# Balance Reduction and Oxidation (Redox) Reactions

Skill to Develop

• Balance reduction-oxidation (redox) equations until you have developed a logical method without having to memorize the steps needed to balance redox equations.

## Balance Oxidation and Reduction Reaction Equations

The skill to balance redox reaction equations 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 balancing 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 $$\ce{H+}$$ and $$\ce{H2O}$$ to balance the charges and other atoms. In a basic solution, use $$\ce{OH-}$$ and $$\ce{H2O}$$ 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 $$\ce{H+}$$, $$\ce{OH-}$$, and $$\ce{H2O}$$ 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:

$$\ce{Cr2O7^2- + 14 H+ + 6 e- \rightarrow 2 Cr^3+ + 7 H2O}$$
$$\ce{H2C2O4 \rightarrow 2 CO2 + 2 H+ + 2 e-}$$

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

$$\ce{Cr2O7^2- + 14 H+ + 6 e- \rightarrow 2 Cr^3+ + 7 H2O}$$
$$\ce{3 H2C2O4 \rightarrow 6 CO2 + 6 H+ + 6 e-}$$

Add the two equations and cancel the electrons to give the overvall equation.

$$\ce{Cr2O7^2- + 8 H+ + 3 H2C2O4 \rightarrow 2 Cr^3+ + 7 H2O + 6 CO2}$$

Discussion
This example illustrates balancing redox equations in acid solutions.

Example 2

Balance the equation from the two half-reactions:

$$\ce{Cd \rightarrow Cd^2+ + 2 e-}$$
$$\ce{4 H+ + NO3- + 3 e- \rightarrow NO + 2 H2O}$$

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

$$\ce{3 Cd \rightarrow 3 Cd^2+ + 6 e-}$$
$$\ce{8 H+ + 2 NO3- + 6 e- \rightarrow 2 NO + 4 H2O}$$

Add the two equations and cancel the electrons to give the overvall equation.

$$\ce{3 Cd + 8 H+ + 2 NO3- \rightarrow 3 Cd^2+ + 2 NO + 4 H2O}$$

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

Example 3

In a basic solution, $$\ce{Fe(OH)2}$$ and $$\ce{Fe(OH)3}$$ are solids. The former may be oxidized by $$\ce{H2O2}$$.

$$\ce{Fe(OH)2 + H2O2 \rightarrow Fe(OH)3 + H2O}$$

Balance this equation.

Solution
The balanced half-reactions are:

$$\ce{Fe(OH)2 + OH- \rightarrow Fe(OH)_{3\large{(s)}} + e-}$$
$$\ce{H2O2 + 2 e- \rightarrow 2 OH-}$$

Thus, the balanced equation is

$$\ce{2 Fe(OH)2 + H2O2 \rightarrow 2 Fe(OH)3}$$

Discussion
Note that the balanced equation does not have an $$\ce{H2O}$$ in it.

Example 4

Balance the following reaction, which is carried out in an acidic solution:

$$\ce{I- + IO3- \rightarrow I2}$$

Solution
The half-reactions are:

$$\ce{2 I- \rightarrow I2 + 2 e- \: (oxidized)}$$
$$\ce{2 IO3- + 10 e- \rightarrow I2 \: (reduced)}$$

The balanced equation is

$$\ce{6 H+ + 5 I- + IO3- \rightarrow 3 I2 + 3H2O}$$

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 skills:

1. Identify oxidation state.
2. Identify reductants and oxidants
3. Add electrons to balance half-reaction equations.

These skills are the foundation on which to build your skill to balance equations.

1. How many electrons should there be in this half-reaction?

$$\ce{Al_{\large{(s)}} \rightarrow Al^3+ +\: ?\: e-}$$

Hint: 3

Skill - Add electrons in a redox half reaction equation.

2. How many electrons should there be in this half-reaction?

$$\ce{2 H2O2 \rightarrow 2 H2O + O2 +\: ?\: e-}$$

This is one of three reaction modes for $$\ce{H2O2}$$.

Hint: 0

Discussion -
This is an example of disproportionation, in which the change of oxidation state of one atom is compensated by the change of oxidation state of another atom.

$$\ce{2 H2O2 \rightarrow 2 H2O + O2}$$

3. How many electrons should there be in this half reaction?

$$\ce{H2O2 \rightarrow 2 H+ + O2 +\: ?\: e-}$$

Note that $$\ce{H2O2}$$ is being oxidized in this reaction.

Hint: 2

Discussion -
In this reaction, $$\ce{H2O2}$$ is a reducing agent.

4. How many electrons should there be in this half reaction?

$$\ce{H2O2 +\: ?\: e- \rightarrow 2 OH-}$$

Note that $$\ce{H2O2}$$ is being reduced in this reaction.

Hint: 2

Note -
The oxidation states of 2 $$\ce{O}$$ atoms change from -1 to -2.

5. How many electrons should there be in this half reaction?

$$\ce{S8 +\: ?\: e- \rightarrow 8 S^2-}$$

Hint: 16

Note -
The charge is also balanced.

6. What is the oxidation state of $$\ce{In}$$ in $$\ce{In(OH)3}$$?

Hint: +3

Note -
Consider $$\ce{OH}$$ in the formula as $$\ce{OH-}$$.

7. What is the oxidation state of $$\ce{Fe}$$ in $$\ce{FeAsO4}$$?

Hint: +3

Note -
The oxidation state for $$\ce{As}$$ is +5, because it is an element in group 5A. Thus, the oxidation state is +3 for $$\ce{Fe}$$. The common oxidation states for $$\ce{Fe}$$ are +2 and +3.

8. In the reaction identify the reducing agent.

$$\ce{2 Cr^3+ + H2O + 6 ClO3- \rightarrow Cr2O7^2- + 6 ClO2 + 2 H+ }$$

Hint:$$\ce{Cr^3+}$$

Discussion -
Since $$\ce{Cr^3+}$$ is oxidized, it is a reducing agent.

9. Identify the reducing agent in the reaction:

$$\ce{Cr2O7^2- + H2C=O \rightarrow HCOOH + Cr^3+ }$$

The formula $$\ce{H2C=O}$$ with a double bond between $$\ce{C}$$ and $$\ce{O}$$ is an aldehyde, and it may also be written as $$\ce{H2CO}$$. The formula $$\ce{HCOOH}$$ is an acid. Write it as $$\ce{H2CO2}$$ to determine the oxidation state of $$\ce{C}$$.

Hint: $$\ce{H2CO}$$

Note -
The aldehyde $$\ce{H2C=O}$$ is a reducing agent. Oxidation of an alcohol gives an aldehyde, oxidation of which gives an acid, which can further be oxidized to give $$\ce{CO2}$$.

10. Identify the reducing agent in the reaction:

$$\ce{7 CN- + 2 OH- + 2 Cu(NH3)4^2+ \rightarrow 2 Cu(CN)3^2- + 8 NH3 + CNO- + H2O}$$

Hint: $$\ce{CN-}$$

Smile -
You have just answered a fairly difficult question.

11. Identify the oxidant in the reaction:

$$\ce{7 CN- + 2 OH- + 2 Cu(NH3)4^2+ \rightarrow 2 Cu(CN)3^2- + 8 NH3 + CNO- + H2O}$$

Hint: The copper ammonia complex is the oxidant.

Note -
The oxidation state for $$\ce{Cu}$$ in $$\ce{Cu(NH3)4^2+}$$ is +2, but its oxidation state in $$\ce{Cu(CN)3^2-}$$ is +1. Thus, $$\ce{Cu(NH3)4^2+}$$ is the oxidant, because it is reduced.