OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Screen Shot 2016-06-20 at 12.37.50.png

Attachment not found

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.

I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

Posted from TSR Mobile

Reply 2781

zirak46

8

Original post by suibster

I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Screen Shot 2016-06-20 at 12.37.50.png

Attachment not found

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.

Think youve misunderstood pal.

It says, as did I when I first did it.

It says 0.01 mole reacts with excess Ag (however much)

First use n = mass/ mr to find mr of the precipitate

2.868/0.01 gives mr of 286.8

Now the precipitate would be AgCl (as learned from f321).

Using rhe formulae of the complexes, its either 3Cl- reacting, 2Cl- or Cl-.

As Ag is in EXCESS, all Chloride ions will react, therefore meaning either :
3AgCl, 2AgCl or AgCl is formed.

Now find Mr of each and match with Mr of 286.6 to get your answer.

As two chloride ions used, it means complex B was used

Reply 2782

zirak46

8

Original post by ImNervous

Haha. Thanks for that. And you are right by the way.

Posted from TSR Mobile

Cheers pal :smile:

Reply 2783

suibster

3

Original post by ImNervous

I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

Posted from TSR Mobile

That does make sense! Thank you.

Reply 2784

RayMasterio

3

Original post by zirak46

Cu2+ + 2e- <-> Cu (s)
That is the eq on the table.
Adding H2O, decreases conc of Cu2+.
As its eq, the eq will shift to left to compensate. Therefore its more:

Cu (s) -> Cu2+ + 2e-

Hence, more oxidiation.

But the thing to understand I think is that, less Cu2+,so eq shifts to the left.

This means, the EP value decreases and so overall cell potential between Ag and Cu will increase due to the larger difference in EP between the two.

Hope that makes sense. (Im literally explaining this from wha I figured from nervous' answer, literally just understood all this from him, thought id give him a break though :wink:

)

Okay, suppose it just asked for the emf of Copper. By this logic, if the equilibrium shifted left then the overall cell emf becomes positive? because it's losing electrons?

Edit: Or is it because it moves to the left, the cu2+ is gaining more electrons and so the emf of the cell is negative?

(edited 7 years ago)

Reply 2785

suibster

3

Original post by ImNervous

I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

Posted from TSR Mobile

I've had a think about the question again and I believe your example is wrong, as you can not simply just change the moles of a reactant.
My conclusion is that since there's two CL- in complex B and there's 0.01 moles of B.
This means if complex b were to dissociate, it would be this:
[Co(nh3)5cl]2+2CL- ---> [Conh3)5cl]2+ + 2Cl-
So the mole ratio is 1:2 so 0.01 : 0.02

When this cl- reacts with ag+, the equation would be ag+ + cl- ---> agcl
mole ratio being 1:1 or (0.02:0.02)

Reply 2786

Ss0

HOW can i revise for this exam?? other than past papers and legacy papers? :frown:

WANT an A! tipss pleasse

Reply 2787

KB_97

10

Can someone please tell me why the answer is -2.31. I don't understand why it's negative.

Reply 2788

suibster

3

Original post by KB_97

Can someone please tell me why the answer is -2.31. I don't understand why it's negative.

Think about the standard cell potential formula

Standard cell potential = positive terminal - negative terminal

Rule: Oxidation= Negative terminal , Reduction=Positive terminal

Identify what's being oxidised and reduced.

O2 is being reduced so ------> Positive terminal

2.71 = 0.4 - (Negative terminal)

Rearrange to get Negative terminal = 0.4 - 2.71

Negative terminal (Al) = -2.31 V

Reply 2789

zirak46

8

Original post by KB_97

Can someone please tell me why the answer is -2.31. I don't understand why it's negative.

Whichever reaction is being reduced is the largest electrode potential value.

Cell potential is Largest Ep - Smallest Ep value

Hence: 0.4 - (Ep of other half cell) = 2.71
Rearrange to get Ep of ther half cell = -2.71

Reply 2790

RayMasterio

3

Original post by tcameron

whenever we do kc calculations the inital moles of the reactants is always known as that's what is added to make the products so the initial moles of the products is always 0
so that's why you needed to use the products as their initial moles were 0
all it was was 0.168x3

i'm confused again :frown:

we already know the moles for the reactant is three?

Reply 2791

ImNervous

2

Original post by suibster

I've had a think about the question again and I believe your example is wrong, as you can not simply just change the moles of a reactant.
My conclusion is that since there's two CL- in complex B and there's 0.01 moles of B.
This means if complex b were to dissociate, it would be this:
[Co(nh3)5cl]2+2CL- ---> [Conh3)5cl]2+ + 2Cl-
So the mole ratio is 1:2 so 0.01 : 0.02

When this cl- reacts with ag+, the equation would be ag+ + cl- ---> agcl
mole ratio being 1:1 or (0.02:0.02)

That's right. That's what I was trying to achieve with my *****y example lol.

Posted from TSR Mobile

Reply 2792

nimoooooooo

6

Hey guys does anyone have the unofficial markscheme for f324? Pleeeeeeaaaassssseeeeeeeeeeee

Reply 2793

Sab_007

how to tackle those redox titrations at the end????????

Reply 2794

tcameron

19

Original post by Sab_007

how to tackle those redox titrations at the end????????

easiest way is to annotate over what it's saying & may draw a diagram to illustrate it better
they'll always be a easy moles calculation in those anyways - no real method but I feel like there won't be a redox titration this year as there were soo many redox questions last year

Reply 2795

lai812matthew

10

Original post by Sab_007

how to tackle those redox titrations at the end????????

do more

Reply 2796

M0n1

For electrode potential, in a full cell
Does reduction happen on the positive terminal or negative terminal of the cell?
For example, silver-copper cell
Ag+ + e- = Ag has E= +0.80v
Cu2+ + 2e- = Cu has E = +0.34v
I get that Ag half cell would be going in forward direction and would be the reduction, so in the full cell would the Ag terminal be the positive or negative terminal?

Reply 2797

Sab_007

Original post by lai812matthew

do more

lol i have done loads, still find it confusing tho

Reply 2798

username2703735

18

Original post by zirak46

Hello peeps, could someone please explain how in electrochemical cells we decide which half cell is negative electrode and which is positive?

So when drawing a diagram of two half cells, which one goes on left or right?

The question is from Jan 12

*not letting me upload pic*