Oxidizing/Reducing Agents

[topdent1]

12. Identify the oxidizing agent in the following reaction:

8H+ (aq) + 6Cl- (aq) + Sn (s) + 4NO3- (aq) SnCl62- (aq) + 4NO2 (g) + 4H2O (l)

What are we supposed to look for? Oxidizing agent...doesn't that just mean its getting reduced. Can we just pick the one that has lost an oxygen. In this case, NO3- would be the oxidizing agent because it becomes NO2?

What if the question asked what is the reducing agent?

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[rcwha]

I think you're right...NO3- is the Ox agent....I believe Sn is the reducing agent

12. Identify the oxidizing agent in the following reaction:

8H+ (aq) + 6Cl- (aq) + Sn (s) + 4NO3- (aq) SnCl62- (aq) + 4NO2 (g) + 4H2O (l)

What are we supposed to look for? Oxidizing agent...doesn't that just mean its getting reduced. Can we just pick the one that has lost an oxygen. In this case, NO3- would be the oxidizing agent because it becomes NO2?

What if the question asked what is the reducing agent?

 

[250rsavage]

also when they ask for agents, they are always found on the reactants side

 

[HoangDDS]

I think you meant :

8H+ (aq) + 6Cl- (aq) + Sn (s) + 4NO3- (aq) ---> SnCl62- (aq) + 4NO2 (g) + 4H2O (l)

oxidizing agent = Cl-
Cl- had an overall charge of -1 in the reactant and an overall charge of -2 in the product or it went from -7 to -8. Therefore it gained one e- / got reduced . Since oxidizing agent get reduced, I think Cl- is the oxidizing agent.

reducing agent = NO3-
NO3- had an overall charge of -1 in the reactant and an overall charge of 0 in the product, therefore it lost one e- . NO3 got oxidized b/c it loss 1e- therefore it is the reducing agent -- since reducing agent gets oxidized.

rwcha: Sn in the reactant is a solid therefore it could not be a reducing or oxidizing agent. -- I think .

Please correct me if I am wrong because I thought I understood redox, until I tried to explain this thread.

FYI: reducing or oxidizing agent is always in the reactant.

 

[HoangDDS]

Can someone confirm my answer or explain to me how I'm wrong. Thanks.

 

[Mstoothlady2012]

I think you meant :

8H+ (aq) + 6Cl- (aq) + Sn (s) + 4NO3- (aq) ---> SnCl62- (aq) + 4NO2 (g) + 4H2O (l)

oxidizing agent = Cl-
Cl- had an overall charge of -1 in the reactant and an overall charge of -2 in the product or it went from -7 to -8. Therefore it gained one e- / got reduced . Since oxidizing agent get reduced, I think Cl- is the oxidizing agent.

reducing agent = NO3-
NO3- had an overall charge of -1 in the reactant and an overall charge of 0 in the product, therefore it lost one e- . NO3 got oxidized b/c it loss 1e- therefore it is the reducing agent -- since reducing agent gets oxidized.

rwcha: Sn in the reactant is a solid therefore it could not be a reducing or oxidizing agent. -- I think .

Please correct me if I am wrong because I thought I understood redox, until I tried to explain this thread.

FYI: reducing or oxidizing agent is always in the reactant.

I agree with you about the reducing agent - NO3- but I think the oxidizing agent is SnCl2

If you look at only Cl, then the charge is same on both sides. On the reactant side Cl has total charge of -6 & it remains the same on the products side as well.

 

[harrygt]

12. Identify the oxidizing agent in the following reaction:

8H+ (aq) + 6Cl- (aq) + Sn (s) + 4NO3- -----> (aq) SnCl62- (aq) + 4NO2 (g) + 4H2O (l)

What are we supposed to look for? Oxidizing agent...doesn't that just mean its getting reduced. Can we just pick the one that has lost an oxygen. In this case, NO3- would be the oxidizing agent because it becomes NO2?

What if the question asked what is the reducing agent?

On the left side, we have Sn(s) which has an oxidation state of zero. On the right side, we have Sn4+ [in SnCl6 2-,Cl is still -1. Let's say oxidation number of Sn = x, then we have x -6 = = -2, so x = +4] . It means Sn got oxidized, going from 0 to +4, so Sn is the reducing agent.

