Lesson

Many chemical reactions involve the loss or gain of electrons by atoms, ions or molecules. The loss or gain of electrons can result in a change in the charge of a species. For example, when sodium atoms lose electrons to chlorine, each species acquires a different net charge. Sodium becomes a 1+ ion and chlorine becomes a 1- ion.

Knowing how to draw Lewis diagrams for atoms, ions and covalent compounds can be an asset in this lesson. If you haven't already done so, you should try the items in the Prerequisites part of this lesson.

Introduction to Oxidation Numbers

The oxidation number - the actual or hypothetical charge of an atom or ion - is a way of keeping track of the electrons in a particular species.

In a redox reaction, two elements undergo a change in oxidation number. The oxidation number of one element will increase as a result of a loss of electrons and the oxidation number of the other element will decrease as a result of a gain of electrons.

Consider the example of sodium reacting with chlorine:

Each sodium atom is electrically neutral. When a sodium atom loses its valence electron, the net charge of sodium changes from 0 to 1+.

Since it is electrically neutral (#protons = #electrons), a sodium atom is assigned an oxidation number of 0. Since the sodium ion has one less electron, it is assigned the oxidation number +1. (Note the location of the + sign; the sign always comes first in an oxidation number.)

When a chlorine atom gains an electron, the net charge of chlorine changes from 0 to 1-.

Since the two atoms in Cl2 have the same number of electrons as a neutral chlorine atom, each chlorine atom is assigned a oxidation number of 0. Since each chloride ion has one extra electron, each is assigned the oxidation number -1.

There are at least two ways to assign oxidation numbers to the elements involved in a chemical reaction. One method involves drawing Lewis diagrams for the species and using electronegativity values to assign oxidation numbers. A second method requires the use of a set of rules. You will explore this second method.

Assigning Oxidation Numbers

There is a well established set of rules for assigning oxidation numbers to elements in a chemical reaction. Mastery of these rules is essential to success. Each rule is presented below. One or more examples of its application is/are provided.

Rule 1: The Elements

The oxidation number of a pure element is zero. Consider these examples.

  • Atoms in pure gold are electrically neutral. The gold atom is assigned the oxidation number of 0.
  • Fluorine atoms in F2 equally share their bonding electron pair. Each atom is assigned the oxidation number of 0.
Rule 2: Simple Ion Charge = Oxidation Number

Assigning oxidation numbers to most monatomic ions is simply a matter of identifying or determining the ion charge. For example,

  • the ions of Group 1 metals are assigned the oxidation number +1.
  • the ions of Group 2 metals are assigned the oxidation number +2.
  • the ions of Group 17 (or Group 7A) nonmetals are assigned the oxidation number -1.
Rule 3: Hydrogen

Hydrogen is assigned the oxidation number +1 in all compounds except in metal hydrides where it is assigned the number -1.

  • In a water molecule, hydrogen is bonded to oxygen by a polar covalent bond. The electron pair is "closer" to oxygen making hydrogen partially positive (δ+) and oxygen partially negative (δ-). Each hydrogen is assigned the oxidation number -1.
  • In calcium hydride, CaH2, hydrogen is in the form of the hydride ion, H-. It is assigned the oxidation number -1. The calcium ion is assigned +2.
Rule 4: Oxygen

Oxygen is assigned the oxidation number -2, except in peroxides like hydrogen peroxide where it is assigned -1, and in OF2 where is is assigned a +2 due to the higher electronegativity of the fluorine atom. Consider these examples:

  • In calcium oxide, CaO, oxide ion has a 2- charge. Its oxidation number is -2.
  • In water, H2O, the oxygen atom strongly attracts the valence electrons of hydrogen and is assigned an oxidation number of -2.
  • In hydrogen peroxide, H2O2, each oxygen equally shares one electron pair but strongly attracts the bonding electron of the covalently bonded hydrogen atom. Each oxygen atom is assigned the oxidation number is -1. 

     
Rule 5: Covalent Compounds

In compounds that do not contain hydrogen or oxygen, the more electronegative element is assigned the oxidation number it would have in an ionic compound. Consider these examples.

  • In sulfur dichloride, SCl2, chlorine has a higher electronegativity so it is assigned -1 and sulfur is assigned +2.
  • In iodine monobromide, IBr, bromine has higher electronegativity than iodine, so it is assigned -1 and iodine is assigned +1.
Rule 6: Compounds

The sum of the oxidation numbers in a compound is zero.

