‌

Programmes/ undefined/ Content/ Chemistry

Navigation

Physical Chemistry

Language of Chemistry Stoichiometry Atomic Structure Oxidation and Reduction Periodic Table Acid and Base Electronic Theory of Valency Equivalent Weight and Atomic Weight

Physical Chemistry-II

Non-Metals

Inorganic Chemistry

Oxidation and Reduction

Oxidation:

Oxidation is a chemical process which can be explained by following four point of views:

In terms of oxygen transfer.
In terms of electron transfer.
In terms of hydrogen transfer.
In terms of oxidation number.

Oxidation in Terms of Oxygen Transfer –

-Oxidation is gain of oxygen or any electronegative radical.

-Example: $2 M g (s) + O_{2} (g) ⟶ 2 M g O (s)$

Oxidation in Terms of Electron Transfer –

-Oxidation is loss of electrons. (i.e., increase in $+ v e$ charge, decrease in $- v e$ charge).

-In the above example magnesium is losing two electrons and getting oxidized to form magnesium oxide.

$M g ⟶ M g^{2 +} + 2 e^{-}$

Oxidation in Terms of Hydrogen Transfer –

-Oxidation can be defined as loss of hydrogen.

Oxidation in Terms of Oxidation Number –

-Oxidation is an increase in the oxidation state or oxidation number of an atom in a reaction.

-Oxidation number defines the degree of oxidation of an atom in a chemical compound.

-In the above example, the oxidation state of sodium is increasing from $0 t o + 1$ . Thus, oxidation is taking place and sodium is getting oxidized.

Reduction:

Reduction is a chemical process which can be explained by following four point of views :

In terms of oxygen transfer.
In terms of electron transfer.
In terms of hydrogen transfer.
In terms of oxidation number.

Reduction in Terms of Oxygen Transfer –

-Reduction is loss of oxygen.

-Example: $2 M g (s) + O_{2} (g) ⟶ 2 M g O (s)$

-In the above example, each atom of oxygen gains two electrons and forms two $O^{- 2}$ anions. Thus, reduction takes place and oxygen gets reduced.

Reduction in Terms of Electron Transfer –

-Reduction is gain of electrons. (i.e. increase in $- v e$ charge, decrease in $+ v e$ charge)

$F e^{3 +} + e^{-} ⟶ F e^{2 +}$

Reduction in Terms of Hydrogen Transfer –

-Reduction can be defined as gain of hydrogen or any electropositive radical or removal of oxygen or any electronegative radical.

Reduction in Terms of Oxidation Number –

-Reduction is decrease in the oxidation state or oxidation number of an atom in a reaction.

-In the above example oxidation state of chlorine is decreasing from $0 t o - 1$ . Thus, reduction is taking place and chlorine is getting reduced.

Oxidizing Agent:

The substance which oxidizes other substance & itself undergoes reduction.
Accept the electron from other.
Its positive charge decreases and negative charge increases.
Examples:

a)Oxygen Supplier:

- $K M n O_{4}, K_{2} C r_{2} O_{7} / H^{+} C r O_{3}, K Cl O_{3}, H N O_{3}, O_{3}, H_{2} O_{2}, C h l or in e w a t er$

b)Electronegative elements:

- $X_{2}, O_{2}, e t c$ .

c) Higher oxidation state of metals:

- $S n C l_{4}, A u C l_{3}, F e C l_{3}, P b C l_{4}, H g O, Z n O, M n O_{2}, S n O_{2}, P b O_{2}, e t c$ .

d)Non-metal oxides of higher oxidation states:

- $S O_{3}, C O_{2}, P_{2} O_{5}, e t c$

Reducing Agent:

The substance which reduce other substance and itself undergoes oxidation.
Donate electron to other.
Its positive charge increases and negative charge decreases.
Examples:

a) Hydrogen Supplier:

- $N a H, M g H_{2}, B e H_{2}, L i A l H_{4}$ , $e t c$

b) Electropositive elements:

- $M e t a l s$

c) Lower oxidation states of metals:

$S n C l_{2}, A u Cl, F e C l_{2} C u_{2} C l_{2}, e t c$

d)Non-metals:

$C, S, P, H_{2}, e t c$

e)Hydracids:

$H_{2} S, H B r, H I H Cl e t c$

f) Others:

- $CO, NO, Hy d roc a r b o n, A l co h o l, A l d e h y d es, e t c$

Oxidation Number (ON):

Rule for determining oxidation number:

