# 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:

Example 2:

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

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

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

$\to$Reduce 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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