Oxidation/Reduction
2Mg(NO3)2(s) --> 2MgO(s) + 4NO2(g) + O2(g)
Use oxidation numbers to show that this reaction involves both oxidation and reduction.
I know this is an old concept but I REALLY don't know how to do this when they involve complex/long compounds. Can anyone break it down for me please?
Use oxidation numbers to show that this reaction involves both oxidation and reduction.
I know this is an old concept but I REALLY don't know how to do this when they involve complex/long compounds. Can anyone break it down for me please?
Scroll to see replies
Well magnesium nitrate is composed of Mg 2+ ions and NO3 - ions
Immediately you can see that Mg is in the +2 oxidation state. The overall charge of the anion is 1-, and O can be assumed to be 2-, which leaves N to be in the +5 oxidation state.
Now looking at the other side of the equation, you have NO2. Assuming O to be 2- means that N must be in the +4 oxidation state.
You can see now that N has gone from +5 to +4, by gaining an electron, so it has been reduced. Consequently, something else must have been oxidised.
The O on the left had side is 2-, but on the right hand side you have O2, which is a pure element so by definition its oxidation state is 0. Hence you have gone from 2- to 0, i.e. two electrons have been lost, which is oxidation.
Immediately you can see that Mg is in the +2 oxidation state. The overall charge of the anion is 1-, and O can be assumed to be 2-, which leaves N to be in the +5 oxidation state.
Now looking at the other side of the equation, you have NO2. Assuming O to be 2- means that N must be in the +4 oxidation state.
You can see now that N has gone from +5 to +4, by gaining an electron, so it has been reduced. Consequently, something else must have been oxidised.
The O on the left had side is 2-, but on the right hand side you have O2, which is a pure element so by definition its oxidation state is 0. Hence you have gone from 2- to 0, i.e. two electrons have been lost, which is oxidation.
Original post by LifeIsGood
2Mg(NO3)2(s) --> 2MgO(s) + 4NO2(g) + O2(g)
Use oxidation numbers to show that this reaction involves both oxidation and reduction.
I know this is an old concept but I REALLY don't know how to do this when they involve complex/long compounds. Can anyone break it down for me please?
Use oxidation numbers to show that this reaction involves both oxidation and reduction.
I know this is an old concept but I REALLY don't know how to do this when they involve complex/long compounds. Can anyone break it down for me please?
So I use a priority list.
In a compound:
F will always be -1
Group 1 will always be +1
Group 2 will always be +2
Then
H will always be +1 (ie in NaH the sodium has priority as +1, so the H has to be -1 to balance it)
Then
O will always be -2 (ie in H2O2 the H has priority as 2x+1, so the O must each be -1)
Then everything else falls into place.
I hope that makes sense!!
2Mg(NO3)2(s) --> 2MgO(s) + 4NO2(g) + O2(g)
So for your example use the above rules to work out the oxidation state of each element in each of the compounds.
I'll let you try first, then let me know if you want more help
Original post by chembob
Well magnesium nitrate is composed of Mg 2+ ions and NO3 - ions
Immediately you can see that Mg is in the +2 oxidation state. The overall charge of the anion is 1-, and O can be assumed to be 2-, which leaves N to be in the +5 oxidation state.
Now looking at the other side of the equation, you have NO2. Assuming O to be 2- means that N must be in the +4 oxidation state.
You can see now that N has gone from +5 to +4, by gaining an electron, so it has been reduced. Consequently, something else must have been oxidised.
The O on the left had side is 2-, but on the right hand side you have O2, which is a pure element so by definition its oxidation state is 0. Hence you have gone from 2- to 0, i.e. two electrons have been lost, which is oxidation.
Immediately you can see that Mg is in the +2 oxidation state. The overall charge of the anion is 1-, and O can be assumed to be 2-, which leaves N to be in the +5 oxidation state.
Now looking at the other side of the equation, you have NO2. Assuming O to be 2- means that N must be in the +4 oxidation state.
You can see now that N has gone from +5 to +4, by gaining an electron, so it has been reduced. Consequently, something else must have been oxidised.
The O on the left had side is 2-, but on the right hand side you have O2, which is a pure element so by definition its oxidation state is 0. Hence you have gone from 2- to 0, i.e. two electrons have been lost, which is oxidation.
Does the molar ratio for the compounds have any effect?
Original post by LifeIsGood
Does the molar ratio for the compounds have any effect?
No
Original post by chembob
No
That is what confused me, thanks so much!!!
