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# Please Help me on a Chemistry F325 Past Paper Titration Question! watch

1. Jan 2013 Last Question on working out it's oxidation number change and then the overall equation. The number of moles are ok to work out- after this it gets confusing!

Question Paper

http://www.ocr.org.uk/Images/144762-...d-elements.pdf

Mark Scheme

http://www.ocr.org.uk/Images/142258-...ts-january.pdf
2. Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses.

I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed.

Spoiler:
Show
(Original post by Puddles the Monkey)
x
3. You say you get the moles but to be sure.

Calc no of moles MnO4- in the 13.2 cm3 = 2.97 x 10^-4
Calc moles V dissolved in the 50 cm3 acid solution = 2.48 x 10^-3

Only 10 cm3 of the 50 cm3 were used so calc moles of V in 10 cm3 by dividing by 5.
2.48 x 10-3 / 5 = 4.96 x 10^-4

All this V metal formed the Vn+ ions so this give you a reacting ration between the MnO4- and Vn+
The ratio is therefore moles Vn+ / moles MnO4- = 4.96 x 10-4 / 2.97 x 10-4 = 1.67 so ratio is 1.67 : 1
Find lowest common multiple. Multiply by 3 = 5:3

To find value of n in Vn+
With the 5:3 ratio the number of electrons gained and lost must be equal.

3MnO4- + 15e --> 3Mn2+ so 15 electrons gained in all
5Vn+ --> 5VO3- + 15e to balance the 15 gained by MnO4- so each Vn+ must lose 3 electrons and end up in +5 oxidation state to Vn+ must be V2+

Equation must be 5V2+ + 3MnO4- + 3H2O --> 5VO3- + 3Mn2+ + 6H=
4. Okay, I have gone through it and i'll try to explain it as best as I can to you, here it goes!.

So, the first thing we need to do is find the number of moles of MnO^4- reacted. Then through some logic we're going to work out the number of moles of V^n+ that reacted with it. Then we're going to divide one by the other to get the molar ratio - that is basically what we're trying to find. Once we have the molar ratio we can deduce the oxidation state that corresponds to n+.

Firstly, lets do as stated and find the number of moles of MnO^4- ; 2.25*10^-2 * 0.0132 = 2.97*10^-4 mol.

Done! Okay, now we need the number of moles of V^n+ reacted, Well we know the mass of Vanadium (0) dissolved and we know the Mr of Vanadium (50.9 gmol^-1) so we can find the number of moles of Vanadium dissolved and since all of the Vanadium is oxidised to +5 oxidation state, the number of moles must equate.i.e. moles of V = moles of V^5+
so, 0.126 / 50.9 = 2.475*10^-3 mol of V^+5

Now we know the solution is reduced, so we know all of those V^+5 ions are all reduced to V^n+, so the moles of V^n+ = 2.475*10^-3 mol.

That is the number of moles of V^n+ in 50 cm^3. We reacted only 10 cm^3 of that solution, so if you like, we removed a fifth of those V^+n, so the number of V^+n reacted =2.475*10^-3 / 5 = 4.95*10^-4 mol.

So now we have the number of moles of MnO4^- and V^+n.to find their ratio; V^+n : MnO4^- we divide their moles, hence the term 'molar ratio'.this ratio is equal to: V^+n / MnO4^- = 4.95*10^-4 / 2.97*10^-4 = 5/3.

So we know now that for every 3 moles of MnO4^- that reacted, 5 moles of V^n+ reacted. The equation for reduction of V^+5 could be laid out as so:V^+5 ----> V^n+ + (5-n)e^-1.
^ study this and make sure it makes sense.Then balance with the molar ratio 5/3 like any other redox reaction (if you don't know what i mean, I suggest you read up on redox)

3MnO4^- + 3*5 e^-1

5V^5+ ----> 5V^n+ + 5*(5-n) e^-1

since we know the equations must balance, the electrons must balance, in the MnO4^- there are 15 electrons, so the 5*(5-n) must equal 15 too!

25 - 5n = 15

10 = 5n

2 = n

so n is +2!I hope that helped, please feel free to ask for clarity if you so need it.

Edit: Grammar.
5. (Original post by Madasahatter)
You say you get the moles but to be sure.

Calc no of moles MnO4- in the 13.2 cm3 = 2.97 x 10^-4
Calc moles V dissolved in the 50 cm3 acid solution = 2.48 x 10^-3

Only 10 cm3 of the 50 cm3 were used so calc moles of V in 10 cm3 by dividing by 5.
2.48 x 10-3 / 5 = 4.96 x 10^-4

All this V metal formed the Vn+ ions so this give you a reacting ration between the MnO4- and Vn+
The ratio is therefore moles Vn+ / moles MnO4- = 4.96 x 10-4 / 2.97 x 10-4 = 1.67 so ratio is 1.67 : 1
Find lowest common multiple. Multiply by 3 = 5:3

To find value of n in Vn+
With the 5:3 ratio the number of electrons gained and lost must be equal.

