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

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Reply 1240
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)



I got 0.399 gdm and 280 cm of o2
Original post by CoventryCity
Yes you need to know the hydrogen fuel cell one they are

1/2 O2 + H20 +2e--------> 2OH- +0.4V
2H2O +2e- ----------> H2 + 2OH- -0.83V

These are the half equations


Do we need to know the half equations? Or just the overall equation??
Reply 1242
raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
June 2010 qeustion 7b)
Original post by Josh0806
Could somebody please explain to me how to balance this equation...

FeO42(–) + H(+) ? Fe(3+)+ O2+ H2O

...to give this...

4FeO42(–) + 20H(+) ? 4Fe(3+)+ 3O2+ 10H2O


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

there's your answer :smile:
Reply 1244
Original post by percy93
I got 0.399 gdm and 280 cm of o2


anyone agree
Has anyone realised that Born Haber cycles have not come in Jan 2011, so they'll defintely come up tommorow?
Reply 1246
Original post by azz92
raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
June 2010 qeustion 7b)


Use the formula n = v/24

Rearrange it, v = n x 24

As you got the n = (1.1725 x 10^–3) x 24 = 0.0281 dm3
Okay, I have read and done the past paper and somethings you people come up with in this forum JUST SCARE ME COMPLETELY , makes me feel like I DONT KNOW ANYTHING
Reply 1248
Original post by azz92
raj16 you never finished the qeustion what about the volume of oxygen gas can someone tell me why they got 0.0281 dm3
June 2010 qeustion 7b)


sorry! i forgot.

.
You knw mols of H202 1.1725x10-3.
ratio 5:5 so no of mols of 02 is the same.
using mols=vol/24
1.1725x10-3 X 24=0.0281
Reply 1249
Original post by SmartFool
ANYONE PLEASE! How do you work out where the equivalence line is on titration curves as Ive seen on a past paper where the line is at 33cm3 and not 25cm3


It isn't always 25 :smile:
It is in the middle of the straight section of the titration curve!
Do we have to remember any equations, such as the redox titrations concerning MnO4 and Fe?
Someone tell me what we need to know about "H for the future"
and "Storage and fuel cell" ! My teacher left it for us to read
Original post by ViralRiver
Do we have to remember any equations, such as the redox titrations concerning MnO4 and Fe?


As far as I remember u can deduce them from the data given! But let me know if there is any :rolleyes:
Reply 1253
Original post by arvin_infinity

Original post by arvin_infinity
Someone tell me what we need to know about "H for the future"
and "Storage and fuel cell" ! My teacher left it for us to read


umm pretty much the advantages and disadvantages of hydrogen fuel cells and storage of hydrogen...i think anyway, well that's all i've came across so far :smile:
Reply 1254
Okay, so in the textbook it gives that table showing what happens when you add NaOH to different solutions.

For example, it says when you add NaOH to Co(H20)6 you get a pink solution and a blue precipitate.

Does the pink solution form from the NaOH as well or is that already there and the NaOH just makes a precipitate?
I really need to do well on this tomorrow, theres no hope in bio. PLZ BE A DECENT PAPER PLZ :sigh:.
Reply 1256
Original post by percy93
anyone agree


Surely you need to divide your 0.399 by 25/1000 because you've figured out the grams in 25cm3 but you need in dm3 so I get 15.9g dm3. which seems big. Hmm maybe I'm wrong, someone correct me.
Original post by Vidja
Okay, so in the textbook it gives that table showing what happens when you add NaOH to different solutions.

For example, it says when you add NaOH to Co(H20)6 you get a pink solution and a blue precipitate.

Does the pink solution form from the NaOH as well or is that already there and the NaOH just makes a precipitate?


nooooo, NaOH just acts as the source of OH- ions, as its when complexes react with OH- ions that precipitation occurs

The pink solution is the Co2+ ions as they are pink in aqueous solutions, so the fact that there is still some pink left suggests that not all of the Co2+ ions have reacted with the OH-
Original post by arvin_infinity
Someone tell me what we need to know about "H for the future"
and "Storage and fuel cell" ! My teacher left it for us to read


You need to know the hydrogen-oxygen fuel cell which is
1/O2 + H2O + 2e- --------> 2OH- +0.4V
2H2O + 2e- --------> H2 + 2OH- -0.83V

These are the half equations for it

Fuel cells are good as more efficient and less pollution
Storage of Hydrogen at ridiculous conditions in liquid form. Or stored on adsorbers or absorbers.
-ves are difficult to transport, adsorbers and absorbers have limited life
Original post by Rocks:)
Surely you need to divide your 0.399 by 25/1000 because you've figured out the grams in 25cm3 but you need in dm3 so I get 15.9g dm3. which seems big. Hmm maybe I'm wrong, someone correct me.


15.9 is correct

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