Original post by clownfish
So I use a priority list.
In a compound:
F will always be -1
Group 1 will always be +1
Group 2 will always be +2
Then
H will always be +1 (ie in NaH the sodium has priority as +1, so the H has to be -1 to balance it)
Then
O will always be -2 (ie in H2O2 the H has priority as 2x+1, so the O must each be -1)
Then everything else falls into place.
I hope that makes sense!!
2Mg(NO3)2(s) --> 2MgO(s) + 4NO2(g) + O2(g)
So for your example use the above rules to work out the oxidation state of each element in each of the compounds.
I'll let you try first, then let me know if you want more help
In a compound:
F will always be -1
Group 1 will always be +1
Group 2 will always be +2
Then
H will always be +1 (ie in NaH the sodium has priority as +1, so the H has to be -1 to balance it)
Then
O will always be -2 (ie in H2O2 the H has priority as 2x+1, so the O must each be -1)
Then everything else falls into place.
I hope that makes sense!!
2Mg(NO3)2(s) --> 2MgO(s) + 4NO2(g) + O2(g)
So for your example use the above rules to work out the oxidation state of each element in each of the compounds.
I'll let you try first, then let me know if you want more help
Hmm stuck on this one:
I'm trying to do FeSO4 ---> Fe2(SO4)3 but am so confused.
I got the FeSO4 bit:
Fe = ?
So4 = -2
therefore Fe = +2
But with the other one I'm not sure:
Fe2 = ?
(S04)3 = -2 (I'm not sure but I assumed it was because the charges swap over when they're neutral). I don't know how to go from there though
Original post by LifeIsGood
Hmm stuck on this one:
I'm trying to do FeSO4 ---> Fe2(SO4)3 but am so confused.
I got the FeSO4 bit:
Fe = ?
So4 = -2
therefore Fe = +2
But with the other one I'm not sure:
Fe2 = ?
(S04)3 = -2 (I'm not sure but I assumed it was because the charges swap over when they're neutral). I don't know how to go from there though
I'm trying to do FeSO4 ---> Fe2(SO4)3 but am so confused.
I got the FeSO4 bit:
Fe = ?
So4 = -2
therefore Fe = +2
But with the other one I'm not sure:
Fe2 = ?
(S04)3 = -2 (I'm not sure but I assumed it was because the charges swap over when they're neutral). I don't know how to go from there though
As there are three "SO4" and each one is -2, the total oxidation number of this part of the compound is -6, so two lots of Fe must equal +6, so each Fe has an oxidation state of +3.
(edited 12 years ago)
Original post by LifeIsGood
But with the other one I'm not sure:
Fe2 = ?
(S04)3 = -2 (I'm not sure but I assumed it was because the charges swap over when they're neutral). I don't know how to go from there though
But with the other one I'm not sure:
Fe2 = ?
(S04)3 = -2 (I'm not sure but I assumed it was because the charges swap over when they're neutral). I don't know how to go from there though
Think of it as 2 Fe not an Fe2 ion
Original post by clownfish
As there are three "SO4" and each one is -3, the total oxidation number of this part of the compound is -6, so two lots of Fe must equal +6, so each Fe has an oxidation state of +3.
The (S03)4 bit is confusing me, how did you exactly know that S04 = -2?
3Fe2(SO4)3
Can you explain how you broke it up because I'm really confused:
If you did 3*(SO4) = -6 I understand but what about that 3 on the outside of the bracket?
Original post by LifeIsGood
The (S03)4 bit is confusing me, how did you exactly know that S04 = -2?
3Fe2(SO4)3
Can you explain how you broke it up because I'm really confused:
If you did 3*(SO4) = -6 I understand but what about that 3 on the outside of the bracket?
3Fe2(SO4)3
Can you explain how you broke it up because I'm really confused:
If you did 3*(SO4) = -6 I understand but what about that 3 on the outside of the bracket?
The sulfate ion SO42- is just one that I was always taught to learn. If you ever see SO4 like that it will be 2-. Think about sulfuric acid H2SO4 - this helps you remember that it has a minus 2 charge.
The 3 after the SO4 tells you there are 3 lots of SO4 in the molecule and so I multiplied the -2 charge by 3.
To confirm the three at the start of the molecule in the balanced equation is completely irrelevant for finding oxidation numbers. Its the three after the bracket that is telling me to multiply the sulfate charge by three.
(edited 12 years ago)
Original post by LifeIsGood
The (S03)4 bit is confusing me, how did you exactly know that S04 = -2?
