Chapter 7: Redox Reactions

Introduction

Overview of Redox Reactions

• Chemistry studies matter and its transformations via reactions.
• Redox reactions involve transfer of electrons, changing matter from one form to another.
• Redox processes occur in physical, biological, industrial, pharmaceutical, metallurgical, and agricultural fields.
• Environmental phenomena like hydrogen fuel use and ozone depletion involve redox reactions.

Solved Examples

• Ex 1: Identify if the reaction 2Mg + O₂ → 2MgO is redox.
  Step 1: Mg gains oxygen → oxidation.
  Step 2: O₂ is reduced.
  Conclusion: Both oxidation and reduction occur simultaneously, so it is a redox reaction.

Practice Set

• Q1: Which of the following is a redox reaction?
  A. NaCl dissolution in water
  B. Combustion of methane
  C. Melting of ice
  D. Boiling of water
• Q2: Redox reactions are important in which of the following fields?
  A. Metallurgy
  B. Agriculture
  C. Pharmaceuticals
  D. All of the above
• Q3: Which environmental issue involves redox reactions?
  A. Acid rain
  B. Ozone hole formation
  C. Global warming
  D. Soil erosion

Answer Key

• Q1: B - Combustion involves oxidation and reduction.
• Q2: D - Redox reactions are used extensively in all listed fields.
• Q3: B - Ozone hole formation is a redox phenomenon.

Memory Aids

• Redox = Reduction + Oxidation happening together.
• Remember: OIL RIG - Oxidation Is Loss, Reduction Is Gain (of electrons).
• Environmental redox: Think hydrogen fuel and ozone layer.

Classical Idea of Redox Reactions-Oxidation and Reduction Reactions

Oxidation

• Oxidation: Addition of oxygen or electronegative element to a substance.
• Or removal of hydrogen or electropositive element from a substance.
• Examples:
  • 2Mg_(s) + O_2(g) → 2MgO_(s)
  • 2H₂S_(g) + O_2(g) → 2S_(s) + 2H₂O_(l)
  • Mg_(s) + F_2(g) → MgF_2(s)
  • 2K₄[Fe(CN)₆]_(aq) + H₂O_2(aq) → 2K₃[Fe(CN)₆]_(aq) + 2KOH_(aq)

Reduction

• Reduction: Removal of oxygen or electronegative element from a substance.
• Or addition of hydrogen or electropositive element to a substance.
• Examples:
  • 

Redox Reactions

• Redox reactions involve simultaneous oxidation and reduction.
• Examples:
  • 
  • Oxidation: Loss of electrons (gain of oxygen or loss of hydrogen).
  • Reduction: Gain of electrons (loss of oxygen or gain of hydrogen).

Solved Examples

• Ex 1: Identify oxidation and reduction in 2Mg + O₂ → 2MgO.
  Step 1: Mg gains oxygen → oxidation.
  Step 2: O₂ loses oxygen → reduction.
• Ex 2: In Fe₃O₄ + Al → Al₂O₃ + Fe, identify oxidized and reduced species.
  Step 1: Fe gains electrons (reduction).
  Step 2: Al loses electrons (oxidation).

Practice Set

• Q1: Which is oxidation in classical terms?
  A. Addition of hydrogen
  B. Removal of oxygen
  C. Addition of oxygen
  D. Removal of electronegative element
• Q2: In the reaction FeCl₃ + H₂ → FeCl₂ + HCl, what happens to Fe?
  A. Oxidized
  B. Reduced
  C. No change
  D. Decomposed
• Q3: Which reaction is not a redox reaction?
  A. 2Mg + O₂ → 2MgO
  B. H₂ + Cl₂ → 2HCl
  C. NaCl dissolution in water
  D. Fe₃O₄ + Al → Al₂O₃ + Fe

Answer Key

• Q1: C - Oxidation is addition of oxygen.
• Q2: B - Fe is reduced by gaining electrons.
• Q3: C - Dissolution is not a redox reaction.