  • The sum of oxidation numbers in NaCl is (+1) + (-1) = 0
  • The sum of oxidation numbers in H2O is 2(+1) + (-2) = 0
Rule 7: Polyatomic Ions

The sum of the oxidation numbers of the elements in a polyatomic ion must equal the ion charge. Consider these examples.

  • Carbonate ion, . Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the carbonate ion has a charge of 2-, the oxidation number of carbon must be +4.
  • Sulfite, . Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the sulfite ion has a charge of 2-, the oxidation number of sulfur must be +4.
  • Sulfate, . Oxygen is assigned the oxidation number -2. There are four oxygen atoms in the formula so the total negative charge is 8-. Since the sulfate ion has a charge of 2-, the oxidation number of sulfur must be +6.
  • Ammonium ion, . Hydrogen is assigned the oxidation number +1. There are four hydrogen atoms in the formula so the total positive charge is 4+. Since the ammonium ion has a charge of 1+, the oxidation number of nitrogen must be -3.
  • Nitrate ion, . Oxygen is assigned the oxidation number -2. There are three oxygen atoms in the formula so the total negative charge is 6-. Since the nitrate ion has a charge of 1-, the oxidation number of nitrogen must be +5.
Rules Summary

Rules 5 to 7 are used to assign oxidation numbers to elements not covered by Rules 1-4. By reviewing the examples above, you should see that an element can be assigned one of several oxidation numbers. Sulfur for example was assigned -2, +4, and +6 in three different examples!

Sample Exercise 1

Determine the oxidation number of the elements in potassium hydrogen sulfate, KHSO4.

Answer

Plan a Strategy

  1. The oxidation numbers of K, H and O can be determined using Rules 1, 2, and 3 respectively.
  2. Apply Rule 5 to find the oxidation number of sulfur.
  3. Communicate the answer.

Step 1: Apply the rules to assign oxidation numbers to K, H and O.

  • potassium is a 1+ ion in this ionic compound, so it is assigned +1
  • hydrogen is assigned +1
  • oxygen is assigned -2

Step 2: Calculate the oxidation number of sulfur.

Let x = oxidation number of S

Step 3: Communicate.

The oxidation numbers of K, H, S, and O are +1, +1, +6, and -2 respectively.

Oxidation Numbers and Redox Reactions

You are probably wondering...."what is the purpose of these rules? What's the point?"

The point is that not all chemical reactions are redox reactions. By assigning oxidation numbers to the elements in a balanced chemical equation, you can tell whether reacting species have been oxidized/reduced. In other words, you can identify redox reactions by applying the oxidation number rules.

Consider the reaction between silver nitrate and sodium chloride:

Now consider it with oxidation numbers assigned to the elements in this total ionic equation:

What do you notice about the oxidation numbers of the elements?

None of them changed. No electrons have been gained or lost by the elements involved in the reaction. When none of the species in a system undergo a change in oxidation number, the reaction is not a redox reaction.

Now consider the reaction between copper metal and silver nitrate:

What do you notice about the oxidation numbers in the total ionic equation for the reaction?

Copper undergoes a change in oxidation number from 0 to +2 by losing two electrons.

Copper is oxidized.

As each silver atom becomes a silver ion, it undergoes a change from +1 to 0 by gaining an electron.

Silver is reduced.

The sum of the half reactions is the redox reaction.

The terms oxidation and reduction can now be defined in another way:

  • oxidation is an increase in oxidation number.
  • reduction is a decrease in oxidation number.

Sample Exercise 2

Determine whether the combustion of ethene, C2H4, is a redox reaction. If it is, identify the oxidizing agent and the reducing agent.

Answer

Plan a Strategy

  1. Write the balanced chemical equation for the combustion reaction. Assume the reaction is complete combustion of ethene in excess oxygen to produce carbon dioxide and water vapour.
  2. Apply the rules for assigning oxidation numbers.
  3. Compare the oxidation numbers of the reactant elements and the product elements to determine whether a redox reaction has occurred.
  4. Identify the species that has been reduced - it is the oxidizing agent.
  5. Identify the species that has been oxidized - it is the reducing agent.
  6. Communicate the answer.

Step 1: Write the balanced chemical equation.

Step 2: Assign the oxidation numbers.

