Understanding Redox Reactions: The Classic View

🔥 Oxidation = Gaining Oxygen or Electronegative Elements

Originally, oxidation meant adding oxygen to a substance. Examples:

  • \(2 \text{Mg} \, (\text{s}) + \, \text{O}_2 \, (\text{g}) \rightarrow 2 \text{MgO} \, (\text{s})\)
  • \(\text{S} \, (\text{s}) + \, \text{O}_2 \, (\text{g}) \rightarrow \, \text{SO}_2 \, (\text{g})\)
  • \(\text{CH}_4 (\text{g}) + 2\text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) + 2\text{H}_2\text{O} (\text{l})\)

Later, chemists expanded the definition:
Oxidation = Adding oxygen/electronegative elements OR Removing hydrogen/electropositive elements.
Example: \(2 \text{H}_2\text{S}(\text{g}) + \text{O}_2 (\text{g}) \rightarrow 2 \text{S} (\text{s}) + 2 \text{H}_2\text{O} (\text{l})\) (H₂S loses hydrogen!).

💫 Reduction = Losing Oxygen or Gaining Hydrogen

Reduction started as “removing oxygen” but now includes:
Reduction = Removing oxygen/electronegative elements OR Adding hydrogen/electropositive elements.
Examples:

  • \(2 \text{HgO} (\text{s}) \xrightarrow{\Delta} 2 \text{Hg} (\text{l}) + \text{O}_2 (\text{g})\) (HgO loses oxygen)
  • \(\text{CH}_2 = \text{CH}_2 (\text{g}) + \text{H}_2 (\text{g}) \rightarrow \text{H}_3\text{C} – \text{CH}_3 (\text{g})\) (Ethene gains hydrogen!)

⚡ The Golden Rule: Oxidation + Reduction = Redox!

They always happen together! 🔄 If one substance oxidizes, another reduces. Examples:

  • In \(2\text{HgCl}_2 (\text{aq}) + \text{SnCl}_2 (\text{aq}) \rightarrow \text{Hg}_2\text{Cl}_2 (\text{s}) + \text{SnCl}_4 (\text{aq})\):
    → SnCl₂ oxidizes (gains electronegative Cl)
    → HgCl₂ reduces (Hg⁺² becomes Hg₂⁺², which is like gaining mercury)

🧲 Redox = Electron Transfer Party!

Modern definition using electrons:

  • Oxidation = Loss of electrons (\(e^-\)) 👋
  • Reduction = Gain of electrons (\(e^-\)) 🤲
  • Oxidizing agent = Accepts \(e^-\) (gets reduced)
  • Reducing agent = Donates \(e^-\) (gets oxidized)

Example: \(2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}\)

\(2\text{Na} \rightarrow 2\text{Na}^+ + 2e^-\) (Oxidation: Na loses \(e^-\))
\(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\) (Reduction: Cl gains \(e^-\))

Even reactions without oxygen fit!
Example: \(2\text{Na} (\text{s}) + \text{H}_2 (\text{g}) \rightarrow 2\text{NaH} (\text{s})\)
→ Na loses \(e^-\) (oxidation), H gains \(e^-\) (reduction).

🎯 NEET Super-Shortlist!

  1. Oxidation/Reduction Definitions: Classic (O₂/H₂ addition/removal) and electron transfer views.
  2. Simultaneity: Redox reactions always involve paired oxidation/reduction.
  3. Agents Identification: Reducing agent = electron donor; Oxidizing agent = electron acceptor.
  4. Half-Reactions: Split redox equations into oxidation/reduction steps to track electrons.
  5. Ionic Compound Clues: Formation of ions (Na⁺Cl⁻, Na⁺H⁻) signals electron transfer.

💡 Practice Power

Problem: Identify oxidation/reduction here:
(i) \(\text{H}_2\text{S} (\text{g}) + \text{Cl}_2 (\text{g}) \rightarrow 2 \text{HCl} (\text{g}) + \text{S} (\text{s})\)
Answer: H₂S oxidizes (loses H, S forms), Cl₂ reduces (gains H to form HCl).

Problem: Is \(2\text{Na} + \text{H}_2 \rightarrow 2\text{NaH}\) redox?
Answer: Yes! Na → Na⁺ (loses \(e^-\)), H → H⁻ (gains \(e^-\)).