Chapter 5: Oxidation and Reduction

Introduction to Redox Chemistry

A large class of reactions of inorganic compounds can be regarded as occurring by the transfer of electrons from one species to another. Electron gain is called reduction and electron loss is called oxidation; the joint process is called a redox reaction.

Key Definitions:

Reducing agent (reductant): The species that supplies electrons

Oxidizing agent (oxidant): The species that removes electrons

Many redox reactions release energy and are exploited in combustion and battery technologies. Examples include:

In gases: 2 NO(g) + O₂(g) → 2 NO₂(g)

In solution: Fe³⁺(aq) + Cr²⁺(aq) → Fe²⁺(aq) + Cr³⁺(aq)

In solids: LiCoO₂(s) + 6 C(s) → LiC₆(s) + CoO₂(s)

A brief illustration: The simplest redox reactions involve the formation of cations and anions from elements. Examples include the oxidation of lithium to Li⁺ ions when it burns in air to form Li₂O and the reduction of chlorine to Cl⁻ when it reacts with calcium to form CaCl₂. Lead is commonly found as Pb(II) in PbO and as Pb(IV) in PbO₂.

The ability to exhibit multiple oxidation numbers is seen at its fullest in d-metal compounds, particularly in Groups 5, 6, 7, and 8; osmium, for instance, forms compounds that span oxidation numbers between −2, as in [Os(CO)₄]²⁻, and +8, as in OsO₄.

Reduction Potentials

Because electrons are transferred between species in redox reactions, electrochemical methods are of major importance. The tendency of an electron to migrate from one species to another is expressed in terms of the differences between their standard potentials.

5.1 Redox Half-Reactions

Key Point: A redox reaction can be expressed as the difference of two reduction half-reactions.

It is convenient to think of a redox reaction as the combination of two conceptual half-reactions:

Reduction half-reaction: 2 H⁺(aq) + 2 e⁻ → H₂(g)

Oxidation half-reaction: Zn(s) → Zn²⁺(aq) + 2 e⁻

The oxidized and reduced species in a half-reaction constitute a redox couple, written with the oxidized species before the reduced, as in H⁺/H₂ and Zn²⁺/Zn.

Example 5.1: Combining Half-Reactions

Problem: Write a balanced equation for the oxidation of Fe²⁺ by permanganate ions (MnO₄⁻) in acid solution.

Systematic approach:

Write the unbalanced half-reactions as reductions
Balance elements other than H and O
Balance O atoms by adding H₂O
Balance H atoms by adding H⁺ (acidic) or OH⁻ (basic)
Balance charge by adding e⁻
Multiply to match electrons
Combine half-reactions

Result:

MnO₄⁻(aq) + 8 H⁺(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 5 Fe³⁺(aq) + 4 H₂O(l)

5.2 Standard Potentials and Spontaneity

Key Point: A reaction is thermodynamically favorable (spontaneous) in the sense K > 1, if E° > 0, where E° is the difference of the standard potentials.

The thermodynamic criterion of spontaneity is that the reaction Gibbs energy change, Δ_rG, is negative:

Δ_rG° = −RT ln K

By convention, the specially chosen reference half-reaction is the reduction of hydrogen ions:

H⁺(aq) + e⁻ → ½ H₂(g) Δ_rG° = 0 at all temperatures

The standard potential E° is related to the standard Gibbs energy by:

Δ_rG° = −νFE°

where ν is the stoichiometric coefficient of electrons and F is Faraday's constant (F = 96.48 kC mol⁻¹).

Galvanic Cell

Interactive Galvanic Cell Diagram

Figure 5.1: A schematic diagram of a galvanic cell. E°_cell is the potential difference when all substances are in their standard states.

Example 5.2: Calculating a Standard Cell Potential

Given:

Cu²⁺(aq) + 2 e⁻ → Cu(s) E°(Cu²⁺,Cu) = +0.34 V

Zn²⁺(aq) + 2 e⁻ → Zn(s) E°(Zn²⁺,Zn) = −0.76 V

Answer: Cu²⁺ is the more oxidizing species (higher potential) and will be reduced by Zn:

E°_cell = E°(Cu²⁺,Cu) − E°(Zn²⁺,Zn) = +0.34 V − (−0.76 V) = +1.10 V

5.3 Trends in Standard Potentials

Key Point: The atomization and ionization of a metal and the hydration enthalpy of its ions all contribute to the value of the standard potential.

