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ocr a f325 revision thread

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could someone please help me with question 4bii, struggling to work out the half equation.

http://ocr.org.uk/Images/65361-question-paper-unit-f325-equilibria-energetics-and-elements.pdf

thanks!

This one is pretty hard to explain but basically what you do is subtract the half equation they give you from the full equation you've written out.

If you can't subtract something because it exists in the half equation but not in the full equation, you add it to the other side.

So the full equation is CH3OH + 3/2O2 -> CO2 + 2H2O

You 'subtract' the O2 from the half equation and you have

CH3OH + 1/2O2 -> CO2 + 2H2O

Then you want to 'subtract' the 4H+ but it doesn't exist, same with 4e- so you add it to the other side

CH3OH + 1/2O2 -> CO2 + 2H2O + 4H+ 4e-

Lastly you want to subtract 2H2O so it becomes

CH3OH + 1/2O2 -> CO2 + 4H+ + 4e-

I think thats the right answer but just double check.

(edited 10 years ago)

Reply 1861

FailedMyExams

Original post by Namod

This method doesn't work because
If I go to OH- instead of Mn(OH)3, this is what happens:
2 mole of I(-) reacts 2 moles of OH- so =2.46x10^-5.
8 moles of OH- is formed by 1 mole of O2, so O2 moles = mole of O2 = 2.46x10^-5/8.
=3.08x10^-6

That's completely different, the OH reacted in the first equation isn't the same OH that's produced in the second. After all, why would you react OH just to get the same amount of OH produced?That's why you're getting a different answer.

Reply 1862

needtosucceed=)

Original post by FailedMyExams

ah thanks for trying to explain but thats not the answer in the mark scheme, this is:

ch3oh + h20 = 6h+ + 6e- + co2

Reply 1863

eggfriedrice

Guys I'm stuck on question 3 e ii on Jan 2012. I haven't looked at the mark scheme because I actually want to know the thought process behind it.

Reply 1864

Pride

Original post by Namod

sorry I didn't reply, I just did the question and left to revise something else without checking.

Had a look at the mark scheme and I've got the same answer. I'll try to explain it.

Original post by Pride

number of moles of sodium thio = 0.001 x 0.00246dm^3 = 2.46x10^-5 mol
half that for no. of moles of I2. Double that back up for no. of moles of Mn(OH)3, divide by 4 for no. of moles of O2, I get 6.15x10^-6 mol of O2.
m = n x Mr = 6.15 x 10^-6 x 32 = 1.968 x 10-4 g = 0.1968mg
DOC = 0.1968mg / 0.025dm^3 = 7.872

right so you calculate the no. of moles of sodium thio. Then you know that half that number of moles of I2 reacted right? Now all of that I2 was produced from the previous reaction. So double the number of moles of Mn(OH)3 must have been reacted.

Again all of that Mn(OH)3 was produced in the previous reaction, so a quarter of that number of moles of O2 must have been reacted.
Then that's m = n x Mr, and x1000 to convert to mg.

Reply 1865

FailedMyExams

Original post by needtosucceed=)

ah thanks for trying to explain but thats not the answer in the mark scheme, this is:

ch3oh + h20 = 6h+ + 6e- + co2

Wtf, if you add that to the half equation they give you, you get

CH3OH + O2 --> H2O + 2H+ + 2e- + CO2

Doesn't balance, neither do charges? Edit: Nvm it does balance I'm blind.

Can someone shed some light on how you do it please?

4bii http://ocr.org.uk/Images/65361-question-paper-unit-f325-equilibria-energetics-and-elements.pdf

(edited 10 years ago)

Reply 1866

Pride

Original post by eggfriedrice

Guys I'm stuck on question 3 e ii on Jan 2012. I haven't looked at the mark scheme because I actually want to know the thought process behind it.

oh yea the silver nitrate question.

Basically they want you to recognise the precipitate is AgCl. So n = m/Mr to find the number of moles of AgCl formed, and it's 0.02mol.

Think about it this way, 0.01 mol of a complex, reacted with excess Ag+ to form 0.02mol AgCl. This must mean that each mole of complex has 2 Cl- ions.

The only one that this is the case for is B.

Reply 1867

Pride

Original post by FailedMyExams

Wtf, if you add that to the half equation they give you, you get

CH3OH + O2 --> H2O + 2H+ + 2e- + CO2

Doesn't balance, neither do charges? Edit: Nvm it does balance I'm blind.

Can someone else take a look at this please?

I got the answer you got earlier...

Reply 1868

ofudge

Are we going to be asked to draw something with the EDTA ligand? Such as cu(EDTA)2-??

Reply 1869

MathsNerd1

Original post by ofudge

Are we going to be asked to draw something with the EDTA ligand? Such as cu(EDTA)2-??

I don't think they could ever ask you to draw it as I'd not even know how to do it! :-/

Posted from TSR Mobile

Reply 1870

needtosucceed=)

Original post by FailedMyExams

Original post by Pride

I got the answer you got earlier...

I dont think the mark schemes wrong because I checked the examiners report and they didnt mention anything.
it was only 1 mark so I suppose it doesnt matter too much :smile:

Reply 1871

eggfriedrice

Original post by Pride

Ah ok! So 1:2 ratio hence the complex with 2 Cl. Why can't it be C and D? In the actual complex ion there are 2 Cl. Also I don't understand how there's a complex and random Cl-ions added.

Reply 1872

Dr00n

Original post by Namod

Didn't get your message.
Are you calling me immature? :mad:

no, i'm calling my self immature because i laughed, your very mature for not seeing how someone could find the diagram a bit humorous

Reply 1873

Pride

Original post by eggfriedrice

Ah ok! So 1:2 ratio hence the complex with 2 Cl. Why can't it be C and D? In the actual complex ion there are 2 Cl. Also I don't understand how there's a complex and random Cl-ions added.

Perhaps think of them as if they were any old ionic compound, a 2+ ion bound to 2 Cl- ions in the ionic lattice. Overall charge of course is 0.

C and D have 2 Cl- ligands, they don't come off of the transition metal ion because they are covalently bonded.

Reply 1874

Pride

Original post by needtosucceed=)

I dont think the mark schemes wrong because I checked the examiners report and they didnt mention anything.
it was only 1 mark so I suppose it doesnt matter too much :smile:

Yep, I'm thrown by that. I might email one of my teachers now, if I can't come up with an explanation myself...

Reply 1875

eggfriedrice

Original post by Pride

Thanks. You've made it really clear. :smile:

Also just to confirm, CuCl4 wouldn't react with aqueous silver nitrate since they're also covalently bonded?

I have another question (not related to this one), if you or anyone else can explain would be awesome ;

So in a fuel cell, say a Ni(s) and Ni2+(aq). Why does the Ni(s) gain or lose mass? I really don't understand it.

I mean surely the only transfer we have are electrons but the mass of electrons are pretty much negligible so how is it possible we even have a measurable change in mass?

(edited 10 years ago)

Reply 1876

needtosucceed=)

Original post by Pride

Yep, I'm thrown by that. I might email one of my teachers now, if I can't come up with an explanation myself...

could you post back on here if you figure it out please :smile:

Reply 1877

chignesh10

Original post by needtosucceed=)

I dont think the mark schemes wrong because I checked the examiners report and they didnt mention anything.
it was only 1 mark so I suppose it doesnt matter too much :smile:

Original post by Pride

Yep, I'm thrown by that. I might email one of my teachers now, if I can't come up with an explanation myself...