# Balance Reduction and Oxidation (Redox) Reactions

### Skill to develop

Balance reduction-oxidation (Redox) equations untill 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 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.

• How many electrons should there be in this half-reaction? Al(s) --> Al3+ + ? e-

Skill - Add electrons in a redox half reaction equation.

• How many electrons should there be in the half-reaction. 2 H2O2 --> 2 H2O + O2 + ? e- This is one of three reaction modes for H2O2.

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.

2 H2O2 --> 2 H2O + O2

• How many electrons should there be in the half reaction
H2O2 --> 2 H+ + O2 + ? e- Note that H2O2 is being oxidized in this reaction.

Discussion -
In this reaction, H2O2 is a reducing agent.

• How many electrons should there be in the half reaction? H2O2 + ? e- --> 2 OH- Note that H2O2 is being reduced in this reaction.

Note -
The oxidation states of 2 O atoms change from -1 to -2.

• How many electrons should there be in the half reaction S8 + ? e- --> 8 S2-

Note -
The charge is also balanced.

• What is the oxidation state of In in In(OH)3?

Note -
Consider OH in the formula as OH-.

• What is the oxidation state of Fe in FeAsO4?

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

• In the reaction 2 Cr3+ + H2O + 6 ClO3- --> Cr2O72- + 6 ClO2 + 2 H+, identify the reducing agent.

Discussion -
Since Cr3+ is oxidized, it is a reducing agent.

• Identify the reducing agent in the reaction
Cr2O72- + H2C=O --> HCOOH + Cr3+. The formula H2C=O with a double bond between C and O is an aldehyde, and it may also be written as H2CO. The formula HCOOH is an acid. Write it as H2CO2 to determine the oxidation state of C.

Note -
The aldehyde 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 CO2.

• Identify the reducing agent in the reaction: 7 CN- + 2 OH- + 2 Cu(NH3)42+ --> 2 Cu(CN)32- + 8 NH3 + CNO- + H2O.

Smile -
You have just answered a fairly difficult question.

• Identify the oxidant in the reaction: 7 CN- + 2 OH- + 2 Cu(NH3)42+ --> 2 Cu(CN)32- + 8 NH3 + CNO- + H2O.

Note -
The oxidation state for Cu in Cu(NH3)42+ is +2, but its oxidation state in Cu(CN)32- is +1. Thus, Cu(NH3)42+ is the oxidant, because itself is reduced.