how come VO2+ isn't the oxidizing agent? Didn't it also gain an electron? Does it have to have kind of the same form on the product side? Because there are 2 O groups on the reactant side which is confusing me I think.
Chad gen chem question about oxidizing agents
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This is a pretty tricky problem, since there are multiple atoms and it isn't immediately clear what is getting oxidized and reduced. To figure that out, we need to look at the oxidation states of the individual atoms.
First let's look at VO 2+. V is a transition metal, which means we have to figure out its oxidation state by context. O is most electronegative, so we assign it its usual oxidation state (-2). Overall charge on VO is 2+, so the V must balance to provide our overall charge - V's oxidation state must be +4.
Now let's take a look at VO2 + on the other side. We use the same technique to figure out oxidation states: O's -2. Since there are two O atoms, O is contributing -4 to the overall oxidation state. Since the overall oxidation state is +1, we know that V's oxidation state must balance that out accordingly. So V's oxidation state is +5.
So what happened to the oxidation state of the individual atoms as we went from VO 2+ to VO2 +?
O's oxidation state: -2 to -2 (no change)
V's oxidation state: +4 to +5 (lost 1 electron)
O's didn't lose or gain electrons, so it wasn't reduced here. V lost an electron. Therefore VO 2+ couldn't be possibly be the oxidizing agent, and in fact must be the reducing agent.
So if VO 2+ didn't act as the oxidizing agent, what did? Taking a look at our reactants, we see O2 (g) in its standard state, which means the O atoms have an oxidation state of 0. The O's on the product side have an oxidation state of -2, so it fits that O2 acted as the oxidizing agent and was itself reduced (gained electrons). Therefore O2 is the oxidizing agent, and VO 2+ is the reducing agent.
First let's look at VO 2+. V is a transition metal, which means we have to figure out its oxidation state by context. O is most electronegative, so we assign it its usual oxidation state (-2). Overall charge on VO is 2+, so the V must balance to provide our overall charge - V's oxidation state must be +4.
Now let's take a look at VO2 + on the other side. We use the same technique to figure out oxidation states: O's -2. Since there are two O atoms, O is contributing -4 to the overall oxidation state. Since the overall oxidation state is +1, we know that V's oxidation state must balance that out accordingly. So V's oxidation state is +5.
So what happened to the oxidation state of the individual atoms as we went from VO 2+ to VO2 +?
O's oxidation state: -2 to -2 (no change)
V's oxidation state: +4 to +5 (lost 1 electron)
O's didn't lose or gain electrons, so it wasn't reduced here. V lost an electron. Therefore VO 2+ couldn't be possibly be the oxidizing agent, and in fact must be the reducing agent.
So if VO 2+ didn't act as the oxidizing agent, what did? Taking a look at our reactants, we see O2 (g) in its standard state, which means the O atoms have an oxidation state of 0. The O's on the product side have an oxidation state of -2, so it fits that O2 acted as the oxidizing agent and was itself reduced (gained electrons). Therefore O2 is the oxidizing agent, and VO 2+ is the reducing agent.
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The oxidant question is incorrect.
Vanadium is going from +4 to +5......it is oxidized (reductant)......02 is at zero to begin with and goes to -2.....it is being reduced,,,and is the oxidant ( oxidizing agent )
Oh ok so you can't treat VO2 as its own separate thing being reduced you have to treat the elements individually. Man forgot some of this stuff from school.
How come we don't choose the O's in H2O instead of the O's in O2 when comparing to the O's in Vo2 +?
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You could have chosen to look at H2O. But what you'd find is that the oxidation state of O in H2O is -2, and the O in VO2 + is also -2. So no transfer of electrons took place there, and we'd have to look elsewhere for the electron transfer.
Will there ever be a question that has O's changing in the product let's say to -4 but have an O with a value of 0 in one compound and an O with a value of -2 in another on the reactant side (like two oxygens in the reactant that each have different oxidation states than one oxygen in the product)?
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I've seen two of the same atom (in different molecules) get reduced in one and oxidized in the other, but I've never seen both get oxidized or both get reduced. I'm not sure if it's possible, but I wouldn't expect anything that complicated or confusing on the DAT.