3Fe2(SO4)3
Can you explain how you broke it up because I'm really confused:
If you did 3*(SO4) = -6 I understand but what about that 3 on the outside of the bracket?
3Fe2(SO4)3
Can you explain how you broke it up because I'm really confused:
If you did 3*(SO4) = -6 I understand but what about that 3 on the outside of the bracket?
If you mean the 3 before the Fe you can ignore it, the 3 after the bracket you have to take into account as the compound itself is Fe2(SO4)3
Original post by chembob
Think of it as 2 Fe not an Fe2 ion
indeed stating that it would be Fe2 would suggest that Fe is diatomic, which it isnt
Original post by clownfish
The sulfate ion SO42- is just one that I was always taught to learn. If you ever see SO4 like that it will be 2-. Think about sulfuric acid H2SO4 - this helps you remember that it has a minus 2 charge.
The 3 after the SO4 tells you there are 3 lots of SO4 in the molecule and so I multiplied the -2 charge by 3.
To confirm the three at the start of the molecule in the balanced equation is completely irrelevant for finding oxidation numbers. Its the three after the bracket that is telling me to multiply the sulfate charge by three.
The 3 after the SO4 tells you there are 3 lots of SO4 in the molecule and so I multiplied the -2 charge by 3.
To confirm the three at the start of the molecule in the balanced equation is completely irrelevant for finding oxidation numbers. Its the three after the bracket that is telling me to multiply the sulfate charge by three.
Got it! Thanks so much for your patience!
Which are the ions I need to know about?
(edited 12 years ago)
Pay attention how much molecules an element have! to fix an oxidation number for an element, you have to see to the molecules, because the oxidation numbers have to neutralized in a (finished) chemical equation.
Example:
potassium permanganate has KMnO4 as chemical formula.
Potassium has one molecule, manganese as well, but oxygen has four molecules in there.
To fix the numbers look at oxygen first. oxygen has -2 in this case as oxidation numbers. As oxygen has also four molecules, you have to multiply this oxidation number by four: (-2)*4 = -8, this is the number for oxygen with four molecules. And now you have to find out oxidation numbers for potassium and manganese which are neutralized -8 to 0. In this case potassium has +1 and manganese +7. Both of them have one molecule, that's why you have to multiply these numbers by one with each other:
For potassium:
+1 * 1 = 1
For manganese:
+7 * 1 = 7
and know you add these numbers together:
7 + 1 = 8
After that you add +8 (both potassium and manganese) to -8 (oxygen):
8 + (-8) = 0, that is neutralized now. The oxidation numbers were right. And that is potassium permanganate with oxidation numbers:
K(+I)Mn(+VII)O4(-II)
Notice: If the oxidation number is above 0 or below, your oxidation numbers for elements are wrong! That's why look always to the molecules first.
Do you have understand me?
Example:
potassium permanganate has KMnO4 as chemical formula.
Potassium has one molecule, manganese as well, but oxygen has four molecules in there.
To fix the numbers look at oxygen first. oxygen has -2 in this case as oxidation numbers. As oxygen has also four molecules, you have to multiply this oxidation number by four: (-2)*4 = -8, this is the number for oxygen with four molecules. And now you have to find out oxidation numbers for potassium and manganese which are neutralized -8 to 0. In this case potassium has +1 and manganese +7. Both of them have one molecule, that's why you have to multiply these numbers by one with each other:
For potassium:
+1 * 1 = 1
For manganese:
+7 * 1 = 7
and know you add these numbers together:
7 + 1 = 8
After that you add +8 (both potassium and manganese) to -8 (oxygen):
8 + (-8) = 0, that is neutralized now. The oxidation numbers were right. And that is potassium permanganate with oxidation numbers:
K(+I)Mn(+VII)O4(-II)
Notice: If the oxidation number is above 0 or below, your oxidation numbers for elements are wrong! That's why look always to the molecules first.
Do you have understand me?
(edited 12 years ago)
Original post by LifeIsGood
Got it! Thanks so much for your patience!
Which are the ions I need to know about?
Which are the ions I need to know about?
All the common ions - nitrate, nitrite, nitride, sulfate, sulfite, sulfide, chlorate, chlorite, chloride, iodate, iodite, iodide, carbonate, peroxide, oxide, manganate, permanganate, chromate, dichromate and some more. At this level you must be familiar with these ions and the charge on them.
You also have to know that Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3.