Memory Aids

• Oxidation: OIL - Oxidation Is Loss (of electrons).
• Reduction: RIG - Reduction Is Gain (of electrons).
• Oxidation = add oxygen or remove hydrogen.
• Reduction = remove oxygen or add hydrogen.

Redox Reactions Electron Transfer

Concepts

• Oxidation: Loss of electrons in half reactions.
• Reduction: Gain of electrons in half reactions.
• Oxidising agent (oxidant): Removes electrons from other substances and gets reduced itself.
• Example: Nitrate ion oxidises Cu to Cu²⁺:

  Cu_(s) + 4H⁺_(aq) + 2NO₃^-_(aq) 1 Cu²⁺_(aq) + 2H₂O_(l) + 2NO_2(g)

• Reducing agent (reductant): Donates electrons to reduce another substance and gets oxidised itself.
• Example: Sn²⁺ reduces Fe³⁺:

  Sn²⁺_(aq) + 2Fe³⁺_(aq) 1 Sn⁴⁺_(aq) + 2Fe²⁺_(aq)

• Example reaction: Sodium (Na) oxidised, Chlorine (Cl₂) reduced forming NaCl.

Solved Examples

• Ex 1: Identify oxidising and reducing agents in Cu + 4H⁺ + 2NO₃^- reaction.
  • Oxidising agent: NO₃^- (gains electrons, reduced)
  • Reducing agent: Cu (loses electrons, oxidised)
• Ex 2: In Sn²⁺ + 2Fe³⁺ reaction, Sn²⁺ is reducing agent, Fe³⁺ is oxidising agent.

Practice Set

• Q1: Which is oxidised in Cu + 4H⁺ + 2NO₃^- reaction?
  • A. Cu
  • B. NO₃^-
  • C. H⁺
  • D. H₂O
• Q2: Reducing agent donates:
  • A. Protons
  • B. Electrons
  • C. Neutrons
  • D. Photons
• Q3: In Sn²⁺ + 2Fe³⁺ reaction, Fe³⁺ is:
  • A. Oxidised
  • B. Reduced
  • C. Neither
  • D. Both

Answer Key

• Q1: A - Cu loses electrons (oxidised).
• Q2: B - Reducing agent donates electrons.
• Q3: B - Fe³⁺ gains electrons (reduced).

Memory Aids

• Oxidation = OIL (Oxidation Is Loss of electrons)
• Reduction = RIG (Reduction Is Gain of electrons)
• Oxidising agent gets reduced; reducing agent gets oxidised.

Competitive Electron Transfer Reactions

Concepts

• Zn reacts with Cu²⁺ solution: Zn oxidised to Zn²⁺, Cu²⁺ reduced to Cu metal.

  Zn_(s) + Cu²⁺_(aq) 1 Zn²⁺_(aq) + Cu_(s)

• Cu reacts with Ag⁺ solution: Cu oxidised to Cu²⁺, Ag⁺ reduced to Ag metal.

  Cu_(s) + 2Ag⁺_(aq) 1 Cu²⁺_(aq) + 2Ag_(s)

• Electron releasing tendency order: Zn > Cu > Ag.
• These reactions demonstrate displacement and electron transfer, basis for Galvanic cells.

Solved Examples

• Ex 1: Write half reactions for Zn + Cu²⁺ reaction.
  • Oxidation: Zn 1 Zn²⁺ + 2e^-
  • Reduction: Cu²⁺ + 2e^- 1 Cu
• Ex 2: Identify oxidising and reducing agents in Cu + 2Ag⁺ reaction.
  • Oxidising agent: Ag⁺
  • Reducing agent: Cu

Practice Set

• Q1: In Zn + Cu²⁺ reaction, which is oxidised?
  • A. Zn
  • B. Cu²⁺
  • C. Cu
  • D. Zn²⁺
• Q2: Electron releasing tendency order is:
  • A. Ag > Cu > Zn
  • B. Zn > Cu > Ag
  • C. Cu > Zn > Ag
  • D. Zn > Ag > Cu
• Q3: In Cu + 2Ag⁺ reaction, silver deposits because:
  • A. Ag is oxidised
  • B. Ag⁺ is reduced
  • C. Cu is reduced
  • D. Cu²⁺ is reduced

Answer Key

• Q1: A - Zn loses electrons (oxidised).
• Q2: B - Zn has highest electron releasing tendency.
• Q3: B - Ag⁺ gains electrons and deposits as Ag metal.

