Chemistry: Post your doubts here!

[Metanoia]

Though 2 years aren't much but I did like from 2010, and the lastest ones are more important, so you'd do fine.
Did you do all variants?

Also any idea, how do we keep reactants separate until the reaction starts?
like maybe tie a thread to crucible and put the bung. then use thread to empty it. I think I read something like that

A thistle funnel can be used

 

[darks]

AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be subject to electrolysis. Thus, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. In this case we have two possible reductions and two possible oxidations, listed here along with their reduction or oxidation potentials:

Reduction:
2H+ + 2e- ==> H2, E = 0.00 V

Ag+ + e- ==> Ag, E = 0.8 V

Oxidation:
2F- ==> F2 + 2e-, E = -2.87 V

2H2O ==> O2 + 4H+ + 4e-, E = -1.23 V

(ALL THE VALUES ARE FROM THE DATA BOOKLET)
The reactions with the *most positive* potentials are most likely to proceed. Therefore, the reduction of Ag+, which occurs at the cathode, is the most favorable reduction reaction and the oxidation of H2O, which occurs at the anode, is the most favorable oxidation reaction. Consequently, the major products of this hydrolysis are Ag and O2, whereas H2 and F2 are not produced in appreciable amounts.

Same principle with all three

Thanks A lot!

 

[darks]

Please explain!! thanks. All parts after E*cell I know it's a lot but this comes a lot in exam and my concepts of this topic are not so good.

 

[HELLNO]

View attachment 62130
Please explain!! thanks. All parts after E*cell I know it's a lot but this comes a lot in exam and my concepts of this topic are not so good.

(ii) At standard conditions the concentration should be 1mol/dm3, in this case the Ag+ concentration as given in the beginning of the question is 2.5×10–2 mol/dm3, which is lower than 1.

Ag+ + e- ==> Ag has an E*cell of 0.8 v the backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode (E*cell=positive electrode-negative electrode)

The Ag+ concentration decreases in the solution, favouring the backward reaction (Le Chatelier's principle) The backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode



(iii) In Fe3+(aq) + Fe2+(aq) no change will take place, because its just gonna increase the sulphate ions in the solution which has zero effect.

While in Ag2SO4(aq) it will decrease the Ecell, thats because the sulphate ions precipitate the the Ag+ forming Ag2SO4, therefore it decreases the Ag+ concentration favouring the backward reaction, as explained above ^^

(iv) Basically what I said in (iii)

 

[areeba240]

9701_w09_qp_42
qtn 5(a)
can someone plzz explain me this part?

 

[areeba240]

can some plzz explain?
9701_w16_qp_41

 

[HELLNO]

9701_w09_qp_42
qtn 5(a)
can someone plzz explain me this part?

Here in this question, it isn’t based on what we studied throughout the year’s course, its more like enough information is given in the question, and based on this you have to make your answers. In the question there are 2 principals we got to know about.
1st is that a delocalised benzene is coplanar, while any other cycloalkane isn’t.
2nd is that a simple chain with no methyl branches is coplanar while a methyl branched one isn’t.

So far nothing we have learned all from the question and you just have to conclude.
For the structures given we have to apply the knowledge we gained.

In A its all benzene rings, so its coplanar (1st principal)

In B there is a cycloalkane attached which makes it not all carbons coplanar. (1st)

In C this is somehow complicated but it is coplanar, thats because both of the side chains attached to the O has no branches and are straight chains making them coplanar (2nd)

In D its just a simple chain without methyl branches so all coplanar (2nd)

And in E Its a benzene ring attached to a non-methylbranched chain, which makes it coplanar. (1st & 2nd)

Get it?

 

[HELLNO]

can some plzz explain?
9701_w16_qp_41

My answer wasn't like the ms's...

Rizwan Javed see this

 

[AnonymousX9]

99/130 in AS in MJ 2016 (A threshold was 81). How much more approximately do I need for an A* in A2?

 

[techgeek]

can some plzz explain?
9701_w16_qp_41

My answer wasn't like the ms's...

Rizwan Javed see this

I think it's because Vanadium exists in multiple oxidation states +2, +3, +4 and +5 .. The V3+ ions are further oxidized by iron to V+4 (i.e. VO2+), therefore the equation is a bit different from one we are expecting

 

[HELLNO]

99/130 in AS in MJ 2016 (A threshold was 81). How much more approximately do I need for an A* in A2?

90-100/130
Depends on the 2017's gt

 

[Rizwan Javed]

My answer wasn't like the ms's...

