Redox Reactions

/Redox reactions (oxidation-reduction reactions) are chemical processes involving the transfer of electrons between chemical species. These reactions are fundamental to:

Energy production and storage (batteries, fuel cells)
Biological processes (respiration, photosynthesis)
Industrial manufacturing (metallurgy, chemical synthesis)
Environmental processes (corrosion, weathering)

1.2 Historical Context

The term "oxidation" originally referred to reactions with oxygen, coined by Antoine Lavoisier. The concept evolved to encompass:

Classical definition: Addition/removal of oxygen or hydrogen
Electronic definition: Transfer of electrons
Modern definition: Change in oxidation state

1.3 Fundamental Principles

All redox reactions involve:

Simultaneous oxidation and reduction
Conservation of charge and mass
Electron transfer between species
Energy changes (ΔG, ΔH)

2. Oxidation and Reduction Concepts {#oxidation-reduction}

2.1 Oxidation

Definition: Process involving loss of electrons or increase in oxidation state.

Characteristics of Oxidation:

Addition of oxygen
- Example: 2Mg + O₂ → 2MgO
- Magnesium gains oxygen atoms
Removal of hydrogen
- Example: H₂S + Cl₂ → 2HCl + S
- Hydrogen sulfide loses hydrogen
Loss of electrons
- Example: Fe²⁺ → Fe³⁺ + e⁻
- Iron(II) loses one electron
Increase in oxidation number
- Example: In SnCl₂ → SnCl₄, Sn goes from +2 to +4

2.2 Reduction

Definition: Process involving gain of electrons or decrease in oxidation state.

Characteristics of Reduction:

Removal of oxygen
- Example: CuO + H₂ → Cu + H₂O
- Copper oxide loses oxygen
Addition of hydrogen
- Example: N₂ + 3H₂ → 2NH₃
- Nitrogen gains hydrogen
Gain of electrons
- Example: Fe³⁺ + e⁻ → Fe²⁺
- Iron(III) gains one electron
Decrease in oxidation number
- Example: In MnO₄⁻ → Mn²⁺, Mn goes from +7 to +2

2.3 Memory Aids

OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)
LEO GER: Loss Electrons Oxidation, Gain Electrons Reduction

3. Oxidation Numbers and States {#oxidation-numbers}

3.1 Definition

Oxidation number (oxidation state): The hypothetical charge an atom would have if all bonds were completely ionic.

3.2 Rules for Assigning Oxidation Numbers

Free elements: Oxidation number = 0
- Examples: Na, O₂, Cl₂, P₄, S₈
Monoatomic ions: Oxidation number = ionic charge
- Examples: Na⁺ = +1, Cl⁻ = -1, Ca²⁺ = +2
Oxygen compounds: Usually -2
- Exceptions:
  - Peroxides (H₂O₂): -1
  - Superoxides (KO₂): -1/2
  - OF₂: +2
Hydrogen compounds: Usually +1
- Exception: Metal hydrides (NaH, CaH₂): -1
Group 1 metals: Always +1 in compounds
Group 2 metals: Always +2 in compounds
Halogens: Usually -1
- Exceptions: In compounds with more electronegative elements
Sum rule:
- Neutral molecules: Sum = 0
- Polyatomic ions: Sum = charge on ion

3.3 Examples and Calculations

Example 1: Find oxidation number of S in H₂SO₄

H: +1 (×2) = +2
O: -2 (×4) = -8
S: +2 + x + (-8) = 0
Therefore, x = +6

Example 2: Find oxidation number of Cr in Cr₂O₇²⁻

O: -2 (×7) = -14
Cr: 2x + (-14) = -2
Therefore, x = +6

Example 3: Fractional oxidation states in Fe₃O₄

Can be written as FeO·Fe₂O₃
Contains Fe²⁺ and Fe³⁺
Average oxidation state: (2 + 6)/3 = +8/3

3.4 Complex Cases

Coordinate bonds:

In H₂SO₄: S forms coordinate bonds with O
S has formal oxidation state +6

Organic compounds:

C-H bonds: C = -1 for each H
C-O bonds: C = +1 for each O
C=O bonds: C = +2

4. Types of Redox Reactions {#types-of-reactions}

4.1 Combination Reactions

Definition: Two or more substances combine to form a single product.
General form: A + B → AB

Examples:

Metal + Nonmetal:
- 2Na + Cl₂ → 2NaCl
- Oxidation: Na → Na⁺ + e⁻
- Reduction: Cl₂ + 2e⁻ → 2Cl⁻
Nonmetal + Nonmetal:
- H₂ + F₂ → 2HF
- Oxidation: H₂ → 2H⁺ + 2e⁻
- Reduction: F₂ + 2e⁻ → 2F⁻
Metal + Oxygen:
- 4Fe + 3O₂ → 2Fe₂O₃
- Oxidation: Fe → Fe³⁺ + 3e⁻
- Reduction: O₂ + 4e⁻ → 2O²⁻

4.2 Decomposition Reactions

Definition: Single compound breaks down into two or more products.
General form: AB → A + B

Examples:

Thermal decomposition:
- 2HgO → 2Hg + O₂
- Oxidation: O²⁻ → O₂ + 4e⁻
- Reduction: Hg²⁺ + 2e⁻ → Hg
Electrolytic decomposition:
- 2H₂O → 2H₂ + O₂
- At cathode: 2H⁺ + 2e⁻ → H₂
- At anode: 2H₂O → O₂ + 4H⁺ + 4e⁻

4.3 Displacement Reactions

Definition: One element displaces another from its compound.
General form: A + BC → AC + B

Types:

Metal displacement:
- Zn + CuSO₄ → ZnSO₄ + Cu
- Oxidation: Zn → Zn²⁺ + 2e⁻
- Reduction: Cu²⁺ + 2e⁻ → Cu
Nonmetal displacement:
- Cl₂ + 2NaBr → 2NaCl + Br₂
- Oxidation: 2Br⁻ → Br₂ + 2e⁻
- Reduction: Cl₂ + 2e⁻ → 2Cl⁻

4.4 Disproportionation Reactions

Definition: Same element simultaneously oxidized and reduced.

Examples:

Chlorine in base:
- 3Cl₂ + 6OH⁻ → ClO₃⁻ + 5Cl⁻ + 3H₂O
- Oxidation: Cl₂ → ClO₃⁻ (0 to +5)
- Reduction: Cl₂ → Cl⁻ (0 to -1)
Hydrogen peroxide:
- 2H₂O₂ → 2H₂O + O₂
- Oxidation: H₂O₂ → O₂ (-1 to 0)
- Reduction: H₂O₂ → H₂O (-1 to -2)

4.5 Comproportionation Reactions

Definition: Same element in different oxidation states forms intermediate state.

Example:

H₂S + SO₂ → 3S + 2H₂O
S(-2) + S(+4) → S(0)

5. Oxidizing and Reducing Agents {#agents}

5.1 Oxidizing Agents (Oxidants)

Definition: Species that cause oxidation by accepting electrons and are themselves reduced.

Characteristics:

Contain elements in high oxidation states
Readily accept electrons
Act as electron acceptors
Are reduced in the process

Common Strong Oxidizing Agents:

Permanganate ion (MnO₄⁻):
- In acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
- In neutral medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
- In basic medium: MnO₄⁻ + e⁻ → MnO₄²⁻
Dichromate ion (Cr₂O₇²⁻):
- Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Concentrated sulfuric acid (H₂SO₄):
- H₂SO₄ + 2H⁺ + 2e⁻ → SO₂ + 2H₂O
Nitric acid (HNO₃):
- HNO₃ + 3H⁺ + 3e⁻ → NO + 2H₂O
Halogens (F₂, Cl₂, Br₂, I₂):
- X₂ + 2e⁻ → 2X⁻

5.2 Reducing Agents (Reductants)

Definition: Species that cause reduction by donating electrons and are themselves oxidized.

Characteristics:

Contain elements in low oxidation states
Readily donate electrons
Act as electron donors
Are oxidized in the process

Common Strong Reducing Agents:

Active metals (Li, K, Na, Ca, Mg):
- M → M^n+ + ne⁻
Metal hydrides (NaH, LiAlH₄, NaBH₄):
- H⁻ → H⁺ + 2e⁻
Hydrogen gas:
- H₂ → 2H⁺ + 2e⁻
Carbon and carbon monoxide:
- C + O₂ → CO₂ + 4e⁻
- CO + H₂O → CO₂ + 2H⁺ + 2e⁻
Sulfur compounds in low oxidation states:
- H₂S → S + 2H⁺ + 2e⁻
- SO₃²⁻ → SO₄²⁻ + 2e⁻

5.3 Substances Acting as Both Oxidizing and Reducing Agents

Some compounds can act as both oxidizing and reducing agents depending on conditions:

Hydrogen peroxide (H₂O₂):
- As oxidizing agent: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O
- As reducing agent: H₂O₂ → O₂ + 2H⁺ + 2e⁻
Sulfur dioxide (SO₂):
- As oxidizing agent: SO₂ + 4H⁺ + 4e⁻ → S + 2H₂O
- As reducing agent: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻
Nitrous acid (HNO₂):
- As oxidizing agent: HNO₂ + H⁺ + e⁻ → NO + H₂O
- As reducing agent: HNO₂ + H₂O → NO₃⁻ + 3H⁺ + 2e⁻

6. Balancing Redox Equations {#balancing}

6.1 Ion-Electron Method (Half-Reaction Method)

Steps for Acidic Solutions:

Write the unbalanced ionic equation
Separate into half-reactions
Balance atoms other than H and O
Balance oxygen by adding H₂O
Balance hydrogen by adding H⁺
Balance charge by adding electrons
Multiply half-reactions to equalize electrons
Add half-reactions and simplify

Example: Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic)

Step 1: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Step 2: Separate half-reactions

Reduction: MnO₄⁻ → Mn²⁺
Oxidation: Fe²⁺ → Fe³⁺

Step 3: Balance atoms (already balanced)

Step 4: Balance oxygen

MnO₄⁻ → Mn²⁺ + 4H₂O

Step 5: Balance hydrogen

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Step 6: Balance charge

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻

Step 7: Multiply to equalize electrons

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×1)
Fe²⁺ → Fe³⁺ + e⁻ (×5)

Step 8: Add and simplify

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Steps for Basic Solutions:

1-6. Same as acidic method
7. Add OH⁻ to neutralize H⁺
8. Combine H⁺ and OH⁻ to form H₂O
9. Simplify by canceling H₂O on both sides

Example: Balance Cr(OH)₃ + IO₃⁻ → CrO₄²⁻ + I⁻ (basic)

Oxidation: Cr(OH)₃ → CrO₄²⁻

Cr(OH)₃ + H₂O → CrO₄²⁻ + 5H⁺ + 3e⁻
Add 5OH⁻: Cr(OH)₃ + H₂O + 5OH⁻ → CrO₄²⁻ + 5H₂O + 3e⁻
Simplify: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻

Reduction: IO₃⁻ → I⁻

IO₃⁻ + 6H⁺ + 6e⁻ → I⁻ + 3H₂O
Add 6OH⁻: IO₃⁻ + 6H₂O + 6e⁻ → I⁻ + 3H₂O + 6OH⁻
Simplify: IO₃⁻ + 3H₂O + 6e⁻ → I⁻ + 6OH⁻

Balanced equation:
2Cr(OH)₃ + IO₃⁻ + 4OH⁻ → 2CrO₄²⁻ + I⁻ + 5H₂O

6.2 Oxidation Number Method

Steps:

Assign oxidation numbers
Identify changes in oxidation numbers
Calculate electron change per atom
Balance electron change
Balance other atoms
Balance charge and atoms

Example: Balance Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O

Step 1: Assign oxidation numbers

Cu: 0 → +2 (change: +2)
N in HNO₃: +5 → +2 in NO (change: -3)

Step 2: Balance electron change

Cu loses 2e⁻, N gains 3e⁻
LCM of 2 and 3 is 6
Need 3 Cu atoms and 2 N atoms

Step 3: Preliminary balance

3Cu + 2HNO₃ → 3Cu(NO₃)₂ + 2NO + H₂O

Step 4: Balance all N atoms

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

7. Electrochemical Series {#electrochemical-series}

7.1 Definition and Significance

The electrochemical series is a list of metals and non-metals arranged in order of their standard electrode potentials (reduction potentials) measured against the standard hydrogen electrode (SHE).

7.2 Standard Hydrogen Electrode (SHE)

Definition: Reference electrode with arbitrarily assigned potential of 0.00 V
Conditions:

[H⁺] = 1 M
P(H₂) = 1 atm
Temperature = 25°C (298 K)

Reaction: 2H⁺ + 2e⁻ ⇌ H₂

7.3 Standard Electrode Potentials

Selected Values (E° in volts):

Element	Half-Reaction	E° (V)
Li⁺/Li	Li⁺ + e⁻ → Li	-3.05
K⁺/K	K⁺ + e⁻ → K	-2.93
Ca²⁺/Ca	Ca²⁺ + 2e⁻ → Ca	-2.87
Na⁺/Na	Na⁺ + e⁻ → Na	-2.71
Mg²⁺/Mg	Mg²⁺ + 2e⁻ → Mg	-2.37
Al³⁺/Al	Al³⁺ + 3e⁻ → Al	-1.66
Zn²⁺/Zn	Zn²⁺ + 2e⁻ → Zn	-0.76
Fe²⁺/Fe	Fe²⁺ + 2e⁻ → Fe	-0.44
Ni²⁺/Ni	Ni²⁺ + 2e⁻ → Ni	-0.25
Sn²⁺/Sn	Sn²⁺ + 2e⁻ → Sn	-0.14
H⁺/H₂	2H⁺ + 2e⁻ → H₂	0.00
Cu²⁺/Cu	Cu²⁺ + 2e⁻ → Cu	+0.34
I₂/I⁻	I₂ + 2e⁻ → 2I⁻	+0.54
Fe³⁺/Fe²⁺	Fe³⁺ + e⁻ → Fe²⁺	+0.77
Ag⁺/Ag	Ag⁺ + e⁻ → Ag	+0.80
Br₂/Br⁻	Br₂ + 2e⁻ → 2Br⁻	+1.09
Cl₂/Cl⁻	Cl₂ + 2e⁻ → 2Cl⁻	+1.36
Au³⁺/Au	Au³⁺ + 3e⁻ → Au	+1.50
F₂/F⁻	F₂ + 2e⁻ → 2F⁻	+2.87

7.4 Applications of Electrochemical Series

Predicting spontaneity of redox reactions:
- E°cell = E°cathode - E°anode
- If E°cell > 0, reaction is spontaneous
Determining relative strength of oxidizing/reducing agents:
- Strong oxidizing agents: High positive E° values
- Strong reducing agents: High negative E° values
Predicting displacement reactions:
- Metal with lower E° displaces metal with higher E°
Corrosion prevention:
- Sacrificial anodes (Zn, Mg) protect Fe

8. Nernst Equation {#nernst-equation}

8.1 Derivation and Expression

The Nernst equation relates electrode potential to concentration and temperature:

General form:
E = E° - (RT/nF) ln Q

At 25°C (298 K):
E = E° - (0.0592/n) log Q

Where:

E = electrode potential under non-standard conditions
E° = standard electrode potential
R = gas constant (8.314 J/mol·K)
T = absolute temperature (K)
n = number of electrons transferred
F = Faraday constant (96,485 C/mol)
Q = reaction quotient

8.2 Applications

Example 1: Calculate potential of Zn²⁺/Zn electrode when [Zn²⁺] = 0.01 M

Given: E°(Zn²⁺/Zn) = -0.76 V

E = E° - (0.0592/n) log (1/[Zn²⁺])
E = -0.76 - (0.0592/2) log (1/0.01)
E = -0.76 - 0.0296 × log(100)
E = -0.76 - 0.0296 × 2
E = -0.82 V

Example 2: Calculate cell potential for Zn|Zn²⁺(0.1 M)||Cu²⁺(1.0 M)|Cu

E°cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = 0.34 - (-0.76) = 1.10 V

Ecell = E°cell - (0.0592/n) log ([Zn²⁺]/[Cu²⁺])
Ecell = 1.10 - (0.0592/2) log (0.1/1.0)
Ecell = 1.10 - 0.0296 × log(0.1)
Ecell = 1.10 - 0.0296 × (-1)
Ecell = 1.13 V

8.3 Equilibrium Constant Calculation

At equilibrium, E = 0 and Q = K:

0 = E° - (0.0592/n) log K
E° = (0.0592/n) log K
log K = nE°/0.0592

Example: Calculate equilibrium constant for Zn + Cu²⁺ → Zn²⁺ + Cu

log K = (2 × 1.10)/0.0592 = 37.16
K = 1.45 × 10³⁷

9. Redox Titrations {#redox-titrations}

9.1 Principles

Redox titrations involve the reaction between oxidizing and reducing agents to determine concentration.

Key features:

Based on electron transfer
Endpoint detected by indicator or potentiometry
Calculations based on stoichiometry

9.2 Common Types

1. Permanganate Titrations:

MnO₄⁻ acts as oxidizing agent
Self-indicating (purple to colorless)
Used for Fe²⁺, C₂O₄²⁻, H₂O₂ analysis

2. Dichromate Titrations:

Cr₂O₇²⁻ acts as oxidizing agent
Requires indicator (diphenylamine)
Used for Fe²⁺ analysis

3. Iodometric Titrations:

I₂ + 2e⁻ → 2I⁻
Starch indicator (blue complex with I₂)
Used for Cu²⁺, ClO⁻ analysis

4. Iodimetric Titrations:

Direct titration with I₂
Used for reducing agents like S₂O₃²⁻

9.3 Calculation Examples

Example 1: Permanganate-Iron Titration

25.0 mL of Fe²⁺ solution requires 20.0 mL of 0.02 M KMnO₄ for complete oxidation in acidic medium. Calculate [Fe²⁺].

Balanced equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Moles MnO₄⁻ = 0.02 M × 0.020 L = 4.0 × 10⁻⁴ mol
Moles Fe²⁺ = 5 × 4.0 × 10⁻⁴ = 2.0 × 10⁻³ mol
[Fe²⁺] = 2.0 × 10⁻³ mol / 0.025 L = 0.08 M

Example 2: Iodometric Determination of Cu²⁺

2Cu²⁺ + 4I⁻ → 2CuI + I₂
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

If 25.0 mL of Cu²⁺ solution liberates I₂ that requires 30.0 mL of 0.1 M Na₂S₂O₃, calculate [Cu²⁺].

Moles S₂O₃²⁻ = 0.1 M × 0.030 L = 3.0 × 10⁻³ mol
Moles I₂ = 3.0 × 10⁻³ / 2 = 1.5 × 10⁻³ mol
Moles Cu²⁺ = 2 × 1.5 × 10⁻³ = 3.0 × 10⁻³ mol
[Cu²⁺] = 3.0 × 10⁻³ mol / 0.025 L = 0.12 M

10. Structures of Oxyacids {#oxyacids}

10.1 Definition and General Structure

Oxyacids (oxoacids) are acids containing hydrogen, oxygen, and at least one other element, with the general formula HₓEOᵧ.

