Balancing Equations Using Oxidation Numbers
So I have this equation here:
MnO2 + FeSO4 + H2SO4 - > Fe2(SO4)3 + MnSO4 + H2O.
What I'd like to do is balance it by using oxidation numbers, and identify the oxidised and reduced spieces.
So on the left hand side I said that magnesium has an oxidation number of +4 because the two oxygen atoms are -2 each.
I said that the sulfur in the H2SO4 has +6 because the two hydrogen are +2 and the four oxygens are -8.
I didn't know how to deal with the middle term on the left hand side so I said S is -2 because it's under oxygen and oxygen is -8 making Fe + 10.
How's that so far? I'm thinking very wrong somehow, but I can't seem to find any rules dealing with S or Fe atoms??
I employed similar methods on the right hand side to get Mn an ox number of +10. Then I said that as oxidized since it increased.
I said (S04)3 is -30 ? Then I said Fe2 should be +30 and each Fe atom +15, so that's also oxidized, which can't be right..
I don't really have a clue to be honest. Any help would be appreciated.
MnO2 + FeSO4 + H2SO4 - > Fe2(SO4)3 + MnSO4 + H2O.
What I'd like to do is balance it by using oxidation numbers, and identify the oxidised and reduced spieces.
So on the left hand side I said that magnesium has an oxidation number of +4 because the two oxygen atoms are -2 each.
I said that the sulfur in the H2SO4 has +6 because the two hydrogen are +2 and the four oxygens are -8.
I didn't know how to deal with the middle term on the left hand side so I said S is -2 because it's under oxygen and oxygen is -8 making Fe + 10.
How's that so far? I'm thinking very wrong somehow, but I can't seem to find any rules dealing with S or Fe atoms??
I employed similar methods on the right hand side to get Mn an ox number of +10. Then I said that as oxidized since it increased.
I said (S04)3 is -30 ? Then I said Fe2 should be +30 and each Fe atom +15, so that's also oxidized, which can't be right..
I don't really have a clue to be honest. Any help would be appreciated.
(edited 11 years ago)
Original post by Freeway Lemonade
So I have this equation here:
MnO2 + FeSO4 + H2SO4 - > Fe2(SO4)3 + MnSO4 + H2O.
What I'd like to do is balance it by using oxidation numbers, and identify the oxidised and reduced spieces.
So on the left hand side I said that magnesium has an oxidation number of +4 because the two oxygen atoms are -2 each.
I said that the sulfur in the H2SO4 has +6 because the two hydrogen are +2 and the four oxygens are -8.
I didn't know how to deal with the middle term on the left hand side so I said S is -2 because it's under oxygen and oxygen is -8 making Fe + 10.
How's that so far? I'm thinking very wrong somehow, but I can't seem to find any rules dealing with S or Fe atoms??
I employed similar methods on the right hand side to get Mn an ox number of +10. Then I said that as oxidized since it increased.
I said (S04)3 is -30 ? Then I said Fe2 should be +30 and each Fe atom +15, so that's also oxidized, which can't be right..
I don't really have a clue to be honest. Any help would be appreciated.
MnO2 + FeSO4 + H2SO4 - > Fe2(SO4)3 + MnSO4 + H2O.
What I'd like to do is balance it by using oxidation numbers, and identify the oxidised and reduced spieces.
So on the left hand side I said that magnesium has an oxidation number of +4 because the two oxygen atoms are -2 each.
I said that the sulfur in the H2SO4 has +6 because the two hydrogen are +2 and the four oxygens are -8.
I didn't know how to deal with the middle term on the left hand side so I said S is -2 because it's under oxygen and oxygen is -8 making Fe + 10.
How's that so far? I'm thinking very wrong somehow, but I can't seem to find any rules dealing with S or Fe atoms??
I employed similar methods on the right hand side to get Mn an ox number of +10. Then I said that as oxidized since it increased.
I said (S04)3 is -30 ? Then I said Fe2 should be +30 and each Fe atom +15, so that's also oxidized, which can't be right..
