# Balancing Redox Equations

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:26

Some redox equations may be balanced using the methods developed in Balancing Chemical Equations, but most are rather difficult to handle. Therefore it is useful to have some rules, albeit somewhat arbitrary ones, to help find appropriate coefficients. These rules depend on whether the reaction occurs in acidic or basic solution. In either situation we must make sure that the number of electrons accepted by the oxidizing agentA chemical species that accepts electrons in order to oxidize another species. In the process the oxidizing agent is itself reduced. exactly equals the number of electrons donated by the reducing agentA chemical species that donates electrons in order to reduce another species. In the process the reducing agent is itself oxidized..

In acid solution       We shall apply the rules to the equation

$\overset{\text{+5 }-\text{2}}{\mathop{\text{IO}_{\text{3}}^{-}}}\,\text{ + }\overset{\text{+4 }-\text{2}}{\mathop{\text{SO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{O}}}\,\text{ }\to \text{ }\overset{\text{0}}{\mathop{\text{I}_{\text{2}}}}\,\text{ + }\overset{\text{+6 }-\text{2}}{\mathop{\text{SO}_{\text{4}}^{\text{2}-}}}\,\text{ + }\overset{\text{+1 }-\text{2 }}{\mathop{\text{H}_{\text{3}}\text{O}^{\text{+}}}}\,\text{ (1)}$

The changes in oxidation numbers verify that it is a redox equation, and the presence of H3O+ indicates that it occurs in acidic solution. The rules are

1 Write unbalanced half-equations for the oxidationThat part of a chemical reaction in which a reactant loses electrons; simultaneous reduction of a reactant must occur. of the reducing agent and for the reductionThat part of a chemical reaction in which a reactant gains electrons; simultaneous oxidation of a reactant must occur. of the oxidizing agent.

Oxidation:         O2 → SO42–

Reduction:      IO3 → I2

2 Balance the elementA substance containing only one kind of atom and that therefore cannot be broken down into component substances by chemical means. reduced or oxidized in each half-equation.

Oxidation:        SO2 → SO42–...S already balanced

Reduction:      2IO3 → I2

3 Balance oxygen atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. by adding water (solventThe substance to which a solute is added to make a solution.) molecules.

Oxidation:      SO2 + 2H2O → SO42–

Reduction:                 IO3 → I2 + 6H2O

4 Balance hydrogen atoms by adding hydrogen ions (available from the acidic solution).

Oxidation:      SO2 + 2H2O → SO42– + 2H+

Reduction:      12H+ + IO3 → I2 + 6H2O

5 Balance electrical charges by adding electrons.

Oxidation:                  SO2 + 2H2O → SO42– + 2H+ + 2e

(The total charge on the left side was 0, but on the right it was – 2 + 4 = +2. Therefore 2e were needed on the right.)

Reduction:      10e + 12H+ + 2IO3 → I2 + 6H2O

(The total charge on the left was 12 - 2 = +10, but on the right it was 0. Therefore 10e were needed on the left.)

6 Use oxidation numbers to check that the number of electrons is correct.

Oxidation: The oxidation number of electrons increases from +4 to +6, corresponding to a loss of 2e.

Reduction: The oxidation number of I falls from +5 to 0, corresponding to a gain of 5e for each I. Since there are 2 I atoms, 10e must be added.

7 Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent. Since only 2 electrons are donated in the oxidation half-equation, while 10 are required by the reduction, the oxidation must occur 5 times for each reduction. That is, both sides of the oxidation half-equation must be multiplied by 5:

Oxidation:               5SO2 + 10H2O → 5SO42– + 20H+ + 10e

Reduction:      10e + 12H+ + 2IO3 → I2 + 6H2O

8 Sum the half-equations. The net equation which result is

5SO2 + 4H2O + 2IO3 → 5SO42– + 8H+ + I2

Note that when the half-equations were summed, the number of electrons was the same on both sides, and so no free electrons (which could not exist in aqueous solution) appear in the final result. It also would be more accurate to write H3O+ instead of H+ for the hydronium ion. This can be done by adding 8H2O to both sides of the equation:

5SO2 + 12H2O + 2IO3 → 5SO42– + 8 H3O+ + I2

(On the right, the 8H2O molecules are protonated to 8H3O+. It is also a good idea at this point to check that all atoms, as well as the electrical charges, balance.

