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Chemistry: Post your doubts here!

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darks

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

HELLNO

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darks said:
View attachment 62130
Please explain!! :) thanks. All parts after E*cell:confused: I know it's a lot but this comes a lot in exam and my concepts of this topic are not so good.
Click to expand...
(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)
 
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areeba240

areeba240

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upload_2017-5-12_17-53-2-png.62126

can some plzz explain?
9701_w16_qp_41
 
HELLNO

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areeba240 said:
9701_w09_qp_42
qtn 5(a)
can someone plzz explain me this part?
Click to expand...

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?(y)
 
Rizwan Javed

Rizwan Javed

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HELLNO said:
My answer wasn't like the ms's...

Rizwan Javed see this:)
Click to expand...
techgeek said:
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
Click to expand...
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

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techgeek said:
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
Click to expand...
Rizwan Javed said:
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+
Click to expand...
Oww okay guys
Thanks!! I never read the excess part though :D
 
areeba240

areeba240

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HELLNO said:
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?(y)
Click to expand...
yeah understood all except for C.....is the chain considered this way
upload_2017-5-13_18-2-27.png
 
Rizwan Javed

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areeba240 said:
can u plzz write and the individual equation from which which we will get that final equation?
Click to expand...
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+
 
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darks

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HELLNO said:
(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)
Click to expand...
Thanks!!!
 
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