You also have to learn the rules for assigning oxidation numbers which are as follows:
1. The oxdn. no. of any uncombined element os 0.
2. The oxdn. no of an uncombined ion is the same as its charge.
3. The sum of all the pxdn. no.s in a molecule is 0 (or in a complicated ion is equal to the charge on the particle).
4. Fluorine always has an oxdn. no. of -1.
5. Hydrogen always has an oxdn. no. of +1, except in metal hydrides (when its oxdn. no. is -1).
6.Oxygen always has an oxdn. no. of -2, except in peroxides (when it has an oxdn. no. of -1), and when it combines with fluorine (when it is positive).
7. Chlorine always has an oxdn. no. of -1, except with oxygen and fluorine when it is positive.
8.Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3.
Hope this helps.
(edited 12 years ago)
Original post by M Kh
All the common ions - nitrate, nitrite, nitride, sulfate, sulfite, sulfide, chlorate, chlorite, chloride, iodate, iodite, iodide, carbonate, peroxide, oxide, manganate, permanganate, chromate, dichromate and some more. At this level you must be familiar with these ions and the charge on them.
You also have to know that Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3 and Halogens (Group 7 elements) always have an oxdn. no. of -1. You also have to learn the rules for assigning oxidation numbers.
Hope this helps.
You also have to know that Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3 and Halogens (Group 7 elements) always have an oxdn. no. of -1. You also have to learn the rules for assigning oxidation numbers.
Hope this helps.
Halogens do not always have an oxidation number of -1. eg in NaClO chlorine is +1.
OP: check your syllabus or ask your teacher which you need to know. I only had to learn sulfate, sulfite, carbonate, nitrate and ammonium.
Original post by M Kh
All the common ions - nitrate, nitrite, nitride, sulfate, sulfite, sulfide, chlorate, chlorite, chloride, iodate, iodite, iodide, carbonate, peroxide, oxide, manganate, permanganate, chromate, dichromate and some more. At this level you must be familiar with these ions and the charge on them.
You also have to know that Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3 and Halogens (Group 7 elements) always have an oxdn. no. of -1. You also have to learn the rules for assigning oxidation numbers.
Hope this helps.
You also have to know that Group 1 elements always have an oxdn. no. of +1, Group 2 elements always have an oxdn. no. of +2, Group 3 elements always have an oxdn. no. of +3 and Halogens (Group 7 elements) always have an oxdn. no. of -1. You also have to learn the rules for assigning oxidation numbers.
Hope this helps.
Wow thanks!
Original post by clownfish
Halogens do not always have an oxidation number of -1. eg in NaClO chlorine is +1.
OP: check your syllabus or ask your teacher which you need to know. I only had to learn sulfate, sulfite, carbonate, nitrate and ammonium.
OP: check your syllabus or ask your teacher which you need to know. I only had to learn sulfate, sulfite, carbonate, nitrate and ammonium.
I'm on OCR
Original post by clownfish
Halogens do not always have an oxidation number of -1. eg in NaClO chlorine is +1.
OP: check your syllabus or ask your teacher which you need to know. I only had to learn sulfate, sulfite, carbonate, nitrate and ammonium.
OP: check your syllabus or ask your teacher which you need to know. I only had to learn sulfate, sulfite, carbonate, nitrate and ammonium.
Oh yes, that's a mistake. I will edit my post now.
Original post by LifeIsGood
Wow thanks!
I'm on OCR
I'm on OCR
Ok so according to the syllabus for OCR A you need to know nitrate (NO3-), sulfate (SO42-), carbonate (CO32-)and ammonium (NH4+). Also hydroxide (OH-)
also the common acids: nitric, sulfuric, hydrochloric.
Quick Reply
Related discussions
- Redox half reactions
- question about a half equation
- Redox reactions between a free element and a compound
- Calculate the oxidation state of the vanadium ion?! URGENT
- How would I write the half equation
- chemistry a level
- A-Level chemistry question
- A level chemistry Redox help please.
- (AS AQA Chemistry) Redox reactions - general question
- A-Level Group 7 Chemistry help
- Need Help on an Electrochem Q
- A-level Chemistry Study Group 2022-2023
- how do i suggest the formula for sodium chlorate (vii) ??
- A Level Chemistry Exam Question-need help!
- Oxidation number format vs ion charge format
- Acidic alkaline basic metals and non metals
- A level chemistry question
- GCSE CHEMISTRY HIGHER AQA PAPER 1 MAY 2018 - Unofficial mark scheme
- Alevel Chemistry HELP
- A level physics nuclear question
Latest
Trending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these productsTrending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these products
