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

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.

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

is the change in the oxidation no. of O from FeO42- to O2 should be +2 ?

there's no need as that is the right answer

the way I did it was to find the concentration of H2O2 in 250 which was 0.0469 mol dm-3

n=mass/molar mass, so mass = nMr

so 0.0469x34 = 1.5946g dm-3 in the 250cm3 solution

the concentration in 250 is going to be 1/10 of the concentration in the original 25cm3

so x10 = 15.9g dm3

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

there's no need as that is the right answer

the way I did it was to find the concentration of H2O2 in 250 which was 0.0469 mol dm-3

n=mass/molar mass, so mass = nMr

so 0.0469x34 = 1.5946g dm-3 in the 250cm3 solution

the concentration in 250 is going to be 1/10 of the concentration in the original 25cm3

so x10 = 15.9g dm3

Hi, here some more recently LEGACY QP and MS, has similar questions to the F325 specification. However there may have some question which is not relevant to our specification.

2816-01 - Unifying Concepts in Chemistry questions - Kc, Ka, Equilibrium, rate- determining step, Orders, Buffers
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions - Born-Haber Cycles, Standard enthalpy, Transiton metal questions as well.

You can also find these question paper on the OCR website.

Could someone help me with Q4,b,ii

2011 Jan.

Thanks

what is this!?!!?
Since when do we need to know this!?
What are you doing to me!?

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

but isn't the 250cm3 solution diluted? they are asking for undiluted
just a thought. correct me if i'm wrong

I get CH3OH + 1/2O2 -------> C02 + 4H+ +4e-

Is this right?

These questions are good! have you got anymore? do you reckon these questions will be similar to what we get tommorow?

I got something like that too but the MS says CH3OH + H2O -------->6H+ +6e- +CO2

I just can't seem to figure out why.

your wrong


if you work out the concentration of H2O2 in the 250, you know the diluted concentration of H2O2

the amount of H2O2 in the dilute sample is 25cm3 in 250cm3, which is the same as 1/10th of the original sample, so if you x10 you get the amount in the original sample

I got something like that too but the MS says CH3OH + H2O -------->6H+ +6e- +CO2

I just can't seem to figure out why.