RedOx Chemistry: How to Balance Reduction and Oxidation Reaction

There are rules you need to learn when you want to balance Oxidation and reduction reaction. Oxidation and reduction reaction is alternatively called Redox reaction.

What is a Redox reaction?

A redox reaction is an abridged name for reduction and oxidation reactions. So it simply means it is a reaction that involves reduction and oxidation.

Oxidation and reduction reaction is a reaction in which oxidation and reduction occur simultaneously.

It means as some elements are undergoing oxidation, some other ones are being reduced.

It will be great we really define what we mean by oxidation and reduction.

There are some basic definitions we can concentrate on.

Definitions of oxidation

1. Oxidation is the loss of electrons

Zn- 2e^–  à Zn

• Oxidation is the addition of oxygen

2Zn + O₂  à 2ZnO

• Oxygen is the removal of hydrogen
• Oxidation is the increase in oxidation number

Definitions of reduction

1. Reduction is the gain of electrons
2. Reduction is the addition of hydrogen
3. Reduction is the removal of oxygen
4. Reduction is the decrease in oxidation number

What is Reducing agent?

A reducing agent can be defined in the following terms

1. Electron donor
2. Substance which is oxidized
3. A substance in which the oxidation number increases

What is an Oxidizing agent?

An oxidizing agent can be defined in the following terms

1. Electron acceptor
2. Substance which is reduced
3. A substance in which oxidation number decreases

Test for the oxidizing agent

An oxidizing agent can be tested using the following reagents:

1. Test with iron II chloride

An oxidizing agent changes the colour of iron II chloride from green to brown Iron III salts.

Fe²⁺  –> Fe³⁺

• Test with Hydrogen sulphide

An oxidizing agent forms a yellow deposit of sulphur when hydrogen sulphide gas is passed over it.

S^2-  –> S⁰

      C.  Test with potassium iodide

           An oxidizing agent will oxidize potassium iodide to iodine which can form a blue-black

           colour with starch.

1. –> I₂

Test for reducing agents

A reducing agent can be tested using the following reagents

1. Test with acidified potassium tetraoxo mangannnate VII solution

A reducing agent will reduce acidified potassium tetraoxomngannate VI solution from purple  KMnO4 to colourless Mn²⁺

MnO₄^– + 8H⁺ +5e^– à Mn²⁺ +4H₂O

• Test with acidified potassium heptaoxodichromate VI solution.

A reducing agent will change the orange colour of potassium heptaoxodichrmate VI solution to green Cr³⁺

Cr₂O₇^2- + 4H₂O +5e^–   –> MnO₄^– +8H⁺

What is an Ionic Equation?

Redox reactions can be written with an ionic equation ad there is a need to learn how to balance the ionic equations involving redox reactions.

Have you heard of the ionic equation? An ionic equation is an equation that involves only ions and is balanced with electrons.

Before I go quickly into ionic equations of redox reactions and their balancing, I will like to start with oxidation and reduction reactions separately.

Redox reaction showing Oxidation reaction

Again, recall that oxidation is the loss of electrons and equally addition of oxygen.

Let’s look at some examples of oxidation reactions and how to balance them..

Mg + O₂ –> MgO

.0        0        +2 -2

This is the oxidation of Magnesium because magnesium was oxidized from zero to two)Mg⁰ to Mg²⁺)

Writing the ionic equation

Mg⁰     à  Mg²⁺

Another example to illustrate oxidation is the reaction of iron II chloride with chlorine.

FeCl₂ + Cl₂   –> FeCl₃

+2 -2      0          +3  -3

Writing the ionic equation

Fe²⁺    –> Fe³⁺

Iron is oxidized from Zero to two.

Redox Reaction showing Reduction Reaction

Let’s look at the reduction of copper II oxide to copper.

CuO + Mg –>Cu + MgO

Cu²⁺  à Cu

How to Identify Oxidizing Agent and Reducing Agent  

I am going to show you how to identify oxidizing and reducing agents in a chemical equation.

Recall that an oxidizing agent is always reduced while a reducing agent is always oxidized at the end of the reaction.

Example 1

Zn + CuSO₄ –> ZnSO₄ + Cu

First, deduce the oxidation states of the elements in the reaction

Zn + CuSO₄ –> ZnSO₄ + Cu

0       +2             +2           0

You can see that zinc increased from 0 to 2

Also copper reduced from 2 to 0.

Therefore Zn is the Reducing agent because it was oxidized while copper is the reducing agent because it was reduced.

Zn + CuSO₄ à ZnSO₄ + Cu

R.A       O.A           

Hint: Your ability to solve or deduce the oxidation numbers of the elements in the reaction will guide you in identifying the oxidizing and reducing agent.

Example 2

The reaction of hydrogen sulphide with Bromine

H₂S  + Br₂ –> 2HBr + S

First, deduce the oxidation numbers of the molecules

H₂S  + Br₂ –> 2HBr + S

   -2      0            -1      0

Sulphur was oxidized from -2 to 0 while Bromine was reduced from 0 to -1.

Consequently, H₂S is the reducing agent while Bromine is the oxidizing agent.

H₂S  + Br₂ –> 2HBr + S

 R.A    O.A

Please take note that we only concentrate on the reactants for oxidizing and reducing agents.

