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

1. (Original post by Rosi M)
I've seen them ask single ions in a few old spec papers so i went and learnt a few but your right, doubt they would ask that for new spec.
hope not, havn't learnt any it will probably be a mark if they do though!
2. (Original post by ruby321)
also cobalt hydroxide is a pink precipitate
Nah cobalt turns from pink to blue and then if it's left it turns back to pink, but markschemes apparently prefer blue.
3. (Original post by iSomething)
Hi, im stuck on this question from the OCR practice question section about redox equations, I'm having trouble balancing it as I cant seem to arrive at the answer they have at the back of the book:

"Aqueous thiosulfate ions, S2O3^2-, are oxidised to sulfate (VI) ions, SO4^2-, by chlorine gas, which is reduced to chloride ions." Use oxidation states to construct a redox equation."

Here's my attempt but im lost as to what I should be doing to the O atoms (I could add 2H2O but as it would solve the 2O atoms and could balance with 4H+ ions on other side but the book says its 5H2O and 10H+ on other side. How is that possible unless im being really stupid.

you've done everything right....so far

you have 3 oxygens on the left and 8 on the right, so you need 5 more oxygens....remember that the S203 is aquous, so water is also involved in the reaction...you need 5O's so 5H2O

so add 5 H2Os on the left....then balance the Hs on the right with 10H+

so the book is right

PS you also need to put a 2 infront of Cl- in the first equation...and then balance oxidation states
4. I'm on my phone so I can't quote more than 1 person, but thank you to everyone who answered
5. (Original post by ruby321)
hope not, havn't learnt any it will probably be a mark if they do though!
Yep doubt we would get a 12 marker on: Outline the colours of transition metal ions
I swear OCR would officially be doomed
6. (Original post by iSomething)
Hi, im stuck on this question from the OCR practice question section about redox equations, I'm having trouble balancing it as I cant seem to arrive at the answer they have at the back of the book:

"Aqueous thiosulfate ions, S2O3^2-, are oxidised to sulfate (VI) ions, SO4^2-, by chlorine gas, which is reduced to chloride ions." Use oxidation states to construct a redox equation."

Here's my attempt but im lost as to what I should be doing to the O atoms (I could add 2H2O but as it would solve the 2O atoms and could balance with 4H+ ions on other side but the book says its 5H2O and 10H+ on other side. How is that possible unless im being really stupid.
You're almost there

Each sulphur from the thio is oxidised from +2 to +4 in the sulphate, so you'll need to lose 2*2e- in total, and you'll need 4 Cl's (2Cl2) to take them.

There's 8 oxygens on the right, 3 on the left, so we need to add 5 O to the left - add 5H2O.

S2O3^2- +2Cl^2 +5H2O --> 2SO4^2- + 4Cl- + 10H+

A couple of things to comment on:
-Work out oxidation states for each atom individually first, adding up the total electron transfer after.
-Its okay to add H2O or H+ as reactants or products in a reaction! Don't be afraid to do it
7. Hey all.
For general interest, does anyone know what topics, broadly speaking, are most likely to come up in tomorrow's exam by way of what's come up in past papers?

I feel like this would be something that a lot of people, not disincluding myself, would find really helpful.
8. (Original post by Rogercbinboy)
Nah cobalt turns from pink to blue and then if it's left it turns back to pink, but markschemes apparently prefer blue.
Oh yeah your right, the text book made a mistake in the summary! thanks
9. (Original post by Kalamari Dave)
You're almost there

Each sulphur from the thio is oxidised from +2 to +4 in the sulphate, so you'll need to lose 2*2e- in total, and you'll need 4 Cl's (2Cl2) to take them.

There's 8 oxygens on the right, 3 on the left, so we need to add 5 O to the left - add 5H2O.

S2O3^2- +2Cl^2 +5H2O --> 2SO4^2- + 4Cl- + 10H+

A couple of things to comment on:
-Work out oxidation states for each atom individually first, adding up the total electron transfer after.
-Its okay to add H2O or H+ as reactants or products in a reaction! Don't be afraid to do it
you need 4Cl2 and 8Cl-

if you look at the overall molecular charges, they need to balance on both sides
so

2- + 0 + 0 --> 2x2- + ?Cl - + 10x1+

= 2- --> 4- + ?Cl0 + 10+
= 2- --> 6+ + ?Cl-

so the quation mark need to be 8 in order to make both sides balance and therefore you have 4Cl2 on the left
10. (Original post by Rosi M)
Yep doubt we would get a 12 marker on: Outline the colours of transition metal ions
I swear OCR would officially be doomed
I know we've spent so long learning pH etc. and then they ask us about colours....
haha
I would cry if that happened :/
11. June 2010 f325
Q7 b)
The student diluted 25cm3 of a solution of hydrogen peroxide with water and made the solution up to 250cm3. The student Titrated 25cm3 of this solution with 0.02 moldm-3 KMno4 under acidic conditions. The volume of KMno4(aq) required to reach the end point was 23.45cm3.

