[StillLearning]

Another oxidation state problem.  This problem is more complex and I'm obviously doing something wrong.  The feedback says i've labeled 3 of 8 targets incorrectly.  I've attached a copy of the problem and the feedback it gave me.  Thanks again. 

[Astrokel]

hey,

your two N and Mn are wrong.

think of it this way,

NO₂^-

oxygen being more electronegative than N(O gets the negative charge instead of N), hence oxygen has an oxidation state of -2 and since the compound overall charge is -1,

you get, x + 2(-2) = -1 where x is the oxidation state of nitrogen. solving x you get + 3.

1) always form this equation and you usually won't go wrong and remeber the exception such as hydride, peroxide...
2) how to form? put the compound overall charge on the right hand side and sum up all the atoms' oxidation states on the left hand side.

i have no idea how you get 1 for the first nitrogen unless you forget about there are two oxygen bonded to it.

hope it helps some

[StillLearning]

hey,

your two N and Mn are wrong.

think of it this way,

NO₂^-

oxygen being more electronegative than N(O gets the negative charge instead of N), hence oxygen has an oxidation state of -2 and since the compound overall charge is -1,

you get, x + 2(-2) = -1 where x is the oxidation state of nitrogen. solving x you get + 3.

1) always form this equation and you usually won't go wrong and remeber the exception such as hydride, peroxide...
2) how to form? put the compound overall charge on the right hand side and sum up all the atoms' oxidation states on the left hand side.

i have no idea how you get 1 for the first nitrogen unless you forget about there are two oxygen bonded to it.

hope it helps some

Ok, let me take this slow.  Looking at the first ion 5NO₂^-.  I thought that the O₂ had a charge of -2.  Are saying that EACH O atom has a charge of -2 and therefore the oxidation state of oxygen alone is -4?  And therefore the oxidation state of N is +3? 

[Astrokel]

ok firstly, you can ignore the stoichiometric coefficient, so just NO₂^-

and you are right, we are looking at the individual atom's oxidation state, so consider Oxygen. Oxgen is in group 6, it tends to form O^2-, therefore an oxygen atom has an oxidation state of -2. Since there are two oxygen atoms in NO₂^-, oxygen atoms has an oxidation state of 2(-2)= -4.

you can try it on the other two which you were wrong.

[StillLearning]

ok firstly, you can ignore the stoichiometric coefficient, so just NO₂^-

and you are right, we are looking at the individual atom's oxidation state, so consider Oxygen. Oxgen is in group 6, it tends to form O^2-, therefore an oxygen atom has an oxidation state of -2. Since there are two oxygen atoms in NO₂^-, oxygen atoms has an oxidation state of 2(-2)= -4.

you can try it on the other two which you were wrong.

Ok, I got that problem right now (thanks to your guidance).  I'm on to the next problem and i'm stumped, once again.  I've attached another PDF that shows the problem.  I understand the properties of oxidizing and reducing agents, but I dont know how to determine which ions posess these properties in a chemical equation.  I hope you're not getting frustrated with my endless questions.  I really appreciate your help. 

[Astrokel]

hey,

you know that oxidizing agent undergo reduction, reducing agents undergoes oxidation. one of the definition of oxidation is "gain in oxidation state" and reduction "loss in oxidation state". can you determine whether the atom gains or loss in oxidation state? for example oxidation state of N in NO₂^- to N in NO₃^-. If there is an increase in oxidation state, it undergoes oxidation and hence it is a reducing agent. vice versa.

[StillLearning]

hey,

you know that oxidizing agent undergo reduction, reducing agents undergoes oxidation. one of the definition of oxidation is "gain in oxidation state" and reduction "loss in oxidation state". can you determine whether the atom gains or loss in oxidation state? for example oxidation state of N in NO₂^- to N in NO₃^-. If there is an increase in oxidation state, it undergoes oxidation and hence it is a reducing agent. vice versa.

Ok, I think i've got that.  Let's take 2MnO₄^- (on the reactant side).  On the product side it's 2Mn²⁺, so Mn has gained electrons and is therefore the oxidizing agent, right?  Where i'm mostly confused is when I have an element that appears more than once in the equation.  For example, oxygen appears 2 times on the reactant side and 2 times on the product side.  So how do I know which ones to compare?

[Astrokel]

hey

you are right, Mn in MnO₄^2- has an oxidation state of +7 and it gets reduced to Mn²⁺, there is a decrease in oxidation state, hence it is reduced, therefore it is oxidizing agent.

you probably don't have to worry about oxygen oxidation state in this case because it is mostly -2, because it is quite an electronegative atom. unless it is bonded to floruine, then oxygen oxidation state will not be -2. And also, take note of peroxide where oxygen has an oxidation state of -1. so in your case, oxygen is neither an oxidizing and reducing agent since its oxidation state remains the same throughout.

once you do more it will be quite intuitive to know which atoms to compare

[StillLearning]

hey

you are right, Mn in MnO₄^2- has an oxidation state of +7 and it gets reduced to Mn²⁺, there is a decrease in oxidation state, hence it is reduced, therefore it is oxidizing agent.

you probably don't have to worry about oxygen oxidation state in this case because it is mostly -2, because it is quite an electronegative atom. unless it is bonded to floruine, then oxygen oxidation state will not be -2. And also, take note of peroxide where oxygen has an oxidation state of -1. so in your case, oxygen is neither an oxidizing and reducing agent since its oxidation state remains the same throughout.

once you do more it will be quite intuitive to know which atoms to compare

Ok, I got the right answer just by analyzing the oxidation state of Nitrogen and Manganese.  I simply ignored the oxygen atoms.  Can I always do this, or do I need to consider the oxidation states of both elements in the compound? 

[Astrokel]

Can I always do this, or do I need to consider the oxidation states of both elements in the compound? 

Once you do more, you will know it, but it is good to consider both elements in the compound. Do you know about the electronegativity series F,O,N. Flourine being the most electronegative species will always get a negative oxidation state. Oxygen will always be negative oxidation state(-1 or -2) unless it is bonded to flourine, then it becomes (+1,+2).

[StillLearning]

Can I always do this, or do I need to consider the oxidation states of both elements in the compound? 

Once you do more, you will know it, but it is good to consider both elements in the compound. Do you know about the electronegativity series F,O,N. Flourine being the most electronegative species will always get a negative oxidation state. Oxygen will always be negative oxidation state(-1 or -2) unless it is bonded to flourine, then it becomes (+1,+2).

No, i've never seen that before.  I dont think we're quite there yet in our class.  Thanks for your help.