3MnO4- + 15e --> 3Mn2+ so 15 electrons gained in all
5Vn+ --> 5VO3- + 15e to balance the 15 gained by MnO4- so each Vn+ must lose 3 electrons and end up in +5 oxidation state to Vn+ must be V2+

Equation must be 5V2+ + 3MnO4- + 3H2O --> 5VO3- + 3Mn2+ + 6H=
(Original post by moontheloon)
Okay, I have gone through it and i'll try to explain it as best as I can to you, here it goes!.

So, the first thing we need to do is find the number of moles of MnO^4- reacted. Then through some logic we're going to work out the number of moles of V^n+ that reacted with it. Then we're going to divide one by the other to get the molar ratio - that is basically what we're trying to find. Once we have the molar ratio we can deduce the oxidation state that corresponds to n+.

Firstly, lets do as stated and find the number of moles of MnO^4- ; 2.25*10^-2 * 0.0132 = 2.97*10^-4 mol.

Done! Okay, now we need the number of moles of V^n+ reacted, Well we know the mass of Vanadium (0) dissolved and we know the Mr of Vanadium (50.9 gmol^-1) so we can find the number of moles of Vanadium dissolved and since all of the Vanadium is oxidised to +5 oxidation state, the number of moles must equate.i.e. moles of V = moles of V^5+
so, 0.126 / 50.9 = 2.475*10^-3 mol of V^+5

Now we know the solution is reduced, so we know all of those V^+5 ions are all reduced to V^n+, so the moles of V^n+ = 2.475*10^-3 mol.

That is the number of moles of V^n+ in 50 cm^3. We reacted only 10 cm^3 of that solution, so if you like, we removed a fifth of those V^+n, so the number of V^+n reacted =2.475*10^-3 / 5 = 4.95*10^-4 mol.

So now we have the number of moles of MnO4^- and V^+n.to find their ratio; V^+n : MnO4^- we divide their moles, hence the term 'molar ratio'.this ratio is equal to: V^+n / MnO4^- = 4.95*10^-4 / 2.97*10^-4 = 5/3.

So we know now that for every 3 moles of MnO4^- that reacted, 5 moles of V^n+ reacted. The equation for reduction of V^+5 could be laid out as so:V^+5 ----> V^n+ + (5-n)e^-1.
^ study this and make sure it makes sense.Then balance with the molar ratio 5/3 like any other redox reaction (if you don't know what i mean, I suggest you read up on redox)

3MnO4^- + 3*5 e^-1

5V^5+ ----> 5V^n+ + 5*(5-n) e^-1

since we know the equations must balance, the electrons must balance, in the MnO4^- there are 15 electrons, so the 5*(5-n) must equal 15 too!

25 - 5n = 15

10 = 5n

2 = n

so n is +2!I hope that helped, please feel free to ask for clarity if you so need it.

Edit: Grammar.
I can't thank you enough for taking your time to do this question and help me!
So for the molar ratio, you always do the larger number / by the smaller?
Also, I now understood upto the way you got a 5:3 ratio. But then putting things together to get an equation and the number of electrons completely threw me off! Please could you explain again the last bit thanks !

Here is a link to another titration question I am struggling with. Any help would be much appreciated, http://www.thestudentroom.co.uk/show....php?t=3330755
6. (Original post by RevisionNad)
I can't thank you enough for taking your time to do this question and help me!
So for the molar ratio, you always do the larger number / by the smaller?
Also, I now understood upto the way you got a 5:3 ratio. But then putting things together to get an equation and the number of electrons completely threw me off! Please could you explain again the last bit thanks !
Sure I can! To answer your first question, no you don't have to do larger moles/smaller moles, I did V+n / MnO4- so that the top number of the fraction would give the 'molar term' for V+n and the bottom number would give MnO4-. Doing it the other way up MnO4- /V+n would equal 3/5 and then the top number would give the 'molar term' for MnO4-, so that would work too. I can't word it correctly but basically, imagine this: a=1, b=2. a/b = 1/2 b/a = 2/1. So in both cases, the numerator (top number) of a fraction will represent the top term in the division. So for a/b, top number is 1 and a is equal to 1, etc. I hope that makes sense.

I'm happy you got why the ratio is 5/3. That is crucial to understanding the equations.
For starters, two 'half reactions' occur when we titrate V+n and MnO4-.
In this case, it states that MnO4- gets reduced (gains electrons) and V+n loses some electrons. On reviewing the half equation I wrote previously ' V^+5 ----> V^n+ + (5-n)e^-1' this would actually be wrong. Let me explain this all again.

So as stated MnO4- is reduced and as a result it has to get its electrons from the V+n it reacted with, so V+n must have been oxidised (lose electrons).

We know the half reaction for MnO4- is:
MnO4- + 8H+ + 5e- -----> Mn2+ + 4H2O.

And we can deduce this equation for the oxidation of V+n
V+n -----> V+5 + (5-n)e-
^ I'm assuming this equation was where some of the confusion occured since I just came out with it (and I wrote it wrong) and in a way I did, but I arrived at it with some logic and it goes like this: If MnO4- reacts with V+n and MnO4- gets reduced, it needs to get electrons from somewhere, and those electrons will be gained from V+n losing them (Redox titrations, read about them if you haven't already). So V+n must be oxidised to V+5. The number of electrons 'given out' by V+n would have to be 5 - n. Think about it, if +n was +3 then to oxidise V+3 to +5 you'd have to give out two electrons, notice: 5 - 3 = 2 so the molar ratio of V+n to e- is 1x(5-n).