Rizwan Javed see this

I think it's because Vanadium exists in multiple oxidation states +2, +3, +4 and +5 .. The V3+ ions are further oxidized by iron to V+4 (i.e. VO2+), therefore the equation is a bit different from one we are expecting

The same explanation as techgeek 's. In the question we are given that Fe3+ is in EXCESS. So the V3+ are futher oxidised to V4+ (VO2+) by Fe3+

 

[HELLNO]

I think it's because Vanadium exists in multiple oxidation states +2, +3, +4 and +5 .. The V3+ ions are further oxidized by iron to V+4 (i.e. VO2+), therefore the equation is a bit different from one we are expecting

The same explanation as techgeek 's. In the question we are given that Fe3+ is in EXCESS. So the V3+ are further oxidised to V4+ (VO2+) by Fe3+

Oww okay guys
Thanks!! I never read the excess part though

 

[areeba240]

Here in this question, it isn’t based on what we studied throughout the year’s course, its more like enough information is given in the question, and based on this you have to make your answers. In the question there are 2 principals we got to know about.
1st is that a delocalised benzene is coplanar, while any other cycloalkane isn’t.
2nd is that a simple chain with no methyl branches is coplanar while a methyl branched one isn’t.

So far nothing we have learned all from the question and you just have to conclude.
For the structures given we have to apply the knowledge we gained.

In A its all benzene rings, so its coplanar (1st principal)

In B there is a cycloalkane attached which makes it not all carbons coplanar. (1st)

In C this is somehow complicated but it is coplanar, thats because both of the side chains attached to the O has no branches and are straight chains making them coplanar (2nd)

In D its just a simple chain without methyl branches so all coplanar (2nd)

And in E Its a benzene ring attached to a non-methylbranched chain, which makes it coplanar. (1st & 2nd)

Get it?

yeah understood all except for C.....is the chain considered this way

 

[areeba240]

The same explanation as techgeek 's. In the question we are given that Fe3+ is in EXCESS. So the V3+ are futher oxidised to V4+ (VO2+) by Fe3+

can u plzz write and the individual equation from which which we will get that final equation?

 

[HELLNO]

yeah understood all except for C.....is the chain considered this way
View attachment 62132

Yess!!

 

[Rizwan Javed]

can u plzz write and the individual equation from which which we will get that final equation?

You need to consider

Fe3+ + e --> Fe2+ E= +0.77
V2+ + 2e --> V

2Fe2+ + V --> 2Fe2+ + V2+

Fe3+ + e --> Fe2+
V3+ + e --> V2+

Fe3+ + V2+ --> V3+ + Fe2+

Fe3+ + e --> Fe2+
VO2+ + 2H+ + e --> V3+ + H2O

Fe3+ + V3+ +H2O --> 2H+ + VO2+ + Fe2+

Now you can see that for the conversion of V to VO2+, you need 4Fe3+ which yield 4Fe2+. Combining the equations, you get this:

4Fe3+ + V3+ H2O --> 2H+ VO2+ + 4Fe2+

 

[darks]

(ii) At standard conditions the concentration should be 1mol/dm3, in this case the Ag+ concentration as given in the beginning of the question is 2.5×10–2 mol/dm3, which is lower than 1.

Ag+ + e- ==> Ag has an E*cell of 0.8 v the backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode (E*cell=positive electrode-negative electrode)

The Ag+ concentration decreases in the solution, favouring the backward reaction (Le Chatelier's principle) The backward reaction means less reduction more oxidation and therefore less Ecell value. The decrease in this, causes and overall decrease as Ag is the positive electrode



(iii) In Fe3+(aq) + Fe2+(aq) no change will take place, because its just gonna increase the sulphate ions in the solution which has zero effect.

While in Ag2SO4(aq) it will decrease the Ecell, thats because the sulphate ions precipitate the the Ag+ forming Ag2SO4, therefore it decreases the Ag+ concentration favouring the backward reaction, as explained above ^^

(iv) Basically what I said in (iii)

Thanks!!!

 

[AnonymousX9]

View attachment 62129
Help me understand this Please!!
ms
anode, O2,O2,Br2
cathode, Ag,H2, H2
Thanks!

For this question at anode, won't it be OH- that gets oxidized? Why H2O?

 

[HELLNO]

For this question at anode, won't it be OH- that gets oxidized? Why H2O?

Both are correct!
If you put in the OH- formula, it will be -0.4 v which is still more positive and is favoured. Both equations produce O2 gas.