Memory Aids

• Electron releasing order: Zn > Cu > Ag (Zinc displaces copper, copper displaces silver).
• Displacement reactions show redox and electron transfer.
• Basis for Galvanic cells producing electrical energy.

Oxidation Number

Rules for Assigning Oxidation Number

• Oxidation number is the charge an atom appears to have in a compound.
• Elements in free state have oxidation number zero.
• For monoatomic ions, oxidation number equals the ion charge.
• Oxygen is usually -2; exceptions: peroxides (-1), superoxides (-1/2), OF₂ (+2), O₂F₂ (+1).
• Hydrogen is +1; in metal hydrides, it is -1.
• Fluorine is always -1 except in HOF (+1).
• Sum of oxidation numbers in a neutral compound is zero.
• Sum of oxidation numbers in a polyatomic ion equals the ion charge.
• Fractional oxidation state: average oxidation state when atoms have different states.
• Stock notation: oxidation state shown by Roman numerals in parentheses, e.g., iron(II) sulfate for FeSO₄.
• Oxidation: increase in oxidation number; Reduction: decrease in oxidation number.
• Oxidising agent: increases oxidation number (oxidant).
• Reducing agent: decreases oxidation number (reductant).
• Redox reactions: involve change in oxidation numbers.

Solved Examples

Ex 1: Find oxidation number of Fe in [Fe(H₂O)₅(NO)]SO₄.

• Let Fe = x.
• 5 waters: 5 × 0 = 0, NO = +1, SO₄ = -2.
• Equation: x + 0 + 1 - 2 = 0 → x - 1 = 0 → x = +1.

Ex 2: Oxidation number of Fe in Fe₃O₄.

• Let Fe = x.
• 3x + 4(-2) = 0 → 3x - 8 = 0 → x = +8/3 (fractional state).
• Average of FeO (+2) and Fe₂O₃ (+3).

Practice Set

• Q1: What is the oxidation number of oxygen in peroxides?
  • A. -2
  • B. -1
  • C. -1/2
  • D. +2
• Q2: In FeSO₄ (iron(II) sulfate), the oxidation state of Fe is:
  • A. +1
  • B. +2
  • C. +3
  • D. 0
• Q3: Which of the following is a reducing agent?
  • A. MnO₄^-
  • B. Zn
  • C. O₂
  • D. F₂

Answer Key

• Q1: B. Oxygen in peroxides has oxidation number -1.
• Q2: B. Fe in FeSO₄ is +2 (iron(II) sulfate).
• Q3: B. Zn is a reducing agent as it lowers oxidation number of other elements.

Memory Aids

• Oxygen exceptions: peroxides (-1), superoxides (-1/2), OF₂ (+2).
• Hydrogen is +1 except in metal hydrides (-1).
• Fluorine always -1 except in HOF (+1).
• Stock notation uses Roman numerals for metal oxidation states.

Oxidation Number of Important Compounds

Key Examples

• C in C₃O₂ (carbon suboxide): Two carbons +2, middle carbon 0.
• Br in Br₃O₄ (tribromooctaoxide): Two Br +6, one Br +4.
• S in S₄O₆^2- (tetrathionate ion): Two S +5, two S 0.
• Cl in Ca(OCl)Cl: Cl in OCl^- is +1; other Cl is -1.
• Cr in CrO₅: Cr is +6; peroxo oxygen -1, normal oxygen -2.

Solved Examples

Ex 1: Calculate oxidation number of Cl in Ca(OCl)Cl.

• In OCl^-, Cl = y; -2 + y = -1 → y = +1.
• Sum: +2 + (-2) + 1 + x = 0 → x = -1 (other Cl).

Practice Set

• Q1: Oxidation state of Cr in CrO₅ is:
  • A. +4
  • B. +5
  • C. +6
  • D. +7
• Q2: In S₄O₆^2-, how many S atoms have zero oxidation state?
  • A. 1
  • B. 2
  • C. 3
  • D. 4
• Q3: In Br₃O₄, the oxidation state of the middle Br atom is:
  • A. +4
  • B. +5
  • C. +6
  • D. 0

Answer Key

• Q1: C. Cr in CrO₅ is +6 due to peroxo bonds.
• Q2: B. Two S atoms have zero oxidation state in S₄O₆^2-.
• Q3: A. Middle Br in Br₃O₄ is +4.

Memory Aids

• CrO₅ has peroxo bonds; peroxo oxygen is -1.
• Tetrathionate ion has mixed S oxidation states (+5 and 0).
• Br in Br₃O₄ varies: outer +6, middle +4.

Types of Redox Reactions

Combination Reactions

• Two atoms/molecules combine to form one compound.
• At least one reactant is in elemental form.
• Example: H₂ + Cl₂ → 2HCl (H: 0 to +1, Cl: 0 to -1).
• Example: H₂ + O₂ → 2H₂O (H: 0 to +1, O: 0 to -2).

Decomposition Reactions

• Compound breaks down into two or more products.
• Reverse of combination reaction.
• For redox, one product must be elemental form.
• Example: 2H₂O → 2H₂ + O₂ (on heating).
• Example: 2LiNO₃ → Li₂O + 2NO₂ + O₂ (on heating).

Displacement Reactions

• One atom/ion replaces another in a compound: AB + C → AC + B.
• Metal displacement: metal replaces metal ion.
• Example: TiCl₄ + Mg → Ti + MgCl₂ (Ti: +4 to 0, Mg: 0 to +2).
• Example: CuSO₄ + Zn → Cu + ZnSO₄ (Cu: +2 to 0, Zn: 0 to +2).

• Non-metal displacement: non-metal replaces metal or non-metal ion.
• Example: Zn + H₂SO₄ → ZnSO₄ + H₂ (Zn oxidized, H reduced).
• Example: Al + H₂O → Al₂O₃ + H₂ (Al oxidized, H reduced).
• Example: Cl₂ + KBr → KCl + Br₂ (Cl reduced, Br oxidized).

Disproportionation Reactions

• Single species is simultaneously oxidized and reduced.
• Element must have at least three variable oxidation states.
• Example: 3ClO₂^- → 2ClO₃^- + Cl^- (Cl: +3 to +5 and -1).
• Reverse is comproportionation: two reactants with different oxidation states form product with same oxidation state.
• Example: Pb + PbO₂ + H₂SO₄ → PbSO₄ + H₂O (lead storage battery).

Practice Set

• Q1: Which reaction is a combination reaction?
  • A. 2H₂ + O₂ → 2H₂O
  • B. 2H₂O → 2H₂ + O₂
  • C. Zn + CuSO₄ → ZnSO₄ + Cu
  • D. 3ClO₂^- → 2ClO₃^- + Cl^-
• Q2: In which reaction does disproportionation occur?
  • A. Zn + H₂SO₄ → ZnSO₄ + H₂
  • B. 3ClO₂^- → 2ClO₃^- + Cl^-
  • C. TiCl₄ + Mg → Ti + MgCl₂
  • D. 2LiNO₃ → Li₂O + 2NO₂ + O₂
• Q3: In the reaction TiCl₄ + Mg → Ti + MgCl₂, the oxidation state of Ti changes from:
  • A. 0 to +4
  • B. +4 to 0
  • C. +2 to +3
  • D. 0 to -2

Answer Key

• Q1: A. Combination reaction forms one product from elements.
• Q2: B. Disproportionation involves simultaneous oxidation and reduction of same species.
• Q3: B. Ti reduces from +4 to 0 by gaining electrons.

Memory Aids

• Combination: elements combine to form compound.
• Decomposition: compound breaks into elements or simpler compounds.
• Displacement: one element replaces another.
• Disproportionation: same element oxidized and reduced.

