CK-12-Chemistry Intermediate

22.2. Oxidation Numbers http://www.ck12.org

Now consider the ionic compound sodium thiosulfate, Na 2 S 2 O 3 (Figure22.5). It contains the thiosulfate polyatomic
ion, S 2 O 32 −. The sodium is not part of the covalently bonded polyatomic ion, and so its oxidation number is the
same as it would be in a binary ionic compound, +1. The sulfur is the atom whose oxidation number is not covered
by one of the rules.

2(+1) + 2(x) + 3(−2) = 0

Solving: x = [0 –2 –(−6)] / 2 = +4 / 2 = +2

Sulfur has an oxidation number of +2 in Na 2 S 2 O 3. Notice how the subscript of 2 for the S atom had to be accounted
for by dividing the result of the subtraction by 2. When assigning oxidation numbers, you do so for each individual
atom. In the above example, the oxidation number of sulfur could also have been determined by looking at just the
thiosulfate ion, S 2 O 32 −.

2(x) + 3(−2) =−2.

x = [−2 –(−6)] / 2 = +2

FIGURE 22.5

Sodium thiosulfate is a white crystalline

compound (left) composed of two sodium

ions (Na+) for every one thiosulfate ion

(S 2 O 32 −) (right).

Oxidation-Number Changes in Reactions

In the previous lesson, oxidation was defined as a loss of electrons while reduction was defined as a gain of electrons.
We can now extend the definition to account for oxidation numbers. Consider the reaction in which methane is
combusted, producing carbon dioxide and water. The reaction is shown below, with each atom’s oxidation number
written above the symbol.

− 4

C

+ 1

H 4 (g)+ 2

0

O 2 (g)→

+ 4

C

− 2

O 2 (g)+

+ 1

H 2

− 2

O(g)

In the course of the reaction, the oxidation number of carbon increases from−4 to +4. The oxidation number
of oxygen decreases from 0 to−2. A loss of negatively charged electrons corresponds to an increase in oxidation
number, while a gain of electrons corresponds to a decrease in oxidation number. Therefore, the atom that is oxidized
undergoes an increase in oxidation number, while the atom that is reduced undergoes a decrease in oxidation number.
In the reaction above, the hydrogen atoms are neither oxidized nor reduced because their oxidation number remains
unchanged. Even though it is the carbon atom in the methane that is oxidized, it is also common to say that the
entire methane molecule has been oxidized by the oxygen. The table below (Table22.1) summarizes the processes
of oxidation and reduction.

TABLE22.1:Redox

Oxidation Reduction