  • C2H4, : hydrogen is assigned +1. Since each carbon is polar covalently bonded to two hydrogens, each carbon atom is assigned -2.

  • O2: each oxygen is nonpolar covalently bonded, so each is assigned 0.

  • CO2: each oxygen is assigned -2, so carbon is assigned +4.

  • H2O: oxygen is assigned -2 and each hydrogen is assigned +1.

Step 3: Compare the oxidation numbers of each element.

The oxidation numbers of carbon and oxygen atoms have changed, so the reaction can be classified as a redox reaction.

Step 4: Identify the oxidizing agent.

The oxidation number of oxygen has changed from 0 to -2; therefore, oxygen was reduced. It is the oxidizing agent.

Step 5: Identify the reducing agent.

The oxidation number of carbon has changed from -2 to +4; therefore, carbon was oxidized. It is the reducing agent.

Step 6: Communicate.

The combustion of ethene is a redox reaction in which carbon is oxidized and oxygen is reduced.

Sample Exercise 3

Determine whether the reaction of zinc metal and lead(II) nitrate is a redox reaction, and if so, identify the oxidizing and reducing agents and their half reactions.

Answer

Plan a Strategy

  1. Write the total ionic equation for the reaction.
  2. Apply the rules for assigning oxidation numbers.
  3. Compare the oxidation numbers of the reactant elements to the product elements to determine whether a redox reaction has occurred.
  4. Identify the species that has been reduced - it is the oxidizing agent.
  5. Identify the species that has been oxidized - it is the reducing agent.
  6. Communicate the answer.

Step 1: Write the total ionic equation.

Step 2: Assign the oxidation numbers.

  • Zn: Zinc metal consists of neutral zinc atoms, so it is assigned 0.

  • Pb2+: Lead(II) ion has a 2+ charge, so it is assigned +2.

  • : is a spectator ion, so it is not necessary to assign oxidation numbers to its elements because they do not change.

  • Pb: Lead metal consists of neutral lead atoms, so it is assigned 0.

  • Zn2+: Zinc ion has a 2+ charge, so it is assigned +2.

Step 3: Compare the oxidation numbers of each element.

The oxidation numbers of zinc and lead have changed, so the reaction can be classified as a redox reaction.

Step 4: Identify the oxidizing agent.

The oxidation number of lead has changed from +2 to 0; therefore, lead was reduced. It is the oxidizing agent.

Step 5: Identify the reducing agent.

The oxidation number of zinc has changed from 0 to +2; therefore, zinc was oxidized. It is the reducing agent.

Step 6: Communicate.

The single replacement reaction involving zinc metal and aqueous lead(II) ions is a redox reaction. Zinc atoms lose electrons to reduce lead(II) ions.

Sample Exercise 4

Given these equations:

(a)  

(b)  

identify the oxidation and reduction half-reactions by balancing the number of electrons lost and gained.

Answer

Plan a Strategy

  1. Rewrite the equations showing the oxidation numbers.
  2. Write the half-reactions.

Step 1: Write the chemical equation showing the assigned oxidation numbers.

(a) 

(b) 

Step 2: Write the half reactions to show the gain/loss of electrons.

(a)    Zinc is oxidized, oxygen is reduced.

 

(b)  Chlorine is oxidized, potassium is reduced.

General Rules for Identifying Redox Reactions

  1. Double replacement reactions are never redox reactions.
  2. All combustion, single replacement, formation and decomposition reactions are redox reactions.
  3. Oxidizing agents and reducing agents are always reactants, never products.

Activity

Textbook Readings

MHR

  • page 721: Oxidation Numbers
  • pages 721-723: Oxidation Numbers from Lewis Diagrams
  • pages 724-725: Using Rules to Find Oxidation Numbers
  • pages 726-728: Applying Oxidation Numbers to Redox Reactions
Textbook Practice Items

MHR

  • page 726: items 9, 10, 11 and 12
  • page 728: items 13, 14 and 15
  • pages 728-729: section review items 1, 2, 4, 5, 7 and 8
  • pages 751-752: items 5, 7, 9, 10, and 12
More Practice Items
  1. Use the rules for assigning oxidation numbers to determine the oxidation number of each element in:
     
    2. H2CO
  2. Write half-reaction equations for each reaction and assign oxidation numbers to the species in the half-reactions.
     

  3. Write balanced half-reaction equations for each reaction.
     

Test Yourself

There is no self test for this lesson.