Example 3

Oxidation of iron II chloride to iron III chloride

2FeCl₂  + Cl₂  –> 2FeCl₃

First deduce the oxidation numbers

2FeCl₂  + Cl₂  –> 2FeCl₃

+2  -2       0          +3  -3

Iron was oxidized from +2 to +3 while chlorine decreased from -2 to -3 as a result FeCl₂ is the reducing agent while Cl₂ is the oxidizing agent.

2FeCl₂  + Cl₂  –> 2FeCl₃

 R.A        O.A

How to Write Ionic Equation from Redox Reaction

Before I go straight to balancing redox reactions or equations, I want to illustrate how to deduce ionic form of the redox reaction.

Example 1

Potassium displaces zinc from zinc chloride

2K + ZnCl₂ à 2KCl + Zn

An ionic equation starts by ionizing the molecules

Hint: Gases will not ionize and also insoluble compounds won’t ionize.

Steps involved in writing ionic Equations

Step 1

Ionize the soluble ionic compounds

2K + ZnCl₂ –> 2KCl + Zn

2K + Zn²⁺ 2Cl^–   –> 2K⁺ 2Cl^–  + Zn

Step 2

Cancel out the spectator ions

Spectator ions are ions that are neither reduced nor oxidized but remain the same.

From the above equation K⁼ is neither reduced nor oxidized, cancel out K⁺

 Zn²⁺ 2Cl^–   –>  2Cl^– + Zn

Example 2

Cl₂ + 2NaBr   –> 2NaCl + Br₂

Step 1

Ionize soluble ionic compounds

Please do not ionize gases

Cl₂ + 2Na⁺   2Br⁺   –> 2Na⁺ 2Cl^– + Br₂

Step 2

Cancel out the spectator ions.

Na⁺ is the spectator ion because it is neither reduced nor oxidized.

Cl₂ + 2Br⁺  –>    2Cl^– + Br₂

Example 3

Reaction of Potassium iodide with iron III sulphate

2K1 + Fe₂(SO₄)₃ –> I₂ + K₂SO₄ + 2FeSO₄

Step 1

Ionize the soluble ionic compounds

2K⁺  2I^–  + 2Fe³⁺  3SO₄^2-   –> I₂ + 2K⁺ SO₄^2- + 2Fe²⁺  2SO₄^2-

Step 2

Cancel out the spectator ions

The spectator ion is SO₄^2- and K⁺, so we will have

2K⁺  2I^–  + 2Fe^{3+  –}   à I₂ + 2K⁺ + 2Fe²⁺ 

Balancing Ionic Equations of Redox reactions

There are two forms of ionic equation we will be balancing; ionic equations that involve oxygen and ionic equations not involving oxygen.

Type 1- ionic equations with no oxygen on both sides

Example 1

Mg + Zn²⁺  à Mg²⁺ + Zn

Solution

We balance charges with electrons

Deduce the oxidation

Recall that oxidation is loss of electrons while reduction is gain of electrons.

Use this Pneumonic to remember

OIL RIG (OIL= oxidation is loss of electrons while RIG =reduction is gain of electrons)

Oxidation half equation

Oxidation is loss of electrons and increase in oxidation number

Mg   à       Mg²⁺

Mg  -2e^–  à  Mg²⁺

Mg  à  Mg²⁺ + -2e^– 

Reduction half equation

Reduction is gain of electrons and decrease in oxidation number

Cu²⁺  + 2e^– à Cu

Example 4

Cl₂ + 2Br^–   –/> 2Cl^– +  Br₂

First deduce the species undergoing oxidation and reduction.

Cl₂ à 2Cl^–   reduction

2Br^– à Br₂  oxidation

Oxidation half equation

 2Br^– -2e^–à Br₂ 

2Br^{– –}à Br₂  + 2e^-,

Reduction half equation

Cl₂ + 2e^–   à 2Cl^–

Type 2- Ionic equations involving oxygen

How to balance ionic equation involving oxygen

Step 1

Add H₂O and H^{+ (}Add water to the side with no oxygen or less oxygen and H⁺ to the side with oxygen or more oxygen)

Step 2

Balance the atoms

Step 3

Balance the charges with electrons

Example 1

MnO₄^– + Fe  à Mn²⁺  + Fe²⁺

First, deduce the oxidation and reduction

Reduction half equation

MnO₄^–    à  Mn²⁺

Add H₂O and H⁺

MnO₄^–   + H⁺  à Mn²⁺  +H₂O

Balance the atoms

MnO₄^–   + 8H⁺  à Mn²⁺  +4H₂O

Balance the charges with electrons

MnO₄^–   + 8H⁺ +5e^– à Mn²⁺  +4H₂O

Oxidation half equation

Fe  -2e^– –> Fe²⁺

Fe –> Fe²⁺ + 2e^–

Example 2

Cr₂O₇^2- + Cu     –> Cr³⁺ + Cu²⁺

First deduce the oxidation and reduction half equations

 Cr₂O₇^2-  –> Cr³⁺

Add H₂O and H⁺

Cr₂O₇^2- + H⁺  –> Cr³⁺ +H₂O

Balance the atoms

Cr₂O₇^2- + 14H⁺  –> 2Cr³⁺ +7H₂O

Balance the charges with electrons

Cr₂O₇^2- + 14H⁺  +6e^– –> 2Cr³⁺ +7H₂O

Oxidation half equation

Cu  -2e^– –> Cu²⁺

Cu–> Cu²⁺ + 2e^–

Balancing redox reactions is very essential in Chemistry because most times we dwell so much on the ionic rquations showing loss orgain of e;lectrons.