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

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Original post by zirak46
Hey guys, hope revision is going well.

Could someone please explaim how to do the following questioms from the June 13 paper?

The part d) question refers to the table attached.



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You need the rest of the question for the photosynthesis Q (look at the equation and units)
1466252948577.jpg
Not sure how to go about this.

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Original post by zirak46
Hey guys, hope revision is going well.

Could someone please explaim how to do the following questioms from the June 13 paper?

The part d) question refers to the table attached.



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Hint for the first Q, DeltaH=E/Mol
Original post by pineneedles
1466252948577.jpg
Not sure how to go about this.

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Okay so it tells you that silver nitrate is used. And all the complexes have chlorine. That means the formula for the precipitate is AgCl. The Mr is 143.3. 2.868 of AgCl is formed. Using moles=mass/Mr you find that it is 0.02 moles which is double 0.01. This means the complex from which it was formed must have 2 chlorine ions. Hope that makes sense.

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Original post by ImNervous
Okay so it tells you that silver nitrate is used. And all the complexes have chlorine. That means the formula for the precipitate is AgCl. The Mr is 143.3. 2.868 of AgCl is formed. Using moles=mass/Mr you find that it is 0.02 moles which is double 0.01. This means the complex from which it was formed must have 2 chlorine ions. Hope that makes sense.

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Meaning it would be complex B. The oons with the cobalt will stay with the cobalt.

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Original post by ImNervous
Meaning it would be complex B. The oons with the cobalt will stay with the cobalt.

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Yeah, that makes a lot of sense, thank you. That's what was confusing me; I wasn't sure whether the silver nitrate reacted with the chloride ions which are part of the complex or not. Why do they only react with the chloride ions which aren't part of the complex, do you know?

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Original post by lai812matthew
it specifies dilute which is a bit akward. is 6M dilute? not really imo.

In the AS textbook it says dilute is anything with a concentration of below 10M.
Original post by pineneedles
Yeah, that makes a lot of sense, thank you. That's what was confusing me; I wasn't sure whether the silver nitrate reacted with the chloride ions which are part of the complex or not. Why do they only react with the chloride ions which aren't part of the complex, do you know?

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No not really
But I guess it would be because the chloride ions that are part of the complex have formed co - ordinate bonds which are essentially dative covalent bonds and are stronger than ionic bonds. I'm not too sure though. And have you checked the answer on the mrkscheme to make sure its complex B? If not could you please check. Thank you very much
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Hi, came across a thing about cells that I have never been able to understand.
In all diagrams, a high-resistance voltmeter is used in the wire between the two electrodes. But electrons need to flow through the wire, from one half cell to another. So why is a high resistance voltmeter used (i.e. why is the flow of electrons impeded)?
Original post by sunsri101
Hi, came across a thing about cells that I have never been able to understand.
In all diagrams, a high-resistance voltmeter is used in the wire between the two electrodes. But electrons need to flow through the wire, from one half cell to another. So why is a high resistance voltmeter used (i.e. why is the flow of electrons impeded)?


When you are measuring standard electrode potentials you don't want current to flow. As soon as it does the concentrations of solutions start to change.

V = IR

So if R is very high, I will be very low.
This was actually a redox titration calc from an old thread- just reposting it as I wanted some help with it.

A solution is made from, 50cm^3 ,iron (ii) chloride, diluted to 500cm^3. the solution was then titrated against both 0.02M potassium manganate, and 0.01677M potassium dichromate (separately). (after adding an excess of dilute sulpuric acid). 25cm^3 solution was used each time, 25.50cm^3 more manganate was needed to be added each time than dichromate. calculate the concentration of iron (ii) chloride in the original solution. before dilution) the answer is 0.51M apparently

Basically, this is my working so far:
1) (Cr2O7)2- + 6Fe2+ + 14H+---->6Fe3+ + 2Cr3+ +7H2O
2) MnO4- +5Fe2+ + 8H+----> 5Fe3+ + Mn2+ +4H2O

Then, I set: volume of (Cr2O72-)= x, and the volume of (MnO4-)= x+25.5
for reaction with Cr2O72- moles of Cr2O72-= 0.01677 X (x/1000)= (0.01677x)/1000
moles of Fe2+ reacted in 25cm^3= 6(0.01677x)/1000= (0.10062x)/1000
moles of Fe2+ in 500cm^3= 20(0.10062x)/1000= (2.0124x)/1000
^^^^This is then the number of moles in 50cm^3 also (before dilution)

Then for the reaction with MnO4-: moles of MnO4-= 0.02 X (x+25.5)/1000
moles of Fe2+ reacted in 25cm^3= moles above X5= 0.1(x+25.5)/1000
moles of Fe2+ in 500cm^3= moles above X 20= 2(x+25.5)/1000
^^^^This is also the number of moles in 50cm^3 (before dilution)

I then equated the two bolded values and came out with like x=4112.9.... Where did I go wrong??? :frown:
(edited 7 years ago)
Hope this is an easy exam😫
What were the highest grade boundaries


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Original post by TeachChemistry
When you are measuring standard electrode potentials you don't want current to flow. As soon as it does the concentrations of solutions start to change.

V = IR

So if R is very high, I will be very low.


But you do want electrons to move between the half cells. Current is the flow of charged particles, so if you want electrons to flow, surely you do want current?
And the solution concentrations do change over time, that's why batteries run out unless they are rechargeable- right?
Any help with this?
1.png79.9KB
Original post by tcameron
Any help with this?


nN=m/Mr = 63.64/14 (Take % as being the mass because it is a ratio.) = 4.55mol nO=m/Mr = 100-63.64/16 = 36.36/16 = 2.27 mol

Hence, nN:nO = 2:1

Hence N2O.
Original post by Timon512
nN=m/Mr = 63.64/14 (Take % as being the mass because it is a ratio.) = 4.55mol nO=m/Mr = 100-63.64/16 = 36.36/16 = 2.27 mol

Hence, nN:nO = 2:1

Hence N2O.


I got that but the question also said about the density being 1.8333gdm-3 so would there need to be a conversion to find the Mr because I don't think N2O is the right answer

don't worry I've figured it out now, you had to x by 24 to find the Mr
(edited 7 years ago)
Original post by tcameron
I got that but the question also said about the density being 1.8333gdm-3 so would there need to be a conversion to find the Mr because I don't think N2O is the right answer

don't worry I've figured it out now, you had to x by 24 to find the Mr


Multiply what by 24? Could you write your working out please? Why did you use V/24=N?
I can't believe how badly I've been taught redox and fuel cells

I need to go over it 100 times :cry:
Original post by sunsri101
But you do want electrons to move between the half cells. Current is the flow of charged particles, so if you want electrons to flow, surely you do want current?
And the solution concentrations do change over time, that's why batteries run out unless they are rechargeable- right?



Yes batteries run down because current has been allowed to flow.

Think about what you are measuring (emf) as a push from the cell. You need to measure the magnitude of that push by, in effect, pushing back using electrical resistance. If you resist enough, current will be negligible.

If you are ok with physics think about the resistance that an ammeter must have, because it goes in series to the other components, and the resistance a voltmeter must have because it goes parallel to the component you are measuring.

I will draw a picture and upload it.
Why is the conc of butanoic acid 0.1?

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