Here is a convenient method for determining oxidation states. Basically, you treat the charges in the compound as a simple algebraic expression. For example, let's determine the oxidation states of the elements in the compound, KMnO_{4}. Applying rule 2, we know that the oxidation state of potassium is +1. We will assign "x" to Mn for now, since manganese may be of several oxidation states. There are 4 oxygens at -2 apiece. The overall charge of the compound is zero:

+1 x 4(-2)

The algebraic expression generated is:

Solving for x gives the oxidation state of manganese:

x = +7

K Mn O_{4}

+1 +7 4(-2)

2x 7(-2)

2x = 12

x = +6

2x 3(-2)

2x - 6 = 0

2x = 6

x = 3

x 3(-2)

x - 6 = -2

x = +4

__Determining the Oxidizing Agent and Reducing Agent__

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-|---|---|---|---|---|---|---|---|---|---|---

-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5

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oxidation number increased, substance oxidized

First, determine the oxidation states for each of the atoms in the reactants and products.

+1 x 3(-2)

1 + x - 6 = 0

x = +5

so, H = +1, N = +5, O = -2

x 3(-2)

x - 6 = -1

x = +5

so, Cu = +2, N = +5, O = -2

x 2(-2)

x - 4 = 0

x = +4

so, N = +4, O = -2

2(+1) -2

so H = +1, O = -2