On the left side, the oxidation number of N in NO3- is +5. It is +4 in NO2. So Nitrogen got reduced, going from +5 to +4. Therefore, nitrogen [or we can say NO3-] is the oxidizing agent.

 

[jdent]

Yes i'm with harrygt on this one. Remember the reducing agent is oxidized and the oxidizing reagent is reduced. You have to look at the individual atoms and not necessarily at the whole molecule.

 

[HoangDDS]

You're right. I really wasn't confident about my answer and I kept changing it in my head. Thanks for double checking me. So the oxidizing agent is NO3- because N was +5 in the reactant and +4 in the product therefore it gained 1 e- thus it was reduced -- ie. oxidizing agent. And Sn went from 0 in the reactant to +4 in the product therefore it lost 4 e- thus it was oxidized and that is why it is the reducing reagent.

Ok, I think I over analyzed it at first.

Do we consider the 4 in 4NO3-, giving N a +23 ox state? I guess it won't really matter because in the product N +8 ox state, thus it gained electrons anyway - got reduced and therefore the oxidant.

And forget about what I said about Sn can't be a oxidant or reductant because it was a solid. I guess I was thinking about thermodynamics when we don't consider solid in entropy.

Thanks Harry.

 

[harrygt]

You're right. I really wasn't confident about my answer and I kept changing it in my head. Thanks for double checking me. So the oxidizing agent is NO3- because N was +5 in the reactant and +4 in the product therefore it gained 1 e- thus it was reduced -- ie. oxidizing agent. And Sn went from 0 in the reactant to +4 in the product therefore it lost 4 e- thus it was oxidized and that is why it is the reducing reagent.

Ok, I think I over analyzed it at first.

Do we consider the 4 in 4NO3-, giving N a +23 ox state? I guess it won't really matter because in the product N +8 ox state, thus it gained electrons anyway - got reduced and therefore the oxidant.

And forget about what I said about Sn can't be a oxidant or reductant because it was a solid. I guess I was thinking about thermodynamics when we don't consider solid in entropy.

Thanks Harry.

Boy, where are you getting those 23+ and 8+? DO NOT ever multiply the coefficients by the oxidation states. Let's look at NO3- . O is 99% of the time 2-. you gotta make an equation now, assuming x = oxidation state of N. so we have x + 3 (-2) = -1 [-1 comes from the overall charge of the polyatomic ioin]
SO you have x -6 = -1 Therefore, x = 5. Do the same thing for other compounds.

 

[HoangDDS]

Boy, where are you getting those 23+ and 8+? DO NOT ever multiply the coefficients by the oxidation states. Let's look at NO3- . O is 99% of the time 2-. you gotta make an equation now, assuming x = oxidation state of N. so we have x + 3 (-2) = -1 [-1 comes from the overall charge of the polyatomic ioin]
SO you have x -6 = -1 Therefore, x = 5. Do the same thing for other compounds.

Ok. haha you don't multiply in the coefficient.

I get the other part. Rule 6 says that oxidation number of Oxygen is -2. 3 *-2 = -6 . the overall charge is -1 therefore N has to be +5. I'm not that lost buddy.

 

[harrygt]

Ok. haha you don't multiply in the coefficient.

I get the other part. Rule 6 says that oxidation number of Oxygen is -2. 3 *-2 = -6 . the overall charge is -1 therefore N has to be +5. I'm not that lost buddy.

I'm glad! No underestimations here! Just trying to make sure you get it.

 

[Mstoothlady2012]

You're right. I really wasn't confident about my answer and I kept changing it in my head. Thanks for double checking me. So the oxidizing agent is NO3- because N was +5 in the reactant and +4 in the product therefore it gained 1 e- thus it was reduced -- ie. oxidizing agent. And Sn went from 0 in the reactant to +4 in the product therefore it lost 4 e- thus it was oxidized and that is why it is the reducing reagent.

Ok, I think I over analyzed it at first.

Do we consider the 4 in 4NO3-, giving N a +23 ox state? I guess it won't really matter because in the product N +8 ox state, thus it gained electrons anyway - got reduced and therefore the oxidant.

And forget about what I said about Sn can't be a oxidant or reductant because it was a solid. I guess I was thinking about thermodynamics when we don't consider solid in entropy.

Thanks Harry.

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