Comprehensive Redox Reactions & Oxidation Numbers Cheat Sheet

Introduction: Understanding Redox Chemistry

Redox (reduction-oxidation) reactions represent one of the most fundamental processes in chemistry, involving the transfer of electrons between chemical species. These reactions drive countless natural and industrial processes, from cellular respiration and photosynthesis to metal refining and battery technology. This cheat sheet provides a comprehensive guide to understanding, identifying, and balancing redox reactions through oxidation numbers, half-reactions, and practical applications.

Core Concepts in Redox Chemistry

Definitions and Terminology

Recognizing Redox Reactions

A reaction is a redox reaction if at least one of these occurs:

1. Change in oxidation numbers of one or more elements
2. Transfer of electrons between reactants
3. Reaction of elements or ions to form compounds
4. Reaction involving free oxygen, hydrogen, or halogens

Non-Redox Reactions Include:

• Acid-base neutralization (no change in oxidation numbers)
• Precipitation reactions (no electron transfer)
• Most double replacement reactions
• Dissolution of ionic compounds

Oxidation Number Rules

Basic Rules for Assigning Oxidation Numbers

1. Free elements have oxidation number of 0

   • Na, Fe, O₂, Cl₂, P₄ all have oxidation number 0

2. Monoatomic ions have oxidation number equal to their charge

   • Na⁺ is +1, Mg²⁺ is +2, Cl⁻ is -1, O²⁻ is -2

3. Oxygen typically has oxidation number -2, except in:

   • Peroxides (O₂²⁻): -1 per oxygen (H₂O₂, Na₂O₂)
   • Superoxides (O₂⁻): -1/2 per oxygen (KO₂)
   • OF₂: +2 (oxygen is less electronegative than fluorine)

4. Hydrogen typically has oxidation number +1, except in:

   • Metal hydrides: -1 (NaH, CaH₂)

5. Fluorine always has oxidation number -1 in compounds

6. Other halogens (Cl, Br, I) usually have oxidation number -1 in binary compounds, but can have positive values when combined with oxygen or more electronegative elements

7. The sum of oxidation numbers in a compound equals 0

8. The sum of oxidation numbers in a polyatomic ion equals the charge of the ion

Common Oxidation States of Elements

Common Polyatomic Ions and Their Element Oxidation Numbers

Calculating Oxidation Numbers: Step-by-Step Process

Example 1: Simple Compound – H₂SO₄

1. Assign known oxidation numbers:

   • Hydrogen: +1 (rule 4)
   • Oxygen: -2 (rule 3)

2. Use the sum rule to find unknown numbers:

   • Let x = oxidation number of S
   • 2(+1) + x + 4(-2) = 0
   • 2 + x – 8 = 0
   • x = +6

3. Verify: H: +1, S: +6, O: -2

   • Sum: 2(+1) + 1(+6) + 4(-2) = 2 + 6 – 8 = 0 ✓

Example 2: Polyatomic Ion – MnO₄⁻

1. Assign known oxidation numbers:

   • Oxygen: -2 (rule 3)

2. Use the charge rule to find unknown numbers:

   • Let x = oxidation number of Mn
   • x + 4(-2) = -1
   • x – 8 = -1
   • x = +7

3. Verify: Mn: +7, O: -2

   • Sum: 1(+7) + 4(-2) = 7 – 8 = -1 ✓

Example 3: Complex Compound – K₂Cr₂O₇

1. Assign known oxidation numbers:

   • Potassium: +1 (rule 2)
   • Oxygen: -2 (rule 3)

2. Use the sum rule to find unknown numbers:

   • Let x = oxidation number of Cr
   • 2(+1) + 2x + 7(-2) = 0
   • 2 + 2x – 14 = 0
   • 2x = 12
   • x = +6

3. Verify: K: +1, Cr: +6, O: -2

   • Sum: 2(+1) + 2(+6) + 7(-2) = 2 + 12 – 14 = 0 ✓

Identifying Redox Reactions Using Oxidation Numbers

Step-by-Step Process

1. Assign oxidation numbers to all elements on both sides of the equation
2. Identify elements that change oxidation numbers
3. The species containing elements that increase in oxidation number are being oxidized
4. The species containing elements that decrease in oxidation number are being reduced

Example: Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

• Cu is oxidized (oxidation number increases from 0 to +2)
• Ag⁺ is reduced (oxidation number decreases from +1 to 0)
• Cu is the reducing agent (gives electrons)
• Ag⁺ is the oxidizing agent (takes electrons)

Balancing Redox Reactions

Half-Reaction Method (Acidic Conditions)

1. Write the unbalanced equation
2. Split into half-reactions (oxidation and reduction)
3. Balance all elements except H and O
4. Balance O by adding H₂O
5. Balance H by adding H⁺
6. Balance charge by adding electrons
7. Multiply half-reactions to equalize electrons transferred
8. Add half-reactions and simplify

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

Step 1: Split into half-reactions

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

Step 2: Balance elements other than H and O

• Reduction: MnO₄⁻ → Mn²⁺ (Mn balanced)
• Oxidation: Fe²⁺ → Fe³⁺ (Fe balanced)

Step 3: Balance O by adding H₂O

• Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O
• Oxidation: Fe²⁺ → Fe³⁺ (no O to balance)

Step 4: Balance H by adding H⁺

• Reduction: 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
• Oxidation: Fe²⁺ → Fe³⁺ (no H to balance)

Step 5: Balance charge with electrons

• Reduction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
  • Left: -1 + 0 + 8 = +7; Right: +2 + 0 = +2; Need 5e⁻
• Oxidation: Fe²⁺ → Fe³⁺ + e⁻
  • Left: +2; Right: +3 + (-1) = +2; Need 1e⁻

Step 6: Multiply to equalize electrons

• Reduction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O (×1)
• Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×5)
• Oxidation (multiplied): 5Fe²⁺ → 5Fe³⁺ + 5e⁻

Step 7: Add half-reactions

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

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

Half-Reaction Method (Basic Conditions)

1. Follow steps 1-6 for acidic conditions
2. Add OH⁻ to both sides to neutralize H⁺
3. Combine H⁺ and OH⁻ to form H₂O
4. Cancel out H₂O appearing on both sides

Example: MnO₄⁻ + I⁻ → MnO₂ + IO₃⁻ (basic solution)

Step 1-6: Balance as if in acidic solution

• 3e⁻ + 4H⁺ + MnO₄⁻ → MnO₂ + 2H₂O
• I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻

Step 7: Multiply to equalize electrons

• 3e⁻ + 4H⁺ + MnO₄⁻ → MnO₂ + 2H₂O (×2)
• I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻ (×1)
• Result: 6e⁻ + 8H⁺ + 2MnO₄⁻ → 2MnO₂ + 4H₂O + I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻

Step 8: Add half-reactions

• 6e⁻ + 8H⁺ + 2MnO₄⁻ + I⁻ + 3H₂O → 2MnO₂ + 4H₂O + IO₃⁻ + 6H⁺ + 6e⁻
• 8H⁺ + 2MnO₄⁻ + I⁻ + 3H₂O → 2MnO₂ + 4H₂O + IO₃⁻ + 6H⁺
• 2H⁺ + 2MnO₄⁻ + I⁻ → 2MnO₂ + H₂O + IO₃⁻

Step 9: Add OH⁻ to neutralize H⁺ (basic solution)

• 2H⁺ + 2MnO₄⁻ + I⁻ + 2OH⁻ → 2MnO₂ + H₂O + IO₃⁻ + 2OH⁻
• 2MnO₄⁻ + I⁻ + 2H₂O → 2MnO₂ + H₂O + IO₃⁻ + 2OH⁻
• 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

Final balanced equation (basic): 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

Oxidation Number Method

1. Assign oxidation numbers to all elements
2. Identify elements that change oxidation numbers
3. Calculate the total increase and decrease in oxidation numbers
4. Add coefficients to equalize the total electron change
5. Balance the remaining elements

Example: Cr₂O₇²⁻ + C₂H₅OH → Cr³⁺ + CO₂ (acidic solution)

Step 1: Assign oxidation numbers

• Cr₂O₇²⁻: Cr = +6, O = -2
• C₂H₅OH: C = -2, H = +1, O = -2
• Cr³⁺: Cr = +3
• CO₂: C = +4, O = -2

Step 2: Identify changes

• Cr: +6 → +3 (change = -3)
• C: -2 → +4 (change = +6)

Step 3: Calculate electron change

• Total Cr change: 2 Cr atoms × (-3) = -6
• Total C change: 2 C atoms × (+6) = +12

Step 4: Add coefficients to balance electron change

• Need Cr₂O₇²⁻ × 2 for -12 change
• Need C₂H₅OH × 1 for +12 change

Step 5: Write partially balanced equation

• 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺ + 2CO₂

Step 6: Balance remaining elements

• 2Cr₂O₇²⁻ + C₂H₅OH + 16H⁺ → 4Cr³⁺ + 2CO₂ + 11H₂O

Final balanced equation: 2Cr₂O₇²⁻ + C₂H₅OH + 16H⁺ → 4Cr³⁺ + 2CO₂ + 11H₂O

Electrochemical Cells and Standard Reduction Potentials

Standard Reduction Potentials Table (Selected Values)