• calculate the concentration in gdm3 , of the undiluted hydrogen peroxide solution
• What volume of oxygen gas, measured at RTP, would be during this Tirtration

THE mark scheme is rubbish HELP!!

oh i think u may need this 2MnO4- + 6H+ + 5H2O2 -> 2Mn2+ + 8H2O + 5O2 this is the equation for 7a)
12. (Original post by azz92)
June 2010 f325
Q7 b)
The student diluted 25cm3 of a solution of hydrogen peroxide with water and made the solution up to 250cm3. The student Titrated 25cm3 of this solution with 0.02 moldm-3 KMno4 under acidic conditions. The volume of KMno4(aq) required to reach the end point was 23.45cm3.

• calculate the concentration in gdm3 , of the undiluted hydrogen peroxide solution
• What volume of oxygen gas, measured at RTP, would be during this Tirtration

THE mark scheme is rubbish HELP!!

oh i think u may need this 2MnO4- + 6H+ + 5H2O2 -> 2Mn2+ + 8H2O + 5O2 this is the equation for 7a)
Go back to page 58, i ave done it on there. near the bottom
13. i've done 3 papers today and have got E, D, D. I need an A in this paper
14. (Original post by Princess_perfect786)
i've done 3 papers today and have got E, D, D. I need an A in this paper
Which papers did you do?
15. (Original post by Princess_perfect786)
i've done 3 papers today and have got E, D, D. I need an A in this paper
and you're going to the same uni as me

ballsed up maths yesterday, so I need an A in chem now, otherwise...bye bye University
16. (Original post by student777)
When you do a redox equation, before you work out oxidation states you must make sure the things being redoxed are balanced. You haven't done that for Cl in the first equation. Try it again, this time balancing Cl

(Original post by viksta1000)
you've done everything right....so far

you have 3 oxygens on the left and 8 on the right, so you need 5 more oxygens....remember that the S203 is aquous, so water is also involved in the reaction...you need 5O's so 5H2O

so add 5 H2Os on the left....then balance the Hs on the right with 10H+

so the book is right

PS you also need to put a 2 infront of Cl- in the first equation...and then balance oxidation states

(Original post by Kalamari Dave)
You're almost there

Each sulphur from the thio is oxidised from +2 to +4 in the sulphate, so you'll need to lose 2*2e- in total, and you'll need 4 Cl's (2Cl2) to take them.

There's 8 oxygens on the right, 3 on the left, so we need to add 5 O to the left - add 5H2O.

S2O3^2- +2Cl^2 +5H2O --&gt; 2SO4^2- + 4Cl- + 10H+

A couple of things to comment on:
-Work out oxidation states for each atom individually first, adding up the total electron transfer after.
-Its okay to add H2O or H+ as reactants or products in a reaction! Don't be afraid to do it
Thanks everyone, much appreciated. Cant believe I forgot the 2 infront of the 2SO4- seems obvious now that there are 8O.
17. (Original post by ruby321)
hmm yeah it probably does, I remember doing the titration and there definitly was no white ppt at the end...
The book fails to say what happens to it!
So im hoping that it wont come up
18. Can anyone help explain question 6 part be please. I got the wrong answer for the second equation because I can't balance the charges.

19. OK simple Qs
Construct the equation for the oxidation of acidified iron(II) ions by oxygen..How do I start to do this?
i get
Fe2+ + O2 + 4H+--------> Fe3+ + 2H2O
But the answer is this, How?
4Fe2+ + O2 + 4H+-------------> 4Fe3+ + 2H2O

Why the 2 in front of the Fe?
20. (Original post by dunnicare)
Can anyone help explain question 6 part be please. I got the wrong answer for the second equation because I can't balance the charges.

Cr2O7^2- + 14H+ + 8e- ------> 2Cr2+ + 7H20

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