Oxidation-Reduction: Reactions, Rules & Applications

What is Oxidation and Reduction?

Oxidation and reduction always happen together in a reaction. They involve the loss or gain of electrons (or change in oxidation state).

Key Definitions

• Oxidation: A substance loses electrons (oxidation state increases).
  • Example: $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$ (Zinc loses 2 electrons).
• Reduction: A substance gains electrons (oxidation state decreases).
  • Example: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$ (Copper ion gains 2 electrons).

Oxidation State (Oxidation Number)

• The charge an atom would have if electrons were transferred completely.
• Rules to find oxidation states:
  1. Pure elements: Oxidation state = 0 (e.g., ${\text{O}}_2,\text{Na},{\text{Cl}}_2$).
  2. Monatomic ions: Oxidation state = charge (e.g., ${\text{Na}}^{+}=+1,{\text{Cl}}^{-}=-1$).
  3. Oxygen (O): Usually -2 (except in ${\text{O}}_2^{-}$ where it’s -1, or in peroxides like ${\text{H}}_2{\text{O}}_2$ where it’s -1).
  4. Hydrogen (H): +1 (except in metal hydrides like $\text{NaH}$ where it’s -1).
  5. Fluorine (F): Always -1 (most electronegative).
  6. Sum of oxidation states in a compound = 0 (neutral molecule).
  7. Sum of oxidation states in a polyatomic ion = charge of the ion.

Example: Find the oxidation state of S in ${\text{SO}}_4^{2-}$.

• Let oxidation state of S = $x$.
• Oxygen = -2 each, total for 4 O = $4\times (-2)=-8$.
• Total charge = -2, so: $x+(-8)=-2$ $x=+6$. Answer: Sulfur has +6 oxidation state in ${\text{SO}}_4^{2-}$.

Types of Redox Reactions

1. Combination Reactions

   • Two substances combine to form one product.
   • Example: $2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$
     • Mg is oxidized (0 → +2), O is reduced (+0 → -2).

2. Decomposition Reactions

   • One compound breaks into two or more substances.
   • Example: $2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2$
     • H is reduced (-1 → 0), O is oxidized (-2 → 0).

3. Displacement Reactions

   • A more reactive element displaces a less reactive one.
   • Example: $\text{Zn}+{\text{CuSO}}_4\to {\text{ZnSO}}_4+\text{Cu}$
     • Zn is oxidized (0 → +2), Cu²⁺ is reduced (+2 → 0).

4. Disproportionation Reactions

   • A single substance is both oxidized and reduced.
   • Example: $2{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}+{\text{O}}_2$
     • In ${\text{H}}_2{\text{O}}_2$, O has -1 oxidation state.
     • In ${\text{H}}_2\text{O}$, O is -2 (reduced).
     • In ${\text{O}}_2$, O is 0 (oxidized).

Balancing Redox Equations (Half-Reaction Method)

To balance redox equations, follow these steps:

graph LR
    A["Skeleton Equation"] --> B["Assign Oxidation States"]
    B --> C["Write Half-Reactions"]
    C --> D["Balance Electrons"]
    D --> E["Balance Atoms"]
    E --> F["Combine Half-Reactions"]
    F --> G["Final Balanced Equation"]

Step 1: Write the skeleton equation

Example: Balance ${\text{Fe}}^{2+}+{\text{Cr}}_2{\text{O}}_7^{2-}+{\text{H}}^{+}\to {\text{Fe}}^{3+}+{\text{Cr}}^{3+}+{\text{H}}_2\text{O}$.

Step 2: Assign oxidation states and identify changes

• Oxidation half: ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+e^{-}$ (Fe loses 1 electron).
• Reduction half: ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$ (Cr gains 3 electrons each).

Step 3: Balance electrons

• Multiply oxidation half by 6 to match electrons: $6{\text{Fe}}^{2+}\to 6{\text{Fe}}^{3+}+6e^{-}$.
• Reduction half remains: ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$.

Step 4: Combine and balance atoms

Add both halves: $6{\text{Fe}}^{2+}+{\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}\to 6{\text{Fe}}^{3+}+2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$.

Final balanced equation:

Oxidizing and Reducing Agents

Example: In $\text{Zn}+{\text{CuSO}}_4\to {\text{ZnSO}}_4+\text{Cu}$:

• Oxidizing agent: ${\text{Cu}}^{2+}$ (reduced to Cu).
• Reducing agent: Zn (oxidized to ${\text{Zn}}^{2+}$).

Applications of Redox Reactions

1. Metallurgy (Extracting Metals)
   • Example: Extraction of iron from ${\text{Fe}}_2{\text{O}}_3$ using coke (C) as a reducing agent. ${\text{Fe}}_2{\text{O}}_3+3\text{CO}\to 2\text{Fe}+3{\text{CO}}_2$.

2. Batteries (Electrochemical Cells)

   • Example: Dry cell battery (Zn-Cu cell):
     • Anode (Oxidation): $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$.
     • Cathode (Reduction): $2{\text{MnO}}_2+2{\text{NH}}_4^{+}+2e^{-}\to {\text{Mn}}_2{\text{O}}_3+2{\text{NH}}_3+{\text{H}}_2\text{O}$.

3. Corrosion (Rusting of Iron)

   • Iron reacts with oxygen and water: $4\text{Fe}+3{\text{O}}_2+6{\text{H}}_2\text{O}\to 4{\text{Fe(OH)}}_3$.
   • Prevention: Painting, galvanization (coating with Zn).

4. Bleaching

   • ${\text{Cl}}_2$ acts as an oxidizing agent to bleach fabrics. ${\text{Cl}}_2+{\text{H}}_2\text{O}\to \text{HCl}+\text{HOCl}$ (HOCl is the bleaching agent).

5. Respiration and Photosynthesis

   • Respiration: Glucose is oxidized to release energy. ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6+6{\text{O}}_2\to 6{\text{CO}}_2+6{\text{H}}_2\text{O}+\text{Energy}$.
   • Photosynthesis: ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ are reduced to form glucose.

Common Oxidizing and Reducing Agents

Exam Tip

1. Memorize oxidation state rules – Always double-check O, H, and F first.
2. Balance redox equations step-by-step – Use the half-reaction method in acidic/basic media.
3. Identify oxidizing/reducing agents – The one that gains electrons is the oxidizing agent; the one that loses electrons is the reducing agent.
4. Practice NEB-style questions – Focus on balancing equations and real-world applications (e.g., corrosion, batteries).
5. Watch for disproportionation – Some reactions involve the same element being oxidized and reduced (e.g., ${\text{H}}_2{\text{O}}_2$).

NEB Board-Style Questions

Short Answer Questions

1. Define oxidation and reduction with examples.
2. What is the oxidation state of sulfur in ${\text{H}}_2{\text{SO}}_4$?
3. Balance the following redox equation in acidic medium: ${\text{MnO}}_4^{-}+{\text{Fe}}^{2+}+{\text{H}}^{+}\to {\text{Mn}}^{2+}+{\text{Fe}}^{3+}+{\text{H}}_2\text{O}$.

Long Answer Questions

4. Explain the role of redox reactions in:
   • Extraction of copper from copper pyrites (${\text{CuFeS}}_2$).
   • Working of a lead-acid battery.
5. How does rusting of iron occur? Suggest two methods to prevent it.
6. Balance the following redox reaction and identify the oxidizing and reducing agents: ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7+{\text{FeSO}}_4+{\text{H}}_2{\text{SO}}_4\to {\text{K}}_2{\text{SO}}_4+{\text{Cr}}_2({\text{SO}}_4{)}_3+{\text{Fe}}_2({\text{SO}}_4{)}_3+{\text{H}}_2\text{O}$.

Summary

• Oxidation = Loss of electrons (increase in oxidation state).
• Reduction = Gain of electrons (decrease in oxidation state).
• Redox reactions involve both oxidation and reduction.
• Balance redox equations using the half-reaction method.
• Oxidizing agents gain electrons; reducing agents lose electrons.
• Applications include metallurgy, batteries, corrosion, and bleaching.

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