Redox Reactions

1. Oxidation State

• Defined as the hypothetical charge held by an atom if all its bonds to other atoms were completely ionic.
• When a bond exists between atoms A and B, the electron pair is shifted towards the more electronegative atom.

Structures and Oxidation States :

2. Special Oxidation States

• Carbon Suboxide (C₃O₂):

• CrO₅: Contains peroxy linkages.

• Caro's Acid (H₂SO₅): S = +6
• Marshal's Acid (H₂S₂O₈): S = +6
• For H₂S₂O₈, Average O.S of S = (+6 + 6) / 2 = +6

3. Rules Governing Oxidation State

i) Fluorine (F) :

• Most electronegative element; always -1 in all compounds.
• In F₂ (free state), it is 0.

ii) Oxygen (O) :

• In general oxides: -2.
• Peroxides (e.g., H₂O₂, Na₂O₂): -1.
• Superoxides (e.g., KO₂): -12.
• Ozonides (e.g., KO₃): -13.
• In O₂F₂: +1; In OF₂: +2.
• In O₂ molecule: 0.

iii) Hydrogen (H) :

• In general: +1.
• In metallic hydrides (e.g., NaH, KH): -1.

iv) Halogens (Cl, Br, I) :

• In general: -1.
• When attached to a more electronegative element, they show positive oxidation states (e.g., KClO₃ (+5), HIO₃ (+5), KBrO₃ (+5), HClO₄ (+7)).

v) Metals :

• Alkali metals (Li, Na, K, Rb, ...): +1.
• Alkaline earth metals (Be, Mg, Ca, ...): +2.
• Aluminium (Al): +3.

vi) General Rules :

• Free State/Allotropes: Oxidation number is 0 (e.g., Fe(s), Na(s), O₂(g)).
• Molecules: Sum of oxidation numbers of all atoms = 0.
• Ions: Sum of oxidation numbers of all atoms = charge on the ion.

4. Exceptions & Maximum Oxidation States

• For Sulphur, the maximum oxidation state is +6.
• H₂SO₅ : direct calculation 2 + x − 10 = 0 gives +8 (wrong, one peroxide O–O linkage); actual S = +6.
• H₂S₂O₈ : direct calculation 2 + 2x − 16 = 0 gives +7 (wrong, one peroxide linkage); actual S = +6.
• CrO₅ : direct calculation x − 10 = 0 gives +10 (wrong, two peroxide linkages); actual Cr = +6.

5. List of Common Ions

6. Metal Cations

7. Redox Reactions

• Redox Reaction : A reaction in which reduction and oxidation simultaneously take place.

8. Identifying Oxidising and Reducing Agents

• Oxidation (Oxⁿ) : Electron loss.
• Reduction (Redⁿ) : Electron gain.
• Substances are identified based on the oxidation state (O.S.) of the central atom:
  • If O.S. = Maximum O.S. → Can only act as an Oxidising Agent (OA).
  • If O.S. = Minimum O.S. → Can only act as a Reducing Agent (RA).
  • If O.S. = Intermediate → Can act as both RA and OA (can show disproportionation).
• Mnemonics:
  • The species that gets oxidised = RA.
  • The species that gets reduced = OA.

9. Combination Reaction

• Two or more substances combine to form a single product.
• One species gets oxidised and the other gets reduced.

Mg = Reducing agent ; O₂ = Oxidising agent

10. Decomposition Reaction

• A single compound breaks down into two or more simpler substances.
• Oxidation and reduction occur simultaneously.

For KClO₃: 1 + x + 3(−2) = 0 ⇒ x = +5

11. Displacement Redox Reaction

• A more reactive element replaces the less reactive element.
• a) Metal displacement:

• b) Non-metal displacement:

12. Disproportionation Reaction

• A single substance acts as both oxidising and reducing agent (the same element undergoes oxidation and reduction simultaneously).

• Note: One of the reactants in a disproportionation reaction always contains an element that can exist in at least three oxidation states.

Examples of disproportionation :

• X₂ + 2OH⁻(dil, cold) → X⁻ + XO⁻ + H₂O
• 3X₂ + 6OH⁻(conc, hot) → 5X⁻ + XO₃⁻ + 3H₂O
• S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O
• 4KClO₃ → 3KClO₄ + KCl

13. Comproportionation Reaction

• Two species containing the same element in different oxidation states react to form a product with an intermediate oxidation state.
• Examples:

NH₄⁺ (x+4=+1 ⇒ x=-3); NO₂⁻ (x-4=-1 ⇒ x=+3)

14. Disproportionation Reactions

• Oxidation state ranges for groups:

• A disproportionation reaction occurs when the same species is simultaneously oxidized and reduced.
• Only a species in an intermediate oxidation state can disproportionate: H₂O₂, ClO₃⁻, P₄, Cl₂, Cu⁺ and NO₂ can; Ag, F₂ and K⁺ cannot.

15. Steps for Balancing Redox Reactions

• Balance all atoms other than O and H.
• Find change in oxidation number for reduction and oxidation half-reactions and equate them.
• Balance the charge on both sides.
• Balance O and H atoms using H₂O.

16. Equivalents and n-Factor

17. Equivalent Concept

• An equivalent is the amount of a substance that actually takes part in a reaction.
• The n-factor depends on the quantity the substance gives or takes:
  • a) electrons
  • b) H⁺ ions
  • c) OH⁻ ions
  • d) charge

1. In Acid-Base Reactions :

• An equivalent is the amount of acid or base that gives 1 H⁺ or 1 OH⁻ ion, respectively.

2. In Redox Reactions :

• An equivalent is the amount of substance that gives or takes 1 electron.
• e.g., Fe²⁺ → Fe³⁺ + e⁻ (loses 1 electron) ⇒ 1 mol Fe²⁺ = 1 equivalent.
• e.g., Sn²⁺ → Sn⁴⁺ + 2e⁻ (loses 2 electrons) ⇒ 1 mol Sn²⁺ = 2 equivalents.

18. n-Factor Calculations for Acids

• For acids, n-factor is the basicity, defined as the total number of replaceable H⁺ ions.

19. Basicity and n-factor of Phosphorus Oxoacids

• For H₃PO₄: H₃PO₄ + NaOH → NaH₂PO₄ (n=1); H₃PO₄ + 2NaOH → Na₂HPO₄ (n=2); H₃PO₄ + 3NaOH → Na₃PO₄ (n=3).
• Note on structures:
  • H₃PO₃: P center bonded to one H (not acidic), one double-bonded O, and two OH groups.
  • H₃PO₂: P center bonded to two H atoms (not acidic), one double-bonded O, and one OH group.

20. n-factor Calculation for Bases

• n-factor corresponds to the number of replaceable OH⁻ ions or accepted H⁺ ions.

21. Salts (No Change in Oxidation State)

• For reactions where no atom undergoes a change in oxidation state, n = total cationic or total anionic charge.

• Example: CaCO₃ → CaO + CO₂. Charge on Ca²⁺ and CO₃²⁻ is 2, so n=2.
• Example: AgNO₃ + KCl → AgCl + KNO₃. n=1 for each.

22. n-factor for KMnO₄ (Redox)

• n = mol of electron transfer per mol of salt.

23. Salts with Two Atoms Undergoing Change in Oxidation State

• For salts where two types of atoms undergo a change in oxidation state (both are either oxidised or reduced):

Example : FeC₂O₄ (Oxidation)

• Fe⁺²C₂O₄^-2 → Fe⁺³ + CO₂
• Fe⁺² → Fe⁺³ (change = 1 × 1 = 1)
• C₂⁺³ → 2C⁺⁴ (change = 1 × 2 = 2)
• Total n-factor = 1 + 2 = 3.

24. n-factor in Redox Reactions

• Within one species, if electrons gained = electrons lost (intramolecular redox), n-factor = gain (= loss).
• If gain ≠ loss, n-factor = |gain − loss| (net electrons given to / taken from the other reactant).
• Example: KClO₃ Δ
  → KCl + O₂

25. n-factor in Disproportionation Reactions

• Example: 2H₂O₂ → 2H₂O + O₂

• 2 mol H₂O₂ transfer 2 mol e^-.
• Per mol transfer = 22 = 1.
• n-factor of H₂O₂ in this disproportionation = 1.

26. Example: Bromine Disproportionation

• Reaction: Br₂ → Br^- + BrO₃^-
• n-factor calculation:

Alternate Method (Net n-factor) :

• Example: 2H₂O₂ (O: −1) → 2H₂O (O: −2) + O₂ (O: 0)
• n₁ = 2, n₂ = 2.
• 1n_net = 12 + 12 = 1. n_net = 1.

27. I₂ and Na₂S₂O₃ Reaction

• Reaction: I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆
• For I₂: I₂ + 2e⁻ → 2I⁻ (n = 2). E₁ = M₁ / 2.
• For Na₂S₂O₃: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻. n = 1. E₂ = M₂ / 1 = M₂.

28. Normality (N)

• Definition: Number of equivalents of solute present in 1 L of solution.

• eq. of solute = Normality × volume of solⁿ (L)
• Normality = Molarity × n-factor

29. Relationship Formulas

30. Normality & Law of Equivalence

• For solutions:

• Law of Equivalence (LOE) : For any reaction A + B → C, according to LOE:

31. Law of Equivalents (LOE)

• In reactions, equivalents (or milliequivalents) react in stoichiometric proportions.
• For Acid 1 + Acid 2 + Base → Products:

• For R.A1 + O.A + R.A2 → Products:

32. General Stoichiometric Ratio

• For reaction: yA + xB → Products
• If n-factor ratio is x : y, then molar ratio is y : x.

33. Equivalent Weight of Metal

• When a metal forms an oxide, the Law of Equivalence states:

Note: Equivalent weight of Oxygen (O₂) = Mol. wtn-factor = 324 = 8.

34. Equivalent Weight of Metal (Chloride)

• When a metal reacts with a non-metal (like Cl₂ or N₂) to form a compound, the law of equivalence (LOE) states:

• For the reaction: metal + Cl₂ → metal chloride

35. Simple Titrations

• The aim is to find the concentration of an unknown solution using a solution of known concentration.
• Types: (a) Acid-base titrations, (b) Redox titrations.

Acid-Base Titration :

• Based on the reaction: HCl + NaOH → NaCl + H₂O
• At the end point, m.Eq of acid = m.Eq of base.

Redox Titration :

• Example: KMnO₄ + FeC₂O₄ → Mn²⁺ + Fe³⁺ + CO₂ (in acidic medium).
• n-factor for KMnO₄ = 5; n-factor for FeC₂O₄ = 3.

Volume of 0.5 M KMnO₄ for 100 mL of 0.1 M FeC₂O₄:

V = 0.1 × 100 × 30.5 × 5 = 12 mL

36. Titration of Copper(II) with Hypo

• Reaction of copper(II) with potassium iodide: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂
• Reaction of liberated iodine with hypo (sodium thiosulphate): I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

37. Double Titration

• Purpose : To find out the composition of an alkali mixture.

• Ph represents Phenolphthalein indicator.
• MeOR represents Methyl Orange indicator.

38. Titration with Phenolphthalein and Methyl Orange

• Phenolphthalein (Ph) indicator: Titration proceeds until Na₂CO₃ is converted to NaHCO₃ (n-factor = 1).
• Methyl Orange (MeOR) indicator: Titration proceeds until NaHCO₃ is converted to H₂CO₃ (n-factor = 1).

Indicator Reactions :

39. Quick Formula Revision