Oxidation/Reduction problem

[MTD52]

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Sn(s) + HNO3 (aq) --> Sn(NO3)2 (aq) + NO (g) + H2O (l)

The answer asks which is being reduced and oxidized, and therefore which is the reducing and oxidizing agent.

I see that Sn is being oxidized and is the reducing agent.

The answer is that Nitrogen is reduced from N(+5) in HNO3 to N(+2), which I have realized is from the NO, not from the Sn(NO3)2. How do you know to use the Nitrogen in NO, and not from the Sn(NO3)2?

 

[nze82]

Sn(s) + HNO3 (aq) --> Sn(NO3)2 (aq) + NO (g) + H2O (l)

The answer asks which is being reduced and oxidized, and therefore which is the reducing and oxidizing agent.

I see that Sn is being oxidized and is the reducing agent.

The answer is that Nitrogen is reduced from N(+5) in HNO3 to N(+2), which I have realized is from the NO, not from the Sn(NO3)2. How do you know to use the Nitrogen in NO, and not from the Sn(NO3)2?

The answer will be the same regardless of which two you compare:
HNO3--> N + 3(-2) = -1 --> N = +5
Sn(NO3)2--> N + 3(-2) = -2 --> N = +4

The oxidation # is lowered from +5 to +4; therefore, N is reduced.

 

[MTD52]

Does that go for any reaction? You can compare either one?

 

[nze82]

Does that go for any reaction? You can compare either one?

Yes!
When you say Nitrogen is reduced you're referring to all the Nitrogen species involved in the reaction. You cant pick a certain group of Nitrogen atoms and assume that they're the only ones that behave in a particular way, while other Nitrogen atoms in the same reaction vessel behave differently.

 

[Shinpe]

The answer will be the same regardless of which two you compare:
HNO3--> N + 3(-2) = -1 --> N = +5
Sn(NO3)2--> N + 3(-2) = -2 --> N = +4

The oxidation # is lowered from +5 to +4; therefore, N is reduced.

That's not right. N is +5 in NO3- whether it's HNO3 or Sn(NO3)2. In the second one you have:
+2 + 2N + 6 (-2) = 0 which still gives N=+5.

Not all of the same element have to oxidized or reduced. Here the N that goes from NO3 to NO3 doesnt change oxidation, but the ones going from NO3 to NO go from +5 to + 2.

Another example: H2O2 -> 1/2 O2 + H2O
The Os going to O2 are oxidized (from -1 to 0) and the ones going to H2O are reduced (from -1 to -2).

 

[MTD52]

That's not right. N is +5 in NO3- whether it's HNO3 or Sn(NO3)2. In the second one you have:
+2 + 2N + 6 (-2) = 0 which still gives N=+5.

Not all of the same element have to oxidized or reduced. Here the N that goes from NO3 to NO3 doesnt change oxidation, but the ones going from NO3 to NO go from +5 to + 2.

Another example: H2O2 -> 1/2 O2 + H2O
The Os going to O2 are oxidized (from -1 to 0) and the ones going to H2O are reduced (from -1 to -2).

That's what I thought! That bolded part is why I got confused because I didn't think that it got reduced. Of course, "none of these" happened to be an answer choice so I went with that. I didn't even think to look at the NO.

However, in a case like what you just said, how would you know whether to say O was getting oxidized or reduced? Since apparently it is both.

 

[Shinpe]

That's what I thought! That bolded part is why I got confused because I didn't think that it got reduced. Of course, "none of these" happened to be an answer choice so I went with that. I didn't even think to look at the NO.

Thanks a lot. I'm glad you cleared that up.

Something to keep in mind is that when something is being oxidized, another thing is being reduced. You can't have either by themselves (well in a balanced full-cell equation, you can in half-cell equations). So when you see that Sn is is oxidized, look to see what is reduced until u find it.

 

[MTD52]

Something to keep in mind is that when something is being oxidized, another thing is being reduced. You can't have either by themselves (well in a balanced full-cell equation, you can in half-cell equations). So when you see that Sn is is oxidized, look to see what is reduced until u find it.

Good point. Thanks 🙂. The timing on these tests really impairs my ability to think lol

Can you answer the question I had about that combustion reaction you said? What do you if something like that comes up where it is oxidized and reduced?

One more question, for assigning the numbers, is this the correct order of rules (I know it's not all of them):

1) Group I is always +1
2) Group II is always +2
3) H is always +1 unless it's paired with metal (Group I)
4) O is always -2 except peroxides (O-O single bond)

 

[Shinpe]

However, in a case like what you just said, how would you know whether to say O was getting oxidized or reduced? Since apparently it is both.

Well O is being both oxidized and reduced. I think it actually had a name when that happens to an element, but I don't remember what right now.

 

[Shinpe]

Good point. Thanks 🙂. The timing on these tests really impairs my ability to think lol

Can you answer the question I had about that combustion reaction you said? What do you if something like that comes up where it is oxidized and reduced?

One more question, for assigning the numbers, is this the correct order of rules (I know it's not all of them):

1) Group I is always +1
2) Group II is always +2
3) H is always +1 unless it's paired with metal (Group I)
4) O is always -2 except peroxides (O-O single bond)

Well it's not a combustion. Destroyer calls it a "disproportionation reaction" which I guess can be confused for free-radical termination reactions.

and those sound about right, also remember all elements in their natural state are 0, aka in O2, H2, Na (s), K (s), whatever.

 

[MTD52]

Well it's not a combustion. Destroyer calls it a "disproportionation reaction" which I guess can be confused for free-radical termination reactions.

and those sound about right, also remember all elements in their natural state are 0, aka in O2, H2, Na (s), K (s), whatever.

Yeah I know the rest of the rules, thanks. Topscore calls it a combustion reaction. I didn't think it was either. Actually, it confused me in the problem because when I read combustion I thought CO2 + H2O as products, but that clearly isn't the case.