I don't really have a clue to be honest. Any help would be appreciated.
I think you mean manganese. Doesn't make any difference, but yeah.
The big mistake you've made is with FeSO4. Forgetting Fe for the time being, think about the sulphate ion, SO42-. The charge on this ion should be able to tell you the oxidation number of S. It should also tell you the oxidation number of Fe when you consider the whole sulphate ion. It's definitely not +10, that's an extremely large oxidation state, I doubt you'll ever come across anything higher than +7.
With this in mind, recalculate the oxidation numbers of iron and sulphur.
Personally, I think it's a lot of effort to go to in order to balance the equation, I'd just use trial and error.
Edit: it took me about 20 seconds to balance in my head, so it's not that difficult.
(edited 11 years ago)
The question insists on using oxidation numbers unfortunately.
Using the fact that SO4 has an oxidation number of -2, I was able to end up with:
MnO2 + 2FeSO4 + 2H2SO4 - > Fe2(SO4)3 + MnSO4 + 2H2O.
which is balanced as far as I can see.
I got Mn to have an ox no of +4 on the LHS and +2 on the RHS, so that's the reduced species? I got Fe to go from +2 to +3, so that's oxidized.
Just one question on the SO4 being -2.
O has two electrons in its outer shell. S has six.
S loses six, and the oxygen atoms gain them, but there's two electrons left between the four O atoms outer shells, and that's why it's -2?
Using the fact that SO4 has an oxidation number of -2, I was able to end up with:
MnO2 + 2FeSO4 + 2H2SO4 - > Fe2(SO4)3 + MnSO4 + 2H2O.
which is balanced as far as I can see.
I got Mn to have an ox no of +4 on the LHS and +2 on the RHS, so that's the reduced species? I got Fe to go from +2 to +3, so that's oxidized.
Just one question on the SO4 being -2.
O has two electrons in its outer shell. S has six.
S loses six, and the oxygen atoms gain them, but there's two electrons left between the four O atoms outer shells, and that's why it's -2?
Original post by Freeway Lemonade
The question insists on using oxidation numbers unfortunately.
Using the fact that SO4 has an oxidation number of -2, I was able to end up with:
MnO2 + 2FeSO4 + 2H2SO4 - > Fe2(SO4)3 + MnSO4 + 2H2O.
which is balanced as far as I can see.
I got Mn to have an ox no of +4 on the LHS and +2 on the RHS, so that's the reduced species? I got Fe to go from +2 to +3, so that's oxidized.
Just one question on the SO4 being -2.
O has two electrons in its outer shell. S has six.
S loses six, and the oxygen atoms gain them, but there's two electrons left between the four O atoms outer shells, and that's why it's -2?
Using the fact that SO4 has an oxidation number of -2, I was able to end up with:
MnO2 + 2FeSO4 + 2H2SO4 - > Fe2(SO4)3 + MnSO4 + 2H2O.
which is balanced as far as I can see.
I got Mn to have an ox no of +4 on the LHS and +2 on the RHS, so that's the reduced species? I got Fe to go from +2 to +3, so that's oxidized.
Just one question on the SO4 being -2.
O has two electrons in its outer shell. S has six.
S loses six, and the oxygen atoms gain them, but there's two electrons left between the four O atoms outer shells, and that's why it's -2?
Yep, you've balanced it correctly - they key is to look at SO42- as a single species.
In SO42- the sulphur is in an oxidation state of +6 and the oxygen atoms each have a oxidation state of -2, so overall there is a -2 charge. I'm not sure exactly how the atoms are bonded covalently and where the electrons are, someone else might be able to tell you, I don't think it's necessary to know for A-level though. Sorry I can't really be of much help with this.
cheers.
Doing this one now:
KIO3 + HCl + KI -> I2 + H2O + KCl
Not sure about KIO3.
On the right hand side I said I was 0, H2 was +2, O was -2, K was +1 and Cl was -1.
On the left hand side I said that K was +1, I was -1, H was+1 and Cl was -1.