A thermodynamic cycle helps identify factors contributing to the standard potential of the couple M⁺/M:

Thermodynamic Cycle for M⁺/M

M(s) + H⁺(aq)

→

M⁺(aq) + ½H₂(g)

Δ_subH°(M) ↑

I(M) ↑

Δ_hydH°(M⁺) ↓

Property	Li	Na	Cs	Ag
Δ_subH° / kJ mol⁻¹	+161	+109	+79	+284
I / kJ mol⁻¹	520	495	376	735
Δ_hydH° / kJ mol⁻¹	−520	−406	−264	−468
E° / V	−3.04	−2.71	−3.03	+0.80

5.4 The Electrochemical Series

Key Point: The oxidized member of a couple is a strong oxidizing agent if E° is positive and large; the reduced member is a strong reducing agent if E° is negative and large.

A negative standard potential (E° < 0) signifies a couple in which the reduced species is a reducing agent for H⁺ ions under standard conditions.

F₂(g) + 2 e⁻ → 2 F⁻(aq) +2.87 V

Ce⁴⁺(aq) + e⁻ → Ce³⁺(aq) +1.76 V

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O +1.51 V

Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) +1.36 V

O₂(g) + 4 H⁺(aq) + 4 e⁻ → 2 H₂O(l) +1.23 V

Fe³⁺(aq) + e⁻ → Fe²⁺(aq) +0.77 V

2 H⁺(aq) + 2 e⁻ → H₂(g) 0.00 V

Zn²⁺(aq) + 2 e⁻ → Zn(s) −0.76 V

Al³⁺(aq) + 3 e⁻ → Al(s) −1.68 V

Ca²⁺(aq) + 2 e⁻ → Ca(s) −2.87 V

Li⁺(aq) + e⁻ → Li(s) −3.04 V

Important: The reduced member of a couple has a thermodynamic tendency to reduce the oxidized member of any couple that lies above it in the series.

5.5 The Nernst Equation

Key Point: The cell potential at an arbitrary composition of the reaction mixture is given by the Nernst equation.

E_cell = E°_cell − (RT/νF) ln Q

At equilibrium E_cell = 0 and Q = K, giving the important relation:

ln K = νFE°_cell / RT

E° / V	K (at 298 K, ν = 1)
+2	10³⁴
+1	10¹⁷
0	1
−1	10⁻¹⁷
−2	10⁻³⁴

Box 5.1: Fuel Cells

A fuel cell converts a chemical fuel directly into electrical power, using O₂ or air as the oxidant. They offer advantages over batteries and combustion engines:

Operates continuously as long as fuel is supplied
More efficient than combustion devices
Less polluting (no nitrogen oxides at low temperatures)

Type	Electrolyte	Temp (°C)	Efficiency (%)
PEMFC	H⁺-conducting polymer	80-100	35-40
AFC	Aqueous alkali	80-250	50-60
SOFC	Solid oxide	800-1000	50-55
DMFC	H⁺-conducting polymer	0-40	20-40

Redox Stability

When assessing the thermodynamic stability of a species in solution, we must consider all possible reactants: the solvent, other solutes, the species itself, and dissolved oxygen.

5.6 The Influence of pH

Key Point: Many redox reactions in aqueous solution involve transfer of H⁺ as well as electrons, and the electrode potential therefore depends on the pH.

For a redox couple involving ν_e electrons and ν_H protons:

E = E' − (ν_H⁺RT ln 10 / ν_eF) × pH

At 25°C:

E = E' − (0.059 V × ν_H⁺/ν_e) × pH

A brief illustration: The half-reaction for the perchlorate/chlorate couple is:

ClO₄⁻(aq) + 2 H⁺(aq) + 2 e⁻ → ClO₃⁻(aq) + H₂O(l)

At pH = 0: E° = +1.201 V

At pH = 7: E = 1.201 − (2/2)(7 × 0.059) V = +0.788 V

The perchlorate anion is a stronger oxidant under acid conditions.

5.7 Reactions with Water

Water may act as an oxidizing agent (reduced to H₂) or a reducing agent (oxidized to O₂):

H⁺(aq) + e⁻ → ½ H₂(g) E = −0.059 V × pH

½ O₂(g) + 2 H⁺(aq) + 2 e⁻ → H₂O(l) E = 1.23 − 0.059 V × pH

(a) Oxidation by Water

Key Point: For metals with large, negative standard potentials, reaction with aqueous acids leads to the production of H₂ unless a passivating oxide layer is formed.

When the standard potential for the reduction of a metal ion to the metal is negative, the metal should undergo oxidation in 1 M acid with evolution of hydrogen.

Passivation: Metals like Mg and Al survive in moist air because they are protected by an impervious film of oxide (passivation).

(b) Reduction by Water

Key Point: Water can act as a reducing agent; that is, it can be oxidized by other species.

Strong oxidizing agents like Co³⁺ (E° = +1.92 V) can oxidize water with evolution of O₂.

(c) The Stability Field of Water

Stability Field of Water

Figure 5.3: The stability field of water showing the range of pH and potential where water is thermodynamically stable.

5.8 Oxidation by Atmospheric Oxygen

Key Point: The O₂ present in air and dissolved in water can oxidize metals and metal ions in solution.

Fe²⁺ in water exposed to air is oxidized to Fe³⁺:

4 Fe²⁺(aq) + O₂(g) + 4 H⁺(aq) → 4 Fe³⁺(aq) + 2 H₂O(l)

5.9 Disproportionation and Comproportionation

Key Point: Standard potentials can be used to define the inherent stability and instability of different oxidation states.

Disproportionation:

A redox reaction in which the oxidation number of an element is simultaneously raised and lowered. The element serves as its own oxidizing and reducing agent.

2 Cu⁺(aq) → Cu²⁺(aq) + Cu(s) E°_cell = +0.36 V, K = 1.3 × 10⁶

Comproportionation:

The reverse of disproportionation—two species with the same element in different oxidation states form a product with an intermediate oxidation state.

Ag²⁺(aq) + Ag(s) → 2 Ag⁺(aq) E°_cell = +1.18 V, K = 1 × 10²⁰

5.10 The Influence of Complexation

Key Point: Formation of a more stable complex in the higher oxidation state favors oxidation (more negative E°); formation of a more stable complex in the lower oxidation state favors reduction (more positive E°).

E°(M) − E°(ML) = (RT/F) ln(K^ox/K^red)

A brief illustration: The standard potential for [Fe(CN)₆]³⁻/[Fe(CN)₆]⁴⁻ is +0.36 V, which is 0.41 V more negative than the aqua couple (+0.77 V). This means CN⁻ has a 10⁷-fold greater affinity for Fe(III) compared to Fe(II).

5.11 Solubility and Standard Potentials

Key Point: The standard cell potential can be used to determine the solubility product.

ln K_sp = νF{E°(MX_ν/M,X⁻) − E°(M^ν+/M)} / RT

Diagrammatic Presentation of Potential Data

Several useful diagrammatic summaries help visualize the relative stabilities of different oxidation states in aqueous solution.

5.12 Latimer Diagrams

Key Point: In a Latimer diagram, oxidation numbers decrease from left to right, and the numerical values of E° in volts are written above the line joining the species.

Latimer Diagram for Chlorine (Acidic Solution, pH = 0)

ClO₄⁻+7

+1.20

ClO₃⁻+5

+1.18

HClO₂+3

+1.67

HClO+1

+1.63

Cl₂0

+1.36

Cl⁻−1

Latimer Diagram for Chlorine (Basic Solution, pH = 14)

ClO₄⁻+7

+0.37

ClO₃⁻+5

+0.30

ClO₂⁻+3

+0.68

ClO⁻+1

+0.42

Cl₂0

+1.36

Cl⁻−1

⚠️ Nonadjacent Species

To derive the standard potential of a nonadjacent couple:

E°(a+b) = [ν(a)E°(a) + ν(b)E°(b)] / [ν(a) + ν(b)]

Disproportionation criterion: A species tends to disproportionate if the potential on the right is more positive than that on the left.

5.13 Frost Diagrams

Key Point: A Frost diagram plots νE° against oxidation number N. The most stable oxidation state lies lowest on the diagram.

Frost Diagram for Nitrogen (pH = 0 and pH = 14)

Figure 5.6: The Frost diagram for nitrogen. Species above the line connecting neighbors tend to disproportionate.

Interpretation rules:

The slope of a line connecting two points equals E° for that couple
A species above the line connecting its neighbors is unstable to disproportionation
Species below the line will form by comproportionation
The most stable oxidation state lies at the lowest point

5.14 Pourbaix Diagrams

Key Point: A Pourbaix diagram (E-pH diagram) indicates the conditions of pH and potential under which a species is thermodynamically stable.

Pourbaix Diagram for Iron

Figure 5.11: Pourbaix diagram for iron showing stability regions of different species.

Interpretation:

Horizontal lines: Separate species related by electron transfer only (not pH dependent)
Vertical lines: Separate species related by proton transfer only (not redox)
Sloped lines: Separate species related by both electron and proton transfer

5.15 Applications in Environmental Chemistry

Key Point: Pourbaix diagrams are useful for predicting the predominant species in natural waters under different conditions of pH and oxygen content.

Natural waters show varying conditions:

Surface waters: Well-oxygenated, higher potential
Deep lake water: Low O₂, reducing conditions
Bog waters: Acidic (pH ~4.5), reducing
Organic-rich soils: Very reducing

Chemical Extraction of the Elements

The extraction of elements involves changing their oxidation number from that in a naturally occurring compound to zero (the element).

5.16 Chemical Reduction

Key Point: An Ellingham diagram summarizes the temperature dependence of the standard Gibbs energies of formation of metal oxides and identifies the temperature at which reduction by carbon becomes spontaneous.

For reduction of a metal oxide by carbon to be favorable, one of these reactions must have a more negative Δ_rG° than the metal oxide formation:

(a) C(s) + ½ O₂(g) → CO(g)

(b) ½ C(s) + ½ O₂(g) → ½ CO₂(g)

(c) CO(g) + ½ O₂(g) → CO₂(g)

Ellingham Diagram (Simplified)

Figure 5.16: Ellingham diagram showing when carbon can reduce metal oxides.

Reading the diagram:

Carbon reduces a metal oxide when its line lies below the metal oxide line
The C/CO line has negative slope (net gas formation)
The C/CO₂ line is flat (no net change in gas amount)
Metal oxide lines have positive slopes (net gas consumption)

Example 5.14: Using an Ellingham Diagram

Problem: What is the lowest temperature at which ZnO can be reduced to zinc by carbon?

Answer: The C/CO line crosses below the ZnO line at approximately 1200°C. The overall reaction is:

C(s) + ZnO(s) → CO(g) + Zn(g)

Note: Zinc boils at 907°C, so it is produced as a gas.

The Blast Furnace

Iron ore reduction is the most important application of carbon pyrometallurgy:

Fe₂O₃(s) + 3 CO(g) → 2 Fe(l) + 3 CO₂(g)

5.17 Chemical Oxidation

Key Point: Elements obtained by chemical oxidation include the heavier halogens, sulfur, and certain noble metals.

The Claus process produces sulfur from H₂S:

2 H₂S + 3 O₂ → 2 SO₂ + 2 H₂O

2 H₂S + SO₂ → 3 S + 2 H₂O (catalyzed)

Gold extraction uses oxidation with cyanide complexation:

Au(s) + 2 CN⁻(aq) → [Au(CN)₂]⁻(aq) + e⁻

5.18 Electrochemical Extraction

Key Point: Elements obtained by electrochemical reduction include aluminium; those obtained by electrochemical oxidation include chlorine.

The Hall-Héroult Process

Aluminium is produced by electrolysis of Al₂O₃ dissolved in molten cryolite (Na₃AlF₆):

2 Al₂O₃ + 3 C → 4 Al + 3 CO₂

The graphite anodes participate by reacting with evolved oxygen atoms.

Chlorine Production

Electrolysis of brine produces Cl₂, H₂, and NaOH:

2 Cl⁻(aq) + 2 H₂O(l) → 2 OH⁻(aq) + H₂(g) + Cl₂(g)

Although O₂ evolution is thermodynamically more favorable, it requires a high overpotential, allowing Cl₂ production to proceed.

Interactive Calculators

🔬 Nernst Equation Calculator

Calculate the cell potential at non-standard conditions using the Nernst equation.

E° (Standard Potential, V)

Temperature (K)

n (electrons transferred)

Q (reaction quotient)

E = E° − (RT/nF) ln Q

⚗️ Equilibrium Constant from E° Calculator

Calculate the equilibrium constant K from the standard cell potential.

E°_cell (V)

n (electrons transferred)

Temperature (K)

ln K = nFE°/RT

📊 pH Effect on Potential Calculator

Calculate how potential changes with pH for a half-reaction involving H⁺.

E° at pH = 0 (V)

ν_H⁺ (H⁺ coefficient)

ν_e (electron coefficient)

pH

E = E° − (0.059 × ν_H⁺/ν_e) × pH

Chapter Summary

Key Equations

Δ_rG° = −νFE°

Δ_rG° = −RT ln K

E_cell = E°_cell − (RT/νF) ln Q

E = E° − (0.059 V × ν_H⁺/ν_e) × pH (at 25°C)

ln K = νFE°_cell/RT

Important Constants

Faraday constant: F = 96,485 C mol⁻¹ ≈ 96.48 kC mol⁻¹
Gas constant: R = 8.314 J K⁻¹ mol⁻¹
At 25°C (298 K): RT/F = 0.0257 V; (RT/F) ln 10 = 0.059 V

Diagram Summary

Diagram	Axes	Use
Latimer	Species with E° values	Quantitative data, calculating potentials
Frost	νE° vs. oxidation number	Stability, disproportionation
Pourbaix	E vs. pH	Stability in aqueous solution
Ellingham	Δ_rG° vs. temperature	Metal extraction conditions

Key Concepts

Oxidation: Loss of electrons, increase in oxidation number
Reduction: Gain of electrons, decrease in oxidation number
E° > 0: Spontaneous reaction (K > 1)
Disproportionation: Species above line connecting neighbors (Frost diagram)
Passivation: Protective oxide layer prevents further reaction
Overpotential: Additional potential needed to drive reaction at practical rate