CBSE Class 11 Chemistry Redox Reactions Notes Set 06

These revision notes on Redox Reactions for CBSE Class 11 Chemistry cover the classical and electronic ideas of oxidation and reduction, oxidising and reducing agents, the rules for finding oxidation numbers, fractional oxidation numbers and the two methods of balancing redox equations.

The reactions that involve oxidation and reduction as their two half reactions are called redox reactions.

1. Classical Idea of Redox Reactions

According to classical concepts, oxidation and reduction are defined as the processes that involve:

Oxidation
(i) Addition of oxygen
(ii) Addition of electronegative element
(iii) Removal of hydrogen
(iv) Removal of electropositive element

Reduction
(i) Removal of oxygen
(ii) Removal of electronegative element
(iii) Addition of hydrogen
(iv) Addition of electropositive element

2. Oxidising and Reducing Agents

In a redox reaction, the substance which oxidises the other species or itself undergoes reduction is called the oxidising agent. The substance that reduces the other species and itself undergoes oxidation is called the reducing agent.

3. Redox Reactions in Terms of Electron Transfer

Loss of electrons or an increase in oxidation state is called oxidation. Gain of electrons or a decrease in oxidation state is called reduction. Because of the simultaneous loss and gain of electrons in oxidation-reduction processes, redox reactions are also called electron transfer reactions.

4. Oxidation Number

It is defined as “the charge that an atom of the element possesses in its ion or appears to have when present in the combined state with other atoms.”

5. Rules for Calculating Oxidation Number

These rules are given below:

Rule 1: The oxidation number of an atom in its free or elementary state or in any of its allotropes is zero, e.g. the oxidation state of H in H₂, S in S₈ and P in P₄ is zero.

Rule 2: In case of ions having only one kind of atoms, the oxidation number of each atom is equal to the charge present on the ion, e.g. in Na⁺, Mg²⁺, Fe³⁺, Cl⁻ and O²⁻, the oxidation state is respectively +1, +2, +3, −1 and −2.

Rule 3: The oxidation state of alkali metals in all their compounds is always +1. Similarly, for alkaline earth metals it is always +2. For aluminium, the oxidation state is always +3.

Rule 4: The oxidation state of oxygen in most of its compounds is −2, with the exception of peroxides and superoxides, in which the oxidation state of oxygen is −1 and −1/2 respectively.

Rule 5: The oxidation state of hydrogen is generally +1, with the exception of metallic hydrides like NaH, CaH₂, etc. In these hydrides, the oxidation state of hydrogen is −1.

Rule 6: The oxidation state of fluorine in all of its compounds is always −1. Other halogens (chlorine, bromine and iodine) also exhibit −1 oxidation state, but not always. In oxoacids and oxoanions, halogens (except fluorine) exhibit positive oxidation states.

Rule 7: The algebraic sum of the oxidation numbers of all the atoms present in a compound must be equal to zero.

Rule 8: In polyatomic ions, the algebraic sum of the oxidation numbers of all the atoms present in the ion must be equal to the charge on the ion, e.g. for the carbonate ion (CO₃²⁻) it is equal to −2.

6. Paradox of Fractional Oxidation Number

1. In C₃O₂ (carbon suboxide):

Two carbon atoms are present in +2 oxidation state each, whereas the third one is in zero oxidation state.
The average oxidation state of the 3 C-atoms in C₃O₂ = \( \frac{2 + 2 + 0}{3} = \frac{4}{3} \)

2. The average oxidation state of Br in Br₃O₈ is 16/3, while the oxidation states of the three Br atoms are +6, +4 and +6, as shown below:

3. The average oxidation state of the four S-atoms in S₄O₆²⁻ is 2.5, while the actual oxidation states of the four S-atoms are +5, 0, 0 and +5, as shown below:

7. Balancing of Redox Reactions

(i) Ion electron method

The method involves the following steps:
(a) Write the redox reaction in ionic form.
(b) Split the redox reaction into oxidation half and reduction half reactions. Balance the atoms of each half reaction by using simple multiples.
For balancing H and O, add H⁺ ions and H₂O to the appropriate sides (acidic medium); similarly add OH⁻ and H₂O to the appropriate sides (basic medium).
Balance the charge on both sides and multiply one or both half reactions by a suitable number to equalise the number of electrons in both equations. Add the two balanced half reactions and cancel the common terms.

(ii) Oxidation number method

The method involves the following steps:
(a) Assign oxidation numbers to the atoms in the equation and write separate equations for the atoms undergoing oxidation and reduction.
(b) Find the change in oxidation number in each equation and make the change equal in both equations by multiplying with suitable integers.
After adding both the equations, complete the balancing (by balancing H and O).