I didn't really know how to deal with KIO3. I figured it had a -6 charge and so I went with O3 being -6, K being +1 and I being -1.
The problem is that the only ox number that changes is I, from -1 to 0.
Doing this one now:
KIO3 + HCl + KI -> I2 + H2O + KCl
Not sure about KIO3.
On the right hand side I said I was 0, H2 was +2, O was -2, K was +1 and Cl was -1.
On the left hand side I said that K was +1, I was -1, H was+1 and Cl was -1.
I didn't really know how to deal with KIO3. I figured it had a -6 charge and so I went with O3 being -6, K being +1 and I being -1.
The problem is that the only ox number that changes is I, from -1 to 0.
Original post by Freeway Lemonade
cheers.
Doing this one now:
KIO3 + HCl + KI -> I2 + H2O + KCl
Not sure about KIO3.
On the right hand side I said I was 0, H2 was +2, O was -2, K was +1 and Cl was -1.
On the left hand side I said that K was +1, I was -1, H was+1 and Cl was -1.
I didn't really know how to deal with KIO3. I figured it had a -6 charge and so I went with O3 being -6, K being +1 and I being -1.
The problem is that the only ox number that changes is I, from -1 to 0.
Doing this one now:
KIO3 + HCl + KI -> I2 + H2O + KCl
Not sure about KIO3.
On the right hand side I said I was 0, H2 was +2, O was -2, K was +1 and Cl was -1.
On the left hand side I said that K was +1, I was -1, H was+1 and Cl was -1.
I didn't really know how to deal with KIO3. I figured it had a -6 charge and so I went with O3 being -6, K being +1 and I being -1.
The problem is that the only ox number that changes is I, from -1 to 0.
It doesn't have a -6 charge, otherwise it would have -6 written above it, the species is neutral.
As a general rule, group I metals always have a oxidation number of +1 (and group II metals always have a oxidation number of +2). Group VII elements on the other hand can have a range of oxidation states. KIO3 is ionic, and made up of K+ and IO3- (iodate) ions. You should be able to do it with this in mind.
Quick Reply
Related discussions
- Etha
- A-Level chemistry question
- really tough balancing equations
- Inorganic Chemistry (AQA) - A level
- Chemistry - yields question
- MCQ from OCR A-level chemistry
- A level chemistry balancing equation
- GCSE CHEMISTRY HIGHER AQA PAPER 1 MAY 2018 - Unofficial mark scheme
- C3.1.6 Moles, need help!
- how to obtain ratio between gases in decomposition reaction
- Ocr a level chem
- AQA Chemistry AS Paper 1 June 2018 Question 07.2
- Chemistry oxidation of alcohol
- Calculating mass
- mistake - dont click
- Catalyst: iodide and peroxodisulphate ions
- Redox and balancing
- What is the ionic charge and formulae for nickel(II) manganate(VII)?
- A level physics nuclear question
- A level chemistry question
Latest
Last reply 1 minute ago
Which of these A-Level combinations would be the best to study a law degree?Posted 1 minute ago
Studying while being stressed!? Participants wanted! :)Last reply 2 minutes ago
Weidenfeld Hoffmann Trust (WHT) Scholarship Notification (2024-2025)Last reply 10 minutes ago
Official UCL Offer Holders Thread for 2024 entryLast reply 14 minutes ago
Durham vs Exeter Finance/AccountingLast reply 21 minutes ago
Official London School of Economics and Political Science 2024 Applicant ThreadLast reply 21 minutes ago
Litigation Caseworker Home Office 2024Last reply 23 minutes ago
BT Degree Apprenticeship 2024Last reply 27 minutes ago
Application approved before evidence was able to be submittedLast reply 32 minutes ago
AQA Level 2 Further Maths 2024 Paper 1 (8365/1) - 11th June [Exam Chat]Trending
Last reply 4 days ago
Im confused about this chemistry question, why does it form these productsTrending
Last reply 4 days ago
Im confused about this chemistry question, why does it form these products