The oxidation number of an element in Free State or Uncombined State is zero. For e.g. $C l_{2}, N a, M g, C, N_{2}, O_{3}, P_{4},$ etc is zero
The oxidation number of ions or radicals is equal to charge carried. For e.g. $F e^{2 +}, F e^{3 +}, C a^{2 +}, K^{+}$ have a oxidationnumber $+ 2, + 3, + 2, + 1$ respectively.
The oxidation number of alkali metals $(e . g . L i, N a, K, R b, C s)$ in combined state is always $+ 1$ .
The oxidation number of alkaline earth metals $(e . g . B e, M g, C a, S, B a, R a)$ in combined state is always $+ 2$ .
The algebraic sum of the oxidation number of all the atoms in a neutral molecule is zero.
The oxidation number of hydrogen in all its compound is $+ 1$ except in metal hydrides $(e . g . L i H, C a H_{2}, M g H_{2})$ wherethe oxidation number is $- 1$ .
The oxidation number of oxygen in combined state is $- 2$ except

i) $- 1$ in peroxides $(e . g . H_{2} O_{2}, N a_{2} O_{2}, B a O_{2}, e t c)$

ii) $- 1/2$ in super oxide $(K O_{2})$

iii) $- 1/3$ in ozonides $(K O_{3})$

iv) $+ 2$ in $O F_{2}$

Redox Reaction:

The term ‘redox’ is a short form of reduction-oxidation.
Such type of a chemical reaction in which both oxidation and reduction occurs simultaneously is called redox reaction.
Example:

i) $Z n + C u S O_{4} ⟶ Z n S O_{4} + C u$

The oxidation half-reaction can be written as: $Z n ⟶ Z n^{+ 2} + + 2 e^{-}$

The reduction half-reaction can be written as: $C u^{2 +} + 2 e^{-} ⟶ C u$

Types:

Decomposition Reaction:

$A B ⟶ A + B$

$2 N a H ⟶ 2 N a + H_{2}$

$2 H_{2} O ⟶ 2 H_{2} + O_{2}$

Combination Reaction:

$A + B ⟶ A B$

$H_{2} + C l_{2} ⟶ 2 H Cl$

$4 F e + 3 O_{2} ⟶ 2 F e_{2} O_{3}$

Displacement Reaction:

$X + Y Z ⟶ XZ + Y$

$C u S O_{4} + Z n ⟶ C u + Z n S O_{4}$

Disproportionation Reactions:

-Reactions in which a single reactant is oxidized and reduced.

$P_{4} + 3 N a O H + 3 H_{2} O ⟶ 3 N a H_{2} P O_{2} + P H_{3}$

Balancing Redox Reaction:

-There are two ways of balancing the redox reaction. One method is by using the change in oxidation number of oxidizing agent and the reducing agent and the other method is based on dividing the redox reaction into two half reactions-one of reduction and other oxidation.

Example 1: Write the balanced redox equation when ferrous sulphate is treated with acidified $(H_{2} S O_{4})$ and $K M n O_{4}$

Solution

$K M n O_{4} + H_{2} S O_{4} + F e S O_{4} ⟶ K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O$

Here, $M n^{+ 7}$ in reactant is reduced to $M n^{+ 2}$ in product.

Some Important Points:

Order of oxidizing character/strength

$F > Cl > B r > I$

$O_{2}^{-} (S u p ero x i d e) > O_{2}^{2 -} (P ero x i d e) > O_{2} (M o n o x i d e)$

$H ClO > H Cl O_{2} > H N O_{2} > H_{2} N_{2} O_{4}$

Only oxidizing agent are: $H_{2} S O_{4}, N_{2} O_{5}, P_{2} O_{5}, S O_{3}, H N O_{3}, C l_{2} O_{7}, H Cl O_{4}, H_{3} P O_{4}, H a l o g e n, K_{2} C r_{2} O_{7}, e t c$ .
Only reducing agent are: $H_{2} S, H I, C l_{2} O, N_{2} O, C a r b o n, I A, II A e l e m e n t, e t c$ .
Both oxidizing and reducing agents are $NO, N O_{2}, N_{2} O_{3}, H_{2} O, O_{3}, N a_{2} S O_{3}, S O_{2}, H_{2} S O_{3}, P_{2} O_{3}, e t c$ .
Reducing property increases going down in group i.e., oxidizing property decrease
Strongest Oxidizing Agent = Fluorine
Strongest Reducing Agent = Lithium (in solution)
Strongest Reducing Agent = Caesium (in absence of water)
Among $H N O_{2}, H_{2} S, H_{2} S O_{3} an d S n C l_{2}$ , the strongest reducing agent is $H_{2} S$ because Sulphur in $H_{2} S$ is in lowest oxidation state.