All About Examples Of Oxidation State Calculation

There is the transfer of electrons during electrochemical reactions. Oxidation and reduction are two important terms used in chemistry to define the loss or gain of electrons. The loss of electrons from a molecule is known as oxidation, whereas the electron’s gain is known as reduction. The oxidation number is the total loss, gain or sharing of electrons in a molecule during chemical interactions. In short, the oxidation number determines the total number of electrons lost or gained by an electron for forming a chemical bond with another atom. The calculation of the average oxidation number of S in H2S4O6 and other examples will be discussed in the topic. 

Rules for Calculating or Assigning Oxidation Numbers 

There are a few rules which are important for calculating the oxidation number. The average oxidation number of S in H2SO4is easily calculated by understanding and implementing the rules. 

• The sum of the oxidation number for all the atoms in an element in its free or uncombined state is zero. For example, in the case of KMnO4 it is zero as the oxidation number of K is +1, Mn is +7, and oxygen is -2. 
• One of the most important conventions in writing formulae is that the cation is written first, followed by anion. For example, KCl, where K is a cation and Cl is an anion. 
• The oxidation number of an atom in its elementary form is always zero. Hence, for example, when atoms like P, N, O, Se etc., are present in element form like P4, N2, O2, Se8 etc., their oxidation number is zero. 
• The oxidation number for monatomic ions is equal to the charge of the ion. This is shown by examples like the oxidation number of N3- is -3.
• The oxidation number of alkali metals within a molecule or compound is +1. For example, Na in the NaCl is an alkali metal with an oxidation number of +1. The oxidation number of alkaline earth metals is +2, and shown by example, Ca in CaCl2 has an oxidation number of +2. 
• In most of the molecules, the oxidation number of oxygen is -2. However, there are two exceptions to which different rules apply: 

1. In the case of peroxides, the oxygen atom has an oxidation number of -1.
2. For superoxide, the oxygen atom has an oxidation number of – (1/2)
3. In ozonides, the oxidation number of each oxygen atom is – (1/3) 

• The oxidation number of H in any compound is always +1 except for metal hydrides. For example, in the case of NaH, the oxidation number of H is -1.
• In the case of an ionic compound, the sum of the oxidation state of all atoms is equal to the total charge of the compound.
• The maximum oxidation number of any element is equal to its group number, but not in the case of oxygen and fluorine.
• In some compounds, all atoms do not have the same oxidation number. Hence, an average value is calculated and obtained for the elements in such compounds. For example, the average oxidation number of S in H2S4O6 is 6. 
• In the case of the organic compound, the oxidation number of carbon varies from -4 to +4. 

The oxidation number of elements in a group

In the periodic table elements from group IA to IV A, the common oxidation number of an element equals the group number. However, the common oxidation number of elements from group V A to VIII-A is shown by the simple formula (Group number – 8)

Here are some examples to help understand better:

For IA group elements, the oxidation number is +1

For III A group elements, the oxidation number is +3

For V A group elements, the oxidation number is -3 

For VII A group elements, the oxidation number is -1

Examples of Oxidation state Calculation 

The calculation of the oxidation state for various compounds is easily calculated by understanding the following examples:

Example 1) Calculation of average oxidation number of S in H2SO4

The following steps calculate the average oxidation number of S in H2SO4:

Let’s assume the oxidation number for S is X

Oxidation number of H = +1

Oxidation number of O = – 2

Hence placing numbers in the H2S04:

2 (+1) + (X) + 4 X (-2) = 0

2 + X – 8 = 0

X = 6

Hence, the average oxidation number of S in H2SO4 is 6. 

Example 2) Oxidation number of S in H2SO5.

Let’s assume the oxidation number of sulphur is X. 

Oxidation number of hydrogen = +1

Oxidation number of oxygen = -2 

Hence, after implementing the values

2 (+1) + (X) + 5 X (-2) = 0

X = 10 – 2 = 8

Hence, the oxidation number of S in H2SO5 is 8

Example 3) Oxidation number of P in H3PO4

Let’s assume the oxidation number of phosphorus is X

Oxidation number of hydrogen = +1 

Oxidation number of oxygen = – 2

Hence, after implementing the values

3 X (+1) + (X) + 4 X (-2)

X = 8 -3 = 5

Hence, the oxidation number of P in H3PO4 is 5

Conclusion

Oxidation is the loss of electrons by an atom or ion during a reaction. The oxidation number is the total electrons lost, gained, or acquired by an atom during an electrochemical reaction. Understanding the concept of oxidation and the calculation of the oxidation number is crucial in chemistry. In summary, the topic gives a comprehensive overview of the rules for calculating the oxidation number of elements. The oxidation number can either be positive, negative or zero, depending on the environment.