Using Standard Reduction Potentials

1. Determining cell potential:

   • E°cell = E°reduction(cathode) – E°reduction(anode)
   • E°cell = E°reduction + E°oxidation

2. Predicting spontaneity:

   • If E°cell > 0, reaction is spontaneous
   • If E°cell < 0, reaction is non-spontaneous

3. Identifying stronger oxidizing/reducing agents:

   • Higher E° value = stronger oxidizing agent
   • Lower E° value = stronger reducing agent

4. Standard reactions:

   • Always write as reduction half-reactions
   • To get oxidation half-reaction, reverse the reaction and change the sign of E°

Example: Cu²⁺ + Zn → Cu + Zn²⁺

Step 1: Identify half-reactions

• Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
• Zn²⁺ + 2e⁻ → Zn (E° = -0.76 V)

Step 2: Determine which reaction is oxidation vs. reduction

• Higher E° will be reduction (Cu²⁺ reduction)
• Lower E° will be oxidation (Zn oxidation: Zn → Zn²⁺ + 2e⁻, E° = +0.76 V)

Step 3: Calculate cell potential

• E°cell = E°reduction + E°oxidation = +0.34 V + 0.76 V = +1.10 V

Step 4: Interpret result

• E°cell > 0, so reaction is spontaneous
• Cu²⁺ is reduced (oxidizing agent)
• Zn is oxidized (reducing agent)

Application of Redox in Common Processes

Biological Redox Processes

Industrial Redox Applications

Common Oxidizing and Reducing Agents

Common Challenges and Troubleshooting

Balancing Complex Redox Reactions

1. For reactions with many elements:

   • Focus on the elements that change oxidation number first
   • Consider splitting complex molecules into simpler components
   • Balance one element at a time

2. For organic redox reactions:

   • Identify the carbon atoms that change oxidation number
   • Handle as separate balancing problems if multiple carbons change differently

3. For redox reactions in different media:

   • For basic solutions, remember to neutralize acids with OH⁻
   • For non-aqueous solutions, consider appropriate electron donors/acceptors

Common Mistakes to Avoid

Problem-Solving Examples

Example 1: Identifying Redox in Complex Reactions

Reaction: CH₄ + 2O₂ → CO₂ + 2H₂O

• Carbon is oxidized: CH₄ is the reducing agent
• Oxygen is reduced: O₂ is the oxidizing agent

Example 2: Balancing Complex Redox in Acidic Solution

Reaction: Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ (acidic solution)

Step 1: Identify half-reactions

• Oxidation: Fe²⁺ → Fe³⁺
• Reduction: Cr₂O₇²⁻ → Cr³⁺

Step 2: Balance oxidation half-reaction

• Fe²⁺ → Fe³⁺ + e⁻

Step 3: Balance reduction half-reaction

• Cr₂O₇²⁻ → 2Cr³⁺
• Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (balance O with H₂O)
• 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (balance H)
• 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (balance charge)

Step 4: Equalize electron transfer

• Fe²⁺ → Fe³⁺ + e⁻ (×6)
• 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (×1)
• 6Fe²⁺ → 6Fe³⁺ + 6e⁻

Step 5: Add half-reactions

• 6Fe²⁺ + 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 6Fe³⁺ + 6e⁻ + 2Cr³⁺ + 7H₂O
• 6Fe²⁺ + 14H⁺ + Cr₂O₇²⁻ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Final balanced equation: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Example 3: Predicting Spontaneity Using Standard Potentials

Question: Will Zn metal reduce Fe²⁺ to Fe metal in standard conditions?

Step 1: Write possible half-reactions

• Fe²⁺ + 2e⁻ → Fe (E° = -0.44 V)
• Zn²⁺ + 2e⁻ → Zn (E° = -0.76 V)

Step 2: Determine which would be oxidation vs. reduction

• For Zn to reduce Fe²⁺, we need:
  • Reduction: Fe²⁺ + 2e⁻ → Fe (E° = -0.44 V)
  • Oxidation: Zn → Zn²⁺ + 2e⁻ (E° = +0.76 V)

Step 3: Calculate cell potential

• E°cell = E°reduction + E°oxidation = -0.44 V + 0.76 V = +0.32 V

Step 4: Interpret result

• Since E°cell > 0, the reaction is spontaneous
• Yes, Zn metal can reduce Fe²⁺ to Fe metal

This cheat sheet provides a comprehensive overview of redox reactions and oxidation numbers. Remember that while these principles apply broadly, specific conditions (pH, temperature, concentration) can affect redox behavior. When solving redox problems, systematic application of these rules and careful tracking of electrons and oxidation numbers are key to success.