In basic solution      Potassium permanganate KMnO4, can be used to oxidize alcohols to carboxylic acids. An example is

$\overset{\text{+7 }-\text{2}}{\mathop{\text{2MnO}_{\text{4}}^{-}}}\,\text{ + }\overset{-\text{2 +1 }-\text{2 +1}}{\mathop{\text{CH}_{\text{3}}\text{OH}}}\,\text{ + }\to \text{ }\overset{\text{+4 }-\text{2}}{\mathop{\text{MnO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{O}}}\,\text{ + }\overset{\text{+1 +2 }-\text{2 }-\text{2 }}{\mathop{\text{HCOO}^{-}}}\,\text{ }\overset{\text{+2}}{\mathop{\text{2Mn}^{\text{2+}}}}\,\text{ + }\overset{-\text{2 +1}}{\mathop{\text{OH}^{-}}}\,\text{ (2)}$

Since OH is produced, the reaction occurs in basic solution. It clearly involves redox.

1 Write unbalanced equations for the oxidation of the reducing agent and the reduction of the oxidizing agent (same as for acid solution).

Oxidation:      CH3OH → HCOO

Reduction:      MnO4 → MnO2

2 Balance the element reduced or oxidized in each half-equation (same as for acid solution). (Both C and Mn are already balanced.)

3 Balance oxygen atoms by adding hydroxide ions (available from the basic solution).

Oxidation:     CH3OH + OH → HCOO

Reduction:               MnO4 → MnO2 + 2OH

4 To the side of each half-equation which lacks hydrogen, add one water molecule for each hydrogen needed. Add an equal number of hydroxide ions to the opposite side.

Oxidation:      CH3OH + 5OH → HCOO + 4H2O

(Four hydrogens were needed on the right, and so 4 water molecules were added on the right and 4 hydroxide ions on the left.)

Reduction:      MnO4 + 2H2O → MnO2 + 4OH

(Two hydrogens were needed on the left, and so 2 water molecules were added on the left and 2 hydroxide ions were added on the right. Note that the added hydroxide ions are to maintain the balance of oxygen atoms.)

5 Balance electrical charges by adding electrons (same as for acid solution).

Oxidation:               CH3OH + 5OH → HCOO + 4H2O + 4e

(The total charge on the left was -5, but on the right it was -1, and so 4e were added on the right.)

Reduction:      MnO4 + 2H2O + 3e → MnO2 + 4OH

(The total charge on the left was -1, but on the right it was -4, and so 3e were added on the left.)

6 Use oxidation numbers to check that the number of electrons is correct (same as for acid solution).

Oxidation: C goes from -2 to +2, corresponding to a loss of 4e.

Reduction: Mn goes from +7 to +4, corresponding to a gain of 3e.

7 Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent (same as for acid solution). Multiplying the oxidation half-equation by 3 and the reduction half-equation by 4 adjusts each so it involves 12e.

Oxidation:               3CH3OH + 15OH → 3HCOO + 12H2O + 12e

Reduction:      4MnO4 + 8H2O + 12e → 4MnO2 + 16OH

8 Sum the half-equations (same as for acid solution). The net equation which results is

3CH3OH +4MnO4 → 3HCOO + 4MnO2 + OH

Again, it is worthwhile to check that all atoms and charges balance.

The rules for balancing redox equations involve adding H+, H2O, and OH to one side or the other of the half-equations. Since these species are present in the solution, they may participate as reactants or products, but usually there is no experiment which can tell whether they do participate. However, the balanced equation derived from our rules does indicate just what role H+, H2O, or OH play in a given redox process.