You are Here: Home

Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

1. (Original post by Schoolio93)
yeah I wanna know that too I divided by 2 so scored only 1/4.

(Original post by susan23)
yes thank you...im so confused on this too.
I thought it was me being tired. I can understand dividing it by 2 but 4?! I can't be this confused 13 hours before the exam.
Jan 2011 paper: I am geting the wrong answer, what is the relationship to get the moles of o2?

A sample of river water is shaken with aqueous Mn2+ and aqueous alkali.
The dissolved oxygen oxidises the Mn2+ to Mn3+, forming a pale brown precipitate of
Mn(OH)3.

O2(aq) + 4Mn2+(aq) + 8OH–(aq) + 2H2O(l) ---->4Mn(OH)3(s)

• The Mn(OH)3 precipitate is then reacted with an excess of aqueous potassium iodide,
which is oxidised to iodine, I2.

2Mn(OH)3(s) + 2I–(aq)---> I2(aq) + 2Mn(OH)2(s) + 2OH–(aq)

• The iodine formed is then determined by titration with aqueous sodium thiosulfate,
Na2S2O3(aq).

2S2O3 + I2–(aq) + I2(aq)---> S4O6 + I2–(aq) + 2I–(aq)

Now to find the conc of 02,
I would find moles of 2S2O3 then Half it to get I2, Now i got I2 i can find out moles of 2Mn(OH)3(s) By X2 the moles of I, as its a 1:2. Then i would times by 2 to get 4moles of 4Mn(OH)3(s). Then Divide by 4 again to get 02.
Now where am i going wrong. I get the moles of 02 being 1.23x10-5.
Please help me, ill give you personal rep today and tomorrow. Serious. Why isnt my way of thinking right?! :O Pleeasee
3. (Original post by BA1)
C1: 99
C2: 100
S1: 98
Total for AS: 297/300

C3: 79
C4: Next week
M1: 82
Total needed for A = 480 - 297 - 79 - 82
= 22 !!! :-)
But it's now categorically impossible for you to get an A*... You need to average 90% in C3/C4
4. (Original post by goldlock)
Listen, to the people who ask why you divide by 4 to get the DOC question on the jan. 11 paper:

you work out the moles of thiosulfate ions they give you (conc. x volume) - those figures are given. Then you see that it's two moles thiosulfate to two moles of Iodide ions so their molar quantities are equal (this is looking at equation 3)

Now look at equation 2: moles of Iodide ions are in equal ratio to Mn 3+ so moles of thiosulfate = moles of Mn 3+

Look at equation 1: It's 4 moles of Mn 3+ to one mole 02 so divide that figure by 4

Hope that helped you!
Thanks a bunch

Thing is I know that I understand it now but come tomorrow under exam conditions I'm not sure if I'd be able to work it out.
5. (Original post by goldlock)
Listen, to the people who ask why you divide by 4 to get the DOC question on the jan. 11 paper:

you work out the moles of thiosulfate ions they give you (conc. x volume) - those figures are given. Then you see that it's two moles thiosulfate to two moles of Iodide ions so their molar quantities are equal (this is looking at equation 3)

Now look at equation 2: moles of Iodide ions are in equal ratio to Mn 3+ so moles of thiosulfate = moles of Mn 3+

Look at equation 1: It's 4 moles of Mn 3+ to one mole 02 so divide that figure by 4

Hope that helped you!
Thanks so much i get it now, they love tricking people dont they!!!
6. (Original post by CoffeeStinks)
Why is oxygen divided by four? What equation are they getting it from?

Q7 a
I thought i was the only one, im stuck on this too :/ I dont understand the molar relationship:/
7. (Original post by viksta1000)
Fe has oxidation state 6+ in FeO4 (2-) and oxidation state 3+ in Fe3+, so has changed by -3

O has oxidation state 2- in FeO4(2-) and oxidation state 0 in O2, so the change is +3, however, 2 oxygen atoms have been used to form O2, so the overall change is +4

to balance the lowest common factor of -3 and +4 is 12, so you multiplu FeO4(2-) by 4 to get:

4FeO4(2-) + n[H+] --> 4Fe(3+) + H2O + 02

now look at the overall molecular charges, as they have to balance on both sides:

we have (4x-2) + n(1) --> (4x3+) + 0 + 0

which is -8 + n1+ --> 12+
we need to get 12+ on both sides, so we need to add 20H+ ions to the left to get:

4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 02

simple balancing from here, 20H+ on right, so you need 10H2O:

4FeO4(2-) + 20[H+] --> 4Fe(3+) + 10H2O + 02

16 oxygens on left, 10 on the right, so we need 6 more, so times O2 by 3

4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 302

Thank you for the help
8. (Original post by goldlock)
Listen, to the people who ask why you divide by 4 to get the DOC question on the jan. 11 paper:

you work out the moles of thiosulfate ions they give you (conc. x volume) - those figures are given. Then you see that it's two moles thiosulfate to two moles of Iodide ions so their molar quantities are equal (this is looking at equation 3)

Now look at equation 2: moles of Iodide ions are in equal ratio to Mn 3+ so moles of thiosulfate = moles of Mn 3+

Look at equation 1: It's 4 moles of Mn 3+ to one mole 02 so divide that figure by 4

Hope that helped you!
but in the 2nd equation its 2Mn(OH)3 and in the 1st equation its 4Mn(OH)3 so how can you say the molar rations are equal??? Is it not 1:2

so wudnt you doulbe the moles of 2Mn(OH)3 to get molesof 4Mn(OH)3 and then divide that by 4 to get moles of O2????
9. can some just go over the redox equation in June 2010 6B....i can't balance it and its 3marks!
10. Someone tell me how to remember which enthalpy change is +ve and -ve!!

Major rep!
11. (Original post by susan23)
but in the 2nd equation its 2Mn(OH)3 and in the 1st equation its 4Mn(OH)3 so how can you say the molar rations are equal??? Is it not 1:2

so wudnt you doulbe the moles of 2Mn(OH)3 to get molesof 4Mn(OH)3 and then divide that by 4 to get moles of O2????
i was gonna write that.
12. (Original post by susan23)
but in the 2nd equation its 2Mn(OH)3 and in the 1st equation its 4Mn(OH)3 so how can you say the molar rations are equal??? Is it not 1:2

so wudnt you doulbe the moles of 2Mn(OH)3 to get molesof 4Mn(OH)3 and then divide that by 4 to get moles of O2????
It's like this: 2 moles of Mn 3+ to everything else in equation 2.
4 moles of Mn 3+ to everything else in equation 1.
Equations 1 and 2 are completely independent of each other, so you can't compare the ratio of moles of Mn 3+ between them

Hope that helped
13. (Original post by arvin_infinity)
Someone tell me how to remember which enthalpy change is +ve and -ve!!

Major rep!
You'd have to think about whether bonds are broken, or bonds are made; if broken, then it's endothermic (+ve) and if made, then exothermic (-ve)
14. Can someone please explain the concept of the feasibility of a reaction, i cant seem to get my head round it
15. (Original post by arvin_infinity)
Someone tell me how to remember which enthalpy change is +ve and -ve!!

Major rep!
exo is making bonds -Ve
endo is breaking bonds +Ve

ex-girlfriend makes *****es negative
enormous butt is breaking positive

LOL thats soooo crap....ahhahahahahaah
so sorry about that they dont even make sense
16. (Original post by goldlock)
It's like this: 2 moles of Mn 3+ to everything else in equation 2.
4 moles of Mn 3+ to everything else in equation 1.
Equations 1 and 2 are completely independent of each other, so you can't compare the ratio of moles of Mn 3+ between them

Hope that helped
i love you goldlock!
17. Can some one help me with the question in the text book module 2 examination questions in the OCR book, question 5 (a) how exactly do they calculate this? looked at the marks scheme and i dont even understand that
18. (Original post by Ayostunner)
Can someone please explain the concept of the feasibility of a reaction, i cant seem to get my head round it
A positive cell potential shows that a reaction is feasible, however it does not give any concept of how fast the rate of the reaction is.
The larger the difference between emf values, the more likely it is that the reaction will take place.
If the difference is less than 0.4V, then a reaction is not likely to happen.
19. (Original post by Ayostunner)
Can someone please explain the concept of the feasibility of a reaction, i cant seem to get my head round it
Someone correct me if i'm wrong but i believe the feasibilty of reaction is the possibility of a reaction occuring. 2 Ways to find this out, First is using delta G which must be <0. This is for spontaneous reaction to occur.

Another way is electrode potentials where the difference must mus 0.4v or larger.
20. Anyone here going to college. Did your teachers give you an idea of what topics may come up tomorrow. I'm an external candidate so have no idea and am revising everything even the AS. I would like to know what areas to particularly focus on. Thankyou

Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 10, 2012
The home of Results and Clearing

1,096

people online now

1,567,000

students helped last year
Today on TSR

Took GCSEs this summer?

Fill in our short survey for Amazon vouchers!

University open days

1. University of Buckingham
Thu, 23 Aug '18
2. University of Glasgow
Tue, 28 Aug '18
3. University of Aberdeen
Tue, 28 Aug '18
Poll
Useful resources

Can you help? Study Help unanswered threadsStudy help rules and posting guidelines

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE