Oxidation or Reduction: NO3- to NO

[Kabloom]

Homework Statement

Is this oxidation or reduction: (NO3)- ---> NO (this is a half reaction I took from the full equation)

Homework Equations

Not necessary

The Attempt at a Solution

At first, I got the oxidation states of each of the atoms

For (NO3)- N has an oxidation state of 5+ and O has an oxidation state of 6-.

for NO N has an oxidation state of 2+ and O has an oxidation state of 2-.


and then I look at the charge of the whole compound.

It goes from - to 0.

therefore, i concluded that this is a oxidation equation.


but it is actually a reduction equation


Did I overlook something?

 

[symbolipoint]

That is a reduction. Look at it this way:

NO3 supscr -1 gives (-2)*3 charge from the oxygens. The charge on the Nitrogen is what? x + (-2)*3 = -1 means x=+5.

Now, what happened in the half reaction? What is the charge that the Nitrogen now carries? How many electrons were gained OR how many electrons were lost? If the Nitrogen had to gain electrons then the half reaction was a reduction. If the Nitrogen had to loose electrons then the half reaction was an oxidation.

 

[Borek]

For (NO3)- N has an oxidation state of 5+

for NO N has an oxidation state of 2+

So, what happened to nitrogen, did it ON went up or down?

Note: while ON for N are OK, you can't write

O has an oxidation state of 6-

There were three O atoms, each with ON -2. It is not the same, as your statement suggests each atom was -6.