Balance Reduction and Oxidation (Redox) Reactions

Skill to develop

Balance Oxidation and Reduction Reaction equations

The skill to balance redox reaction equation is rather complicated, but it can be broken down into some steps.

The first step is the skill to identify Oxidation states of any element in a chemical formula. This link helps you to acquire that skill, if you have not already done so.

The second step is to Balance Half Reaction Equations. a skill to acquire from this link.

As a review, here are the guidelines for balance reaction equations:

  1. Identify the elements that are oxidized and reduced by examining their Oxidation states
  2. Write the oxidation and reduction half-reactions and Balance Half Reaction Equations. In an acid solution, use H+ and H2O to balance the charges and other atoms. In a basic solution, use OH- and H2O to balance the charges and other atoms.
  3. Add the two half-reactions algebraically such that the electrons in the two half-reaction equations cancel completely. Cancel other species such as H+, OH-, and H2O common to the two sides, if necessary.
  4. Check your equation and make certain that numbers of atoms and charge are equal on both sides.

Some examples are given here to illustrate the last step for balancing a reaction equation.

Example 1

Balance the equation from the following two half-reactions: Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
H2C2O4 --> 2 CO2 + 2 H+ + 2 e-

Solution
Multiply the second equation by 3 and then add them algebraically so that the electrons in the two half-reaction equations cancel completely.

Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
3 H2C2O4 --> 6 CO2 + 6 H+ + 6 e-
add the two equations and cancel the electrons to give the overvall equation
Cr2O72- + 8 H+ + 3 H2C2O4 --> 2 Cr3+ + 7 H2O + 6 CO2

Discussion
This example illustrate balancing redox equations in acid solutions.

Example 2

Balance the equation from the two half-reactions: Cd --> Cd2+ + 2 e-
4 H+ + NO3- + 3 e- --> NO + 2 H2O

Solution
The first half-reaction has 2 electrons, whereas the second one has 3. The lowest common multiple of 2 and 3 is 6. Thus, you multiply the first equation by 3 and the second one by 2. The half-reaction equations become

3 Cd --> 3 Cd2+ + 6 e-
8 H+ + 2 NO3- + 6 e- --> 2 NO + 4 H2O
add the two equations and cancel the electrons to give the overvall equation
3 Cd + 8 H+ + 2 NO3- --> 3 Cd2+ + 2 NO + 4 H2O

Discussion
The last step in balancing oxidation and reduction reactions is simple.

Example 3

In a basic solution, Fe(OH)2 and Fe(OH)3 are solids. The former may be oxidized by H2O2. Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O. Balance this equation.

Solution
The balanced half-reactions are:

Fe(OH)2 + OH- --> Fe(OH)3 (s) + e-
H2O2 + 2 e- --> 2 OH-
Thus, the balanced equation is 2 Fe(OH)2 + H2O2 --> 2 Fe(OH)3

Discussion
Note that the balanced equation does not have an H2O in it.

Example 4

Balance the following reaction, which is carried out in an acidic solution: I- + IO3- --> I2

Solution
The half-reactions are:

2 I- --> I2 + 2 e- (oxidized)
2 IO3- + 10 e- --> I2 (reduced)
The balanced equation is 6 H+ + 5 I- + IO3- ® 3 I2 + 3H2O

Discussion
You may study Electrochemistry to gain better insight to the oxidation reduction process.

Confidence Building Questions

The last step to eliminate electrons in the two half reactions is the easy one. The questions below give you a chance to test your skill in
  • Identify oxidation state.
  • Identify reductants and oxidants
  • Add electrons to balance half-reaction equations.
  • These skills are the fundation on which to build your skill to balance equations.