2.1 Oxidation-Reduction Concepts | Electrochemistry | Chemistry 12

Oxidation and Reduction

Oxidation and reduction always occur simultaneously in a redox reaction (short for reduction-oxidation reaction). Neither process can occur without the other.

Definitions in Terms of Electron Transfer

Term	Definition
Oxidation	Loss of electrons by a species. The oxidation state increases.
Reduction	Gain of electrons by a species. The oxidation state decreases.
Redox reaction	A reaction in which both oxidation and reduction occur simultaneously.

Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).

Definitions in Terms of Oxidation Number (Oxidation State)

Oxidation occurs when the oxidation number of an atom increases.
Reduction occurs when the oxidation number of an atom decreases.

Example: In the reaction $Z n + C u S O_{4} \to Z n S O_{4} + C u$

$Z n$ : oxidation state goes from $0 \to + 2$ → oxidised
$C u$ : oxidation state goes from $+ 2 \to 0$ → reduced

Oxidizing and Reducing Agents

Oxidizing agent (oxidant): The species that causes oxidation of another substance. It gains electrons and is itself reduced (its oxidation state decreases).
Reducing agent (reductant): The species that causes reduction of another substance. It loses electrons and is itself oxidised (its oxidation state increases).

In the example above:

$C u S O_{4}$ is the oxidizing agent (Cu²⁺ gains electrons)
$Z n$ is the reducing agent (Zn loses electrons)

Disproportionation

Disproportionation is a special type of redox reaction in which a single element in one oxidation state is simultaneously oxidised and reduced to give two different products.

Example: Decomposition of hydrogen peroxide:

$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$

In $H_{2} O_{2}$ , oxygen is in the $- 1$ oxidation state.

In $H_{2} O$ , oxygen is $- 2$ → reduced
In $O_{2}$ , oxygen is $0$ → oxidised

The same element (oxygen) is both oxidised and reduced, so this is a disproportionation reaction.

Another example: Reaction of chlorine with cold dilute NaOH:

$C l_{2} + 2 N a O H \to N a C l + N a O C l + H_{2} O$

$C l_{2}$ starts at oxidation state $0$
In $N a C l$ , Cl is $- 1$ → reduced
In $N a O C l$ , Cl is $+ 1$ → oxidised

Rules for Assigning Oxidation Numbers

The oxidation state of any element in its free (elemental) state is zero (e.g., $H_{2}$ , $O_{2}$ , $N a$ , $S_{8}$ ).
The oxidation state of a monoatomic ion equals its charge (e.g., $N a^{+}$ is $+ 1$ , $C l^{-}$ is $- 1$ ).
In compounds, oxygen is usually $- 2$ (except in peroxides where it is $- 1$ , and in $O F_{2}$ where it is $+ 2$ ).
In compounds, hydrogen is usually $+ 1$ (except in metal hydrides where it is $- 1$ ).
The sum of oxidation states in a neutral compound equals zero.
The sum of oxidation states in a polyatomic ion equals the charge of the ion.

Example — finding oxidation state of Mn in $K M n O_{4}$ :

$K = + 1, O = - 2 \times 4 = - 8$ $+ 1 + x + (- 8) = 0 ⟹ x = + 7$

So Mn is in the $+ 7$ oxidation state in $K M n O_{4}$ .

Oxidation and Reduction

Oxidation and reduction always occur simultaneously in a redox reaction (short for reduction-oxidation reaction). Neither process can occur without the other.

Definitions in Terms of Electron Transfer

Term	Definition
Oxidation	Loss of electrons by a species. The oxidation state increases.
Reduction	Gain of electrons by a species. The oxidation state decreases.
Redox reaction	A reaction in which both oxidation and reduction occur simultaneously.

Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).

Definitions in Terms of Oxidation Number (Oxidation State)

Oxidation occurs when the oxidation number of an atom increases.
Reduction occurs when the oxidation number of an atom decreases.

Example: In the reaction $Z n + C u S O_{4} \to Z n S O_{4} + C u$

$Z n$ : oxidation state goes from $0 \to + 2$ → oxidised
$C u$ : oxidation state goes from $+ 2 \to 0$ → reduced

Oxidizing and Reducing Agents

Oxidizing agent (oxidant): The species that causes oxidation of another substance. It gains electrons and is itself reduced (its oxidation state decreases).
Reducing agent (reductant): The species that causes reduction of another substance. It loses electrons and is itself oxidised (its oxidation state increases).

In the example above:

$C u S O_{4}$ is the oxidizing agent (Cu²⁺ gains electrons)
$Z n$ is the reducing agent (Zn loses electrons)

Disproportionation

Disproportionation is a special type of redox reaction in which a single element in one oxidation state is simultaneously oxidised and reduced to give two different products.

Example: Decomposition of hydrogen peroxide:

$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$

In $H_{2} O_{2}$ , oxygen is in the $- 1$ oxidation state.

In $H_{2} O$ , oxygen is $- 2$ → reduced
In $O_{2}$ , oxygen is $0$ → oxidised

The same element (oxygen) is both oxidised and reduced, so this is a disproportionation reaction.

Another example: Reaction of chlorine with cold dilute NaOH:

$C l_{2} + 2 N a O H \to N a C l + N a O C l + H_{2} O$

$C l_{2}$ starts at oxidation state $0$
In $N a C l$ , Cl is $- 1$ → reduced
In $N a O C l$ , Cl is $+ 1$ → oxidised

Rules for Assigning Oxidation Numbers

The oxidation state of any element in its free (elemental) state is zero (e.g., $H_{2}$ , $O_{2}$ , $N a$ , $S_{8}$ ).
The oxidation state of a monoatomic ion equals its charge (e.g., $N a^{+}$ is $+ 1$ , $C l^{-}$ is $- 1$ ).
In compounds, oxygen is usually $- 2$ (except in peroxides where it is $- 1$ , and in $O F_{2}$ where it is $+ 2$ ).
In compounds, hydrogen is usually $+ 1$ (except in metal hydrides where it is $- 1$ ).
The sum of oxidation states in a neutral compound equals zero.
The sum of oxidation states in a polyatomic ion equals the charge of the ion.

Example — finding oxidation state of Mn in $K M n O_{4}$ :

$K = + 1, O = - 2 \times 4 = - 8$ $+ 1 + x + (- 8) = 0 ⟹ x = + 7$

So Mn is in the $+ 7$ oxidation state in $K M n O_{4}$ .