1.1 What is Chemistry?

22.3. Balancing Redox Equations http://www.ck12.org

S^2 −→S^6 ++8e−

O^0 +2e−→O^2 −

In the complete reaction, both of these processes occur simultaneously. However, free electrons are not actually a
reactant or product of this reaction. In order for the electrons to cancel out, we would need the number of electrons
lost by oxidation to be equal to the number gained during reduction. This can be accomplished by multiplying the
entire second reaction by 4:

S^2 −→S^6 ++8e−

4O^0 +8e−→4O^2 −

For the electron flow in this reaction to be balanced, four oxygen atoms are reduced for each sulfur atom that is
oxidized. Therefore, for each unit of ZnS (which contains one sulfur atom), we would need two molecules of O 2
(four oxygen atoms total). The ratio of reactants must be the following:

ZnS(aq)+2O 2 (g)→

We can then find the coefficient of the product (in this case, 1) by inspection:

ZnS(aq)+2O 2 (g)→ZnSO 4 (aq)

Now that we have seen the basic procedure, let’s try a more difficult problem.

Example 22.3

Balance the following equation using the half-reaction method:

KMnO 4 (aq)+KNO 2 (aq)+H 2 SO 4 (aq)→MnSO 4 (aq)+H 2 O(l)+KNO 3 (aq)+K 2 SO 4 (aq)

Answer:

First, determine which atoms are being oxidized and which are being reduced. In order to assign oxidation numbers,
we should convert this molecular equation into an ionic one (no coefficients are used, since we are not starting with
a balanced equation anyway):

K+(aq)+MnO− 4 (aq)+NO− 2 (aq)+H+(aq)+SO^24 −(aq)→Mn^2 ++SO^24 −(aq)+H 2 O(l)+K+(aq)+NO− 3 (aq)

There are 6 elements in this reaction. All of the monatomic ions have oxidation numbers equal to their charges.
Additionally, we can assign an oxidation state of -2 to each oxygen atom and +1 to each hydrogen atom. Now, we
just need to assign the oxidation states of manganese in MnO 4 −, nitrogen in the nitrite and nitrate ions, and sulfur in
the sulfate ion. These can be determined by choosing a value for which all of the oxidation states in the ion add up
to its overall charge. For example, in the nitrite ion:

N + 2(O) = -1

N + 2(-2) = -1

N - 4 = -1

N = +3

Nitrogen has an oxidation number of +3 in the nitrite ion. Similar reasoning shows us that manganese is +7 in
MnO 4 −, nitrogen is +5 in NO 3 −, and sulfur is +6 in SO 42 −. The only two elements that change oxidation numbers
over the course of the reaction are Mn (from +7 in MnO 4 −to +2 in Mn^2 +) and N (from +3 in NO 2 −to +5 in NO 3 −).
Thus, we can write the two half-reactions as follows: