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.
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calculate the concentration in gdm3 , of the undiluted hydrogen peroxide solution
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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)
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
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
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 It is in the middle of the straight section of the titration curve!
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
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.
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-
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
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.