Redox reaction

A reaction in which both oxidation and reduction take place simultaneously is called redox (reduction + oxidation) reaction.

E. g.

(i) H_2 + S \to H_2S

In this reaction Ox. No. of hydrogen increases from zero to +1 (oxidation) and that of sulphur decreases from zero to -2 (reduction). In this reaction hydrogen is reducing agent and S is oxidising agent.

(ii) 2FeCl_2 + Cl_2 \to 2FeCl_3

In this reaction Ox. No. of Fe increases from +2 to +3 (oxidation) and that of chlorine decreases from zero to (reduction). Hence FeCl_2 is reducing agent and Cl_2 is oxidising agent.


H_2O_2 + H_2O_2 \to 2H_2O + O_2 \uparrow


In this reaction Ox. No. of hydrogen remains the same, i. e., +1, but the Ox. No. of oxygen in H_2O_2 , H_2O \text{and} O_2 are -1, -2 and 0 respectively. Hence, in this reaction oxygen is undergoing oxidation as well as reduction. Therefore H_2O_2 acts both as oxidising and reducing agent. Such a molecule is said to auto-oxidise and the process (reaction) as auto-oxidation. Decomposition of Pb(NO_3)_2 is also an example of auto-oxidation.


2Pb(NO_3)_2 \to 2PbO + 4NO_2 \uparrow + O_2 \uparrow


In this reaction oxygen is oxidised while nitrogen is reduced. Therefore NO^-_3 is acting as both an oxidising and reducing agent.

Such reactions are often called disproportion reactions.

Example 1:

Redox reaction

Example 2:

Redox Reaction


all type of cannizzaro reactions are e. g. , of disproportion.


Redox Reaction


Intramolecular Redox Reaction: Here in any compound one element is oxidised while the other element is reduced.


Redox Reaction


Don’t forget these

Loss of electrons = Oxidation

Increase in Ox. No. = Oxidation

Gain of electrons = Reduction

Decrease in Ox. No. = Reduction

Reduction + Oxidation = Redox

Oxidants \to Oxidize other

\toReduce themselves

\to Gain electrons

Reductants \to Reduce other

\to Oxidize themselves

\to Loss electrons

\text{Equivalent wt. of redox reagent} = \dfrac{\text{Molecular weight}}{\text{Change in Ox. No.}}


Calculation of Oxidation Number: With the help of the rules mentioned above the oxidation number of any one element in a compound can be calculated. For example;

(i) Oxidation number of chromium in K_2Cr_2O_7:  Let the oxidation number of Cr be x in K_2Cr_2O_7 .Now we know that oxidation numbers of K and O are +1 and -2 respectively, thus

Or      2(+1) + 2 (x) +7 (-2) = 0

Or       2 + 2x -14 = 0

Or        x = 6

(ii) Oxidation number of Ta in K_2TaF_7: Let the oxidation number of Ta be x in K_2 TaF_7. Now we know that oxidation numbers of K and F are +1 and -1 respectively, thus

Or         (2)(+1) + x + 7 (-1) = 0

Or          2 + x – 7 = 0

Or          x = 5

(iii) Oxidation number of carbon in its various compounds may be determined as follows:

\underset{\underset{\text{or}x = -4} x +4 = 0}{CH_4} \hspace{20mm} \underset{\underset{\text{or} x =-2} 2 + x- 2 = 0}{CH_3Cl} \hspace{20mm} \underset{\underset{\text{or} x = 0}x + 2- 2 = 0}{CH_2Cl_2} \\ \underset{\underset{\text{or}x = 2} x + 1- 3 = 0}{CHCl_3} \hspace{20mm} \underset{\underset{\text{or} x = 4} x- 4 = 0}{CCl_4}


(iv) Oxidation number of nitrogen in its various compounds may be determined as follows:

\underset{\underset{x = 2} x- 2 = 0}{NO} \hspace{20mm} \underset{\underset{\text{or} x = 3} 2x- 6 = 0}{N_2O_3} \hspace{20mm} \underset{\underset{\text{or} x = 4} 2x- 8 = 0}{N_2O_4} \hspace{20mm} \underset{\underset{\text{or}x = 5} 2x- 10 = 0}{N_2O_5} \\ \underset{\underset{x = 5} x- 6 = -1}{NO^-_3} \hspace{20mm} \underset{\underset{\text{or}x = -2} 2x + 5 = +1}{N_2H^+_5} \hspace{20mm} \underset{\underset{\text{or} x = -3}x + 3 = 0}{NH_3}

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