Now we get to balancing these
we know the molar ratios were 5:3. This means that for every 3 moles of MnO4- that reacted, 5 moles of V+n reacted.

so, the MnO4- equation becomes:
3x MnO4- + 8H+ + 5e- -----> Mn2+ + 4H2O.
which equals: 3MnO4- + 24H+ + 15e- -----> 3Mn2+ + 12H2O.

and for the V+n reaction:
5x V+n -----> V+5 + (5-n)e-
which equals: 5V+n -----> 5V+5 + 5*(5-n)e-

When you combine the two half equations like you would for any redox reaction, once you've balanced by taking into account the molar ratios, this means the both reactions must have the same number of electrons in their equation. (I know I keep banging on about this but unless you've read up on and understood redox this won't make sense, so be sure to do that) then we can solve for n with algebra.

In the MnO4- reaction there is 15 electrons, so there has to be 15 electrons in the Vn+ equation. We know in the Vn+ reation we have 5*(5-n) electrons and this amount has to equal 15! so...
5*(5-n) = 25 - 5n
25 - 5n = 15
-5n = -10
n = 2

I hope that is clearer to you!
7. (Original post by moontheloon)
Sure I can! To answer your first question, no you don't have to do larger moles/smaller moles, I did V+n / MnO4- so that the top number of the fraction would give the 'molar term' for V+n and the bottom number would give MnO4-. Doing it the other way up MnO4- /V+n would equal 3/5 and then the top number would give the 'molar term' for MnO4-, so that would work too. I can't word it correctly but basically, imagine this: a=1, b=2. a/b = 1/2 b/a = 2/1. So in both cases, the numerator (top number) of a fraction will represent the top term in the division. So for a/b, top number is 1 and a is equal to 1, etc. I hope that makes sense.

I'm happy you got why the ratio is 5/3. That is crucial to understanding the equations.
For starters, two 'half reactions' occur when we titrate V+n and MnO4-.
In this case, it states that MnO4- gets reduced (gains electrons) and V+n loses some electrons. On reviewing the half equation I wrote previously ' V^+5 ----> V^n+ + (5-n)e^-1' this would actually be wrong. Let me explain this all again.

So as stated MnO4- is reduced and as a result it has to get its electrons from the V+n it reacted with, so V+n must have been oxidised (lose electrons).

We know the half reaction for MnO4- is:
MnO4- + 8H+ + 5e- -----> Mn2+ + 4H2O.

And we can deduce this equation for the oxidation of V+n
V+n -----> V+5 + (5-n)e-
^ I'm assuming this equation was where some of the confusion occured since I just came out with it (and I wrote it wrong) and in a way I did, but I arrived at it with some logic and it goes like this: If MnO4- reacts with V+n and MnO4- gets reduced, it needs to get electrons from somewhere, and those electrons will be gained from V+n losing them (Redox titrations, read about them if you haven't already). So V+n must be oxidised to V+5. The number of electrons 'given out' by V+n would have to be 5 - n. Think about it, if +n was +3 then to oxidise V+3 to +5 you'd have to give out two electrons, notice: 5 - 3 = 2 so the molar ratio of V+n to e- is 1x(5-n).

Now we get to balancing these
we know the molar ratios were 5:3. This means that for every 3 moles of MnO4- that reacted, 5 moles of V+n reacted.

so, the MnO4- equation becomes:
3x MnO4- + 8H+ + 5e- -----> Mn2+ + 4H2O.
which equals: 3MnO4- + 24H+ + 15e- -----> 3Mn2+ + 12H2O.

and for the V+n reaction:
5x V+n -----> V+5 + (5-n)e-
which equals: 5V+n -----> 5V+5 + 5*(5-n)e-

When you combine the two half equations like you would for any redox reaction, once you've balanced by taking into account the molar ratios, this means the both reactions must have the same number of electrons in their equation. (I know I keep banging on about this but unless you've read up on and understood redox this won't make sense, so be sure to do that) then we can solve for n with algebra.

In the MnO4- reaction there is 15 electrons, so there has to be 15 electrons in the Vn+ equation. We know in the Vn+ reation we have 5*(5-n) electrons and this amount has to equal 15! so...
5*(5-n) = 25 - 5n
25 - 5n = 15
-5n = -10
n = 2

I hope that is clearer to you!
THANK YOU SOOO MUCH! You are amazing at chemistry. You have helped me a lot, I hope you get what you want and more, seriously thanks a ton!
8. (Original post by RevisionNad)
THANK YOU SOOO MUCH! You are amazing at chemistry. You have helped me a lot, I hope you get what you want and more, seriously thanks a ton!
Pleasure to be of service, if you have any more questions, etc, feel free to ask!
9. (Original post by moontheloon)
Pleasure to be of service, if you have any more questions, etc, feel free to ask!
Will do so thanks! Oh, can you please check out the link above for another Q I was having issues with?

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