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# How to balance the following equation using oxidation states? watch

1. Sn+HNO3-------> SnO2 + NO2 + H2O

How do I balance this using oxidation stateS?
2. (Original post by xXxiKillxXx)
Sn+HNO3-------> SnO2 + NO2 + H2O

How do I balance this using oxidation stateS?
Can't you use the example I showed you earlier and apply it to this problem?

The only species changing oxidation state are:

Sn(zero) --> Sn4+ + 4e
N5+ + 1e --> N4+

now take it from there...
3. (Original post by charco)
Can't you use the example I showed you earlier and apply it to this problem?

The only species changing oxidation state are:

Sn(zero) --> Sn4+ + 4e
N5+ + 1e --> N4+

now take it from there...
Ohh I understand! but one question; in the AQA book it says Oxygen's oxidation state is always -2, except for in Peroxides where it is -1.

Isn't SnO2 a peroxide? so therefore it should be -2 for O2 and thus +2 for Sn?

Or am I missing something?

Thanks, much appreciated
4. (Original post by xXxiKillxXx)
Ohh I understand! but one question; in the AQA book it says Oxygen's oxidation state is always -2, except for in Peroxides where it is -1.

Isn't SnO2 a peroxide? so therefore it should be -2 for O2 and thus +2 for Sn?

Or am I missing something?

Thanks, much appreciated
No, SnO2 isn't a peroxide.

The oxygen is in the usual -2 state.
5. (Original post by charco)
No, SnO2 isn't a peroxide.

The oxygen is in the usual -2 state.
How would you know this though?
6. (Original post by xXxiKillxXx)
How would you know this though?
Good question. Experience I guess.

Peroxides only appear in group 1 group 2 and organic compounds. The most common peroxides in inorganic chemistry are sodium and barium peroxides, Na2O2 and BaO2
7. (Original post by charco)
Good question. Experience I guess.

Peroxides only appear in group 1 group 2 and organic compounds. The most common peroxides in inorganic chemistry are sodium and barium peroxides, Na2O2 and BaO2
ohh thanks, hopefully a compound that I am used to will come up in the exam.

Also one last question, sorry for bothering you, but doesn't Oxygen's oxidation state change?

Sn+HNO3-------> SnO2 + NO2 + H2O

LHS: -2x3=-6 RHS: (-2x2)+(-2x2)+-2=-10
8. (Original post by xXxiKillxXx)
ohh thanks, hopefully a compound that I am used to will come up in the exam.

Also one last question, sorry for bothering you, but doesn't Oxygen's oxidation state change?

Sn+HNO3-------> SnO2 + NO2 + H2O

LHS: -2x3=-6 RHS: (-2x2)+(-2x2)+-2=-10
no, the equation isn't balanced yet, so you can't 'count up'
9. (Original post by charco)
no, the equation isn't balanced yet, so you can't 'count up'
I meant as in shouldn't I list oxygen the same way these 2 were listed:

Sn(zero) --> Sn4+ + 4e
N5+ + 1e --> N4+

and then see if oxygen has been reduced/oxidised?
10. (Original post by xXxiKillxXx)
I meant as in shouldn't I list oxygen the same way these 2 were listed:

Sn(zero) --> Sn4+ + 4e
N5+ + 1e --> N4+

and then see if oxygen has been reduced/oxidised?
You can tell the oxidation state of each oxygen by considering the compound it's in - however as it is always -2 (except for peroxides and the element) there is no need...
11. (Original post by charco)
Can't you use the example I showed you earlier and apply it to this problem?

The only species changing oxidation state are:

Sn(zero) --> Sn4+ + 4e
N5+ + 1e --> N4+

now take it from there...
uve made a mistake geez, u forgot to do hydrogen.

Sn+HNO3-------> SnO2 + NO2 + H2O

+1 and +1x2 = 2+

becos hydrogens oxidaton nr has gone from 1 to 2, it means electron has been lost, so its oxidised
12. (Original post by jamesrune75)
uve made a mistake geez, u forgot to do hydrogen.

Sn+HNO3-------> SnO2 + NO2 + H2O

+1 and +1x2 = 2+

becos hydrogens oxidaton nr has gone from 1 to 2, it means electron has been lost, so its oxidised
OP, please disregard this post.
13. (Original post by charco)
OP, please disregard this post.
I'm sure you have more experience, but isn't he right ? lol
14. (Original post by xXxiKillxXx)
I'm sure you have more experience, but isn't he right ? lol
It's the same as the oxygen question. You cannot compare oxidation states in this way. If the equation is not balanced there is no way you can add up species on both sides.

This is the idea of balancing the equation in the first place.

You have to follow a methodology and balance the equation by considering the INDIVIDUAL oxidation states of the components.

Sn + HNO3 -----> SnO2 + NO2 + H2O

Left hand side
------------------

Sn (element) = zero
------------------------------

HNO3

H = +1
N = +5
O = -2

-----------------------------
Right hand side
------------------

SnO2
Sn = +4
O = -2

-----------------------------

NO2
N = +4
O = -2

-----------------------------

H2O
H = +1
O = -2

----------------------------

So, conclusion: The only elements that change their oxidation state are Sn and N

Now you follow a system to balance the equation using this information (as previously outlined)
15. (Original post by charco)
it's the same as the oxygen question. You cannot compare oxidation states in this way. if the equation is not balanced there is no way you can add up species on both sides.

This is the idea of balancing the equation in the first place.

You have to follow a methodology and balance the equation by considering the individual oxidation states of the components.

Sn + hno3 -----> sno2 + no2 + h2o

left hand side
------------------

sn (element) = zero
------------------------------

hno3

h = +1
n = +5
o = -2

-----------------------------
right hand side
------------------

sno2
sn = +4
o = -2

-----------------------------

no2
n = +4
o = -2

-----------------------------

h2o
h = +1
o = -2

----------------------------

so, conclusion: The only elements that change their oxidation state are sn and n

now you follow a system to balance the equation using this information (as previously outlined)
thank you so much, i fully understand now!!

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