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:

Identify the elements that are oxidized and reduced by examining their Oxidation states
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.
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.
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: Cr₂O₇^2-+ 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.

Cr₂O₇^2- + 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
Cr₂O₇^2- + 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, H₂O₂is 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 S^2-

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^- --> Cr₂O₇^2- + 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
Cr₂O₇^2- + 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)₃^2- + 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)₃^2- + 8 NH₃ + CNO^- + H₂O.

Note -
The oxidation state for Cu in Cu(NH₃)₄²⁺ is +2, but its oxidation state in Cu(CN)₃^2- is +1. Thus, Cu(NH₃)₄²⁺ is the oxidant, because itself is reduced.