General structure: H-O-E-O (where E is the central atom)

10.2 Factors Affecting Acid Strength

1. Electronegativity of central atom:

Higher electronegativity → stronger acid
Order: HClO > HBrO > HIO

2. Oxidation state of central atom:

Higher oxidation state → stronger acid
Order: HClO₄ > HClO₃ > HClO₂ > HClO

3. Number of oxygen atoms:

More oxygen atoms → stronger acid
Oxygen atoms withdraw electron density

10.3 Common Oxyacids and Their Structures

Halogen Oxyacids:

Hypochlorous acid (HClO):
```
text
```
H-O-Cl
- Oxidation state of Cl: +1
- Weak acid (Ka = 3.0 × 10⁻⁸)
Chlorous acid (HClO₂):
```
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```
H-O-Cl=O
- Oxidation state of Cl: +3
- Unstable, disproportionates
Chloric acid (HClO₃):
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text
```
O ‖ H-O-Cl=O
- Oxidation state of Cl: +5
- Strong acid
Perchloric acid (HClO₄):
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text
```
O ‖ H-O-Cl=O ‖ O
- Oxidation state of Cl: +7
- Strongest acid (completely ionized)

Sulfur Oxyacids:

Sulfurous acid (H₂SO₃):
```
text
```
H-O-S-O-H ‖ O
- Oxidation state of S: +4
- Weak acid (Ka1 = 1.5 × 10⁻²)
Sulfuric acid (H₂SO₄):
```
text
```
O ‖ H-O-S-O-H ‖ O
- Oxidation state of S: +6
- Strong acid (first ionization complete)

Nitrogen Oxyacids:

Nitrous acid (HNO₂):
```
text
```
H-O-N=O
- Oxidation state of N: +3
- Weak acid (Ka = 4.5 × 10⁻⁴)
Nitric acid (HNO₃):
```
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```
O ‖ H-O-N=O
- Oxidation state of N: +5
- Strong acid

Phosphorus Oxyacids:

Phosphorous acid (H₃PO₃):
```
text
```
H-P-O-H ‖ ‖ O O-H
- Oxidation state of P: +3
- Diprotic acid (one H directly bonded to P)
Phosphoric acid (H₃PO₄):
```
text
```
O ‖ H-O-P-O-H ‖ O-H
- Oxidation state of P: +5
- Triprotic acid

10.4 Acid Strength Trends

Pauling's Rules:

For acids HₙXOₘ, if (m-n) = 0, Ka ≈ 10⁻⁸ (very weak)
If (m-n) = 1, Ka ≈ 10⁻³ (weak)
If (m-n) = 2, Ka ≈ 10² (strong)
If (m-n) = 3, Ka ≈ 10⁷ (very strong)

Examples:

HClO (m-n = 1-1 = 0): Very weak
HClO₂ (m-n = 2-1 = 1): Weak
HClO₃ (m-n = 3-1 = 2): Strong
HClO₄ (m-n = 4-1 = 3): Very strong

11. Example Reactions and Calculations {#examples}

11.1 Balancing Complex Reactions

Example 1: Balance in acidic medium
Cr₂O₇²⁻ + C₂H₅OH → Cr³⁺ + CH₃COOH

Solution:
Oxidation: C₂H₅OH → CH₃COOH + 4H⁺ + 4e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Multiply oxidation by 3 and reduction by 2:
3C₂H₅OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Example 2: Balance in basic medium
MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₃²⁻

Solution:
Oxidation: C₂O₄²⁻ + 4OH⁻ → 2CO₃²⁻ + 2H₂O + 2e⁻
Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Multiply oxidation by 3 and reduction by 2:
2MnO₄⁻ + 3C₂O₄²⁻ + 4OH⁻ → 2MnO₂ + 6CO₃²⁻ + 2H₂O

11.2 Equivalent Weight Calculations

Formula: Equivalent weight = Molecular weight / Number of electrons transferred

Example 1: KMnO₄ in different media

In acidic medium: MnO₄⁻ → Mn²⁺ (5e⁻ change)
Equivalent weight = 158/5 = 31.6 g/equiv

In neutral medium: MnO₄⁻ → MnO₂ (3e⁻ change)
Equivalent weight = 158/3 = 52.7 g/equiv

In basic medium: MnO₄⁻ → MnO₄²⁻ (1e⁻ change)
Equivalent weight = 158/1 = 158 g/equiv

Example 2: H₂O₂ as oxidizing and reducing agent

As oxidizing agent: H₂O₂ → 2H₂O (2e⁻ gained)
Equivalent weight = 34/2 = 17 g/equiv

As reducing agent: H₂O₂ → O₂ (2e⁻ lost)
Equivalent weight = 34/2 = 17 g/equiv

11.3 Stoichiometric Calculations

Example: A sample containing Fe²⁺ and Fe³⁺ weighing 1.0 g is dissolved in acid. After reduction with Zn, the solution requires 25.0 mL of 0.02 M KMnO₄ for oxidation. Calculate % Fe in the sample.

Solution:
After reduction, all Fe becomes Fe²⁺
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Moles MnO₄⁻ = 0.02 × 0.025 = 5.0 × 10⁻⁴ mol
Moles Fe²⁺ = 5 × 5.0 × 10⁻⁴ = 2.5 × 10⁻³ mol
Mass of Fe = 2.5 × 10⁻³ × 55.85 = 0.140 g
% Fe = (0.140/1.0) × 100 = 14.0%

12. Applications and Industrial Processes {#applications}

12.1 Metallurgy

Extraction of metals involves redox reactions:

Iron extraction (Blast furnace):
- Fe₂O₃ + 3CO → 2Fe + 3CO₂
- C + O₂ → CO₂
- CO₂ + C → 2CO
Aluminum extraction (Hall-Héroult process):
- Al₂O₃ + 3C → 2Al + 3CO (at cathode)
- C + O₂ → CO₂ (at anode)
Copper purification (Electrolysis):
- At anode: Cu → Cu²⁺ + 2e⁻
- At cathode: Cu²⁺ + 2e⁻ → Cu

12.2 Energy Storage and Conversion

1. Batteries:

Lead-acid battery:

Anode: Pb + HSO₄⁻ → PbSO₄ + H⁺ + 2e⁻
Cathode: PbO₂ + HSO₄⁻ + 3H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Lithium-ion battery:

Anode: Li → Li⁺ + e⁻
Cathode: Li⁺ + e⁻ + CoO₂ → LiCoO₂

2. Fuel Cells:

Hydrogen fuel cell:

Anode: H₂ → 2H⁺ + 2e⁻
Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O

12.3 Environmental Applications

1. Water treatment:

Chlorination: Cl₂ + H₂O → HClO + HCl
Ozonation: O₃ + organics → CO₂ + H₂O

2. Corrosion prevention:

Galvanization (Zn coating)
Cathodic protection
Sacrificial anodes

12.4 Biological Processes

1. Cellular respiration:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP

2. Photosynthesis:
6CO₂ + 6H₂O + light → C₆H₁₂O₆ + 6O₂

3. Nitrogen fixation:
N₂ + 8H⁺ + 8e⁻ → 2NH₃ + H₂

12.5 Industrial Chemical Production

1. Chlor-alkali process:
2NaCl + 2H₂O → Cl₂ + H₂ + 2NaOH

2. Aluminum production:
2Al₂O₃ + 3C → 4Al + 3CO₂

3. Steel production:
Fe₂O₃ + 3CO → 2Fe + 3CO₂