend the IO- in the second equation as one mole of hydrogen peroxide and one mole of iodide ion and we also have hydrogen ion in the second equation so a= 1 b = 1 and c = 1 ... now for the third equation, we have HOI which is made from H+ AND IO- which is made from one mole of hydrogen peroxide and one mole of iodide but take care u also have got another mole of iodide ion in the third equation so in total u have 2 moles of iodide and 2 moles of hydrogen ion and one mole of hydrogen peroxide so a = 1 b = 2 c = 2

c) initial rate means u draw a tangent at the start of the graph the second the reaction started and calculate the gradient at that point and that would be ur rate

d) half life means the time the reactants reach its half concentration so 0.001/2 = 0.0005 this will be at 90s and 0.0005/2 = 0.00025 at 180s...180-90 = 90s so see the time is constant it always reaches its half concentration after 90s that proves its a first order reaction

e) the way i do this part of the question is i look at 2 lines and see which one has same values so when i divide the 2 lines over each other one reactant cancels and i would be able to calculate the power the reactant raised to for example...

look at line 1 and 2 when u divide them the concentration of the hydrogen ion will cancel each other so u have (0.05/0.07)^x = 1/1.4 do the math and find x and it will be 1 so the order with respect to h2o2 is 1 and now pick line 2 and 3 and divide them over each other (0.07/0.09)^1(0.05/0.07)^y = 1.4/1.8 do the maths and find y and it will be equal to zero so its order with respect to H ion is zero which means our calculation is consistent with step 1! thats it

i hope u got it!

 

[freakybandi]

hey there..

a) iodine is purple so the colourless solution will turn purple..

b) only reactants of the slow reaction gets written down in our rate equation so for reaction one we got one mole of hydrogen peroxide and one mole of iodide ion but no moles of hydrogen ions so it will not be written in the rate equation so we have a = 1 b = 1 and c = o .... now for the second equation, IO- is made from H2O2 and iodide ion so u can pretend the IO- in the second equation as one mole of hydrogen peroxide and one mole of iodide ion and we also have hydrogen ion in the second equation so a= 1 b = 1 and c = 1 ... now for the third equation, we have HOI which is made from H+ AND IO- which is made from one mole of hydrogen peroxide and one mole of iodide but take care u also have got another mole of iodide ion in the third equation so in total u have 2 moles of iodide and 2 moles of hydrogen ion and one mole of hydrogen peroxide so a = 1 b = 2 c = 2

c) initial rate means u draw a tangent at the start of the graph the second the reaction started and calculate the gradient at that point and that would be ur rate

d) half life means the time the reactants reach its half concentration so 0.001/2 = 0.0005 this will be at 90s and 0.0005/2 = 0.00025 at 180s...180-90 = 90s so see the time is constant it always reaches its half concentration after 90s that proves its a first order reaction

e) the way i do this part of the question is i look at 2 lines and see which one has same values so when i divide the 2 lines over each other one reactant cancels and i would be able to calculate the power the reactant raised to for example...

look at line 1 and 2 when u divide them the concentration of the hydrogen ion will cancel each other so u have (0.05/0.07)^x = 1/1.4 do the math and find x and it will be 1 so the order with respect to h2o2 is 1 and now pick line 2 and 3 and divide them over each other (0.07/0.09)^1(0.05/0.07)^y = 1.4/1.8 do the maths and find y and it will be equal to zero so its order with respect to H ion is zero which means our calculation is consistent with step 1! thats it

i hope u got it!

thanku soooooooo much!!

 

[iKhaled]

thanku soooooooo much!!

welcome!!