A-level Chemistry/WJEC/Module 3/Redox Reactions

Oxidation and Reduction

Redox reactions were studied extensively at AS-level. The key points are summarised here:

The gain and loss of electrons can be shown by means of full equations;

Oxidation is the loss of electrons. When a species loses electrons it is said to be oxidised.

E.g. Fe²⁺ → Fe³⁺ + e^-

Reduction is the gain of electrons. When a species gains electrons it is said to be reduced.

E.g. MnO₄^- + 8 H⁺ + 5 e^- → Mn²⁺ + 4 H₂O

Electrons can never be created or destroyed; they can only be transferred from one species to another. Reactions which involve the transfer of electrons are known as redox reactions.

Overall redox equations can be created by combining the half-equations for the oxidation process and reduction processes, after multiplying all the coefficients of the species in one of the half-equations by a factor which ensures that the number of electrons gained is equal to the number of electrons lost.

E.g. Fe²⁺ → Fe³⁺ + e^- oxidation MnO₄^- + 8 H⁺ + 5 e^- → Mn²⁺ + 4 H₂O reduction Multiplying all coefficients in the oxidation reaction by 5: 5 Fe²⁺ → 5 Fe³⁺ + 5 e^- means that 5 electrons are gained and five are lost overall equation: MnO₄^- + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺

A species which can accept electrons from another species is an oxidising agent. Oxidising agents are reduced during redox reactions.

E.g. MnO₄^- is the oxidising agent in the above reaction.

A species which can donate electrons to another species is a reducing agent. Reducing agents are oxidised during redox reactions.

E.g. Fe²⁺ is the reducing agent in the above reaction.

The oxidation number of an atom is the charge that would exist on the atom if the bonding were completely ionic.

In simple ions, the oxidation number of the atom is the charge on the ion: Na⁺, K⁺, H⁺ all have an oxidation number of +1. O^2-, S^2- all have an oxidation number of -2.

In molecules or compounds, the sum of the oxidation numbers on the atoms is zero: E.g. SO₃; oxidation number of S = +6, oxidation number of O = -2. +6 + 3(-2) = 0

In complex ions, the sum of the oxidation numbers on the atoms is equal to the overall charge on the ion. E.g. MnO₄^-; oxidation number of Mn = +7, oxidation number of O = -2. +7 + 4(-2) = -1 E.g. Cr₂O₇^2-; oxidation number of Cr = +6, oxidation number of O = -2. 2(+6) + 7(-2) = -2 E.g. VO₂⁺; oxidation number of V = +5, oxidation number of O = -2. +5 + 2(-2) = +1

In elements in their standard states, the oxidation number of each atom is zero: In Cl₂, S, Na and O₂ all atoms have an oxidation number of zero.

Many elements, including most d-block elements, can have different oxidation numbers. In complex ions or molecules, the oxidation number of these elements can be calculated by assuming that the oxidation number of the other atom in the species is fixed.

Oxidation numbers are useful for writing half-equations:

The number of electrons gained or lost can be deduced from the formula: No of electrons gained/lost = change in oxidation number x number of atoms changing oxidation number

The oxygen atoms are balanced by placing an appropriate number of water molecules on one side.

The hydrogen atoms are balanced by placing an appropriate number of H⁺ ions on one side.

Disproportionation is the simultaneous oxidation and reduction of the same species.

There are many d-block species which readily undergo both oxidation and reduction, and which can therefore behave as both oxidising agents and reducing agents. Cu⁺, Mn³⁺ and MnO₄^2- are all examples:

E.g. Cu⁺ → Cu²⁺ + e^- oxidation Cu⁺ + e^- → Cu reduction

E.g. Mn³⁺ + 2 H₂O → MnO₂ + 4 H⁺ + e^- oxidation Mn³⁺ + e^- → Mn²⁺ reduction

E.g. MnO₄^2- → MnO₄^- + e^- oxidation MnO₄^2- + 2 H⁺ + 2 e^- → MnO₂ + 2 H₂O reduction

Species such as these are capable of undergoing oxidation and reduction simultaneously. Disproportionation reactions are special examples of redox reactions.