Balancing of Redox Reactions

Oxidation Number Method

• Write correct formulas of reactants and products.
• Assign oxidation numbers to all atoms.
• Identify atoms with change in oxidation number.
• Calculate total increase and decrease in oxidation numbers; equalize by multiplying.
• Add H⁺ (acidic) or OH^- (basic) to balance charges.
• Balance hydrogen by adding H₂O; check oxygen atoms.

Half Reaction Method

• Write ionic form of reaction.
• Separate into oxidation and reduction half reactions.
• Balance atoms except O and H in each half.
• In acidic medium: add H₂O for O, H⁺ for H; in basic medium: add H₂O and OH^- accordingly.
• Add electrons to balance charge; multiply half reactions to equalize electrons.
• Add half reactions; cancel electrons to get balanced equation.

Practice Set

• Q1: In acidic medium, to balance oxygen atoms in half reaction, add:
  • A. H₂O
  • B. OH^-
  • C. H⁺
  • D. Electrons
• Q2: In half reaction method, electrons are added to:
  • A. Oxidation half only
  • B. Reduction half only
  • C. Both halves to balance charge
  • D. Neither half
• Q3: Which step is NOT part of oxidation number method?
  • A. Assign oxidation numbers
  • B. Add H⁺ or OH^-
  • C. Separate into half reactions
  • D. Balance hydrogen and oxygen

Answer Key

• Q1: A. H₂O is added to balance oxygen in acidic medium.
• Q2: C. Electrons are added to balance charge in both half reactions.
• Q3: C. Separating into half reactions is not part of oxidation number method.

Memory Aids

• Oxidation number method focuses on changes in oxidation states.
• Half reaction method splits reaction into oxidation and reduction parts.
• Use H⁺ in acidic and OH^- in basic medium for balancing.

Redox Reactions as the Basis for Titrations

Redox Titrations

• Used to determine strength of oxidants or reductants.
• Indicators show end point by color change.
• Some reagents (e.g., MnO₄^-) act as self-indicators due to intense color.
• Other indicators (e.g., diphenylamine) change color when oxidized at equivalence point.
• Starch is used as indicator for reactions involving I^- or I₂, giving intense blue color.

Redox Reactions and Electrode Processes

Key Concepts

• Redox couple: Oxidised and reduced forms of a substance in a half-reaction, e.g., Zn²⁺/Zn, Cu²⁺/Cu in Daniell cell.
• Salt bridge: Completes circuit, allows ion flow without mixing solutions, maintains electrical neutrality.
• Electrode potential: Potential difference at metal/ion interface; tendency to gain or lose electrons.
• Standard electrode potential (EB0): Electrode potential at 1 M concentration, 298 K; hydrogen electrode EB0 = 0.00 V.
• Electrochemical series: List of reducing agents in decreasing strength; negative EB0 means stronger reducing agent than H⁺/H₂.

Setup of Daniell cell: Zn rod in ZnSO₄, Cu rod in CuSO₄. Zn oxidises to Zn²⁺, electrons flow to Cu cathode where Cu²⁺ reduces to Cu metal. Salt bridge maintains neutrality. Electric current flows through external circuit.

Standard reduction potentials table ranks substances by electron gain ability. Negative EB0: stronger oxidising agent; positive EB0: stronger reducing agent.

Solved Examples

• Ex 1: Identify anode and cathode in Daniell cell.
  - Anode: Zn (oxidation)
  - Cathode: Cu (reduction)
• Ex 2: Calculate equivalent mass of oxidising agent if change in oxidation state is 2 and molar mass is 80.
  - Equivalent mass = 80 / 2 = 40

Practice Set

• Q1: What is the role of salt bridge in an electrochemical cell?
  A. Prevents electron flow
  B. Maintains electrical neutrality
  C. Acts as anode
  D. Acts as cathode
• Q2: Which has the highest standard electrode potential?
  A. Zn²⁺/Zn
  B. Cu²⁺/Cu
  C. H⁺/H₂
  D. Li⁺/Li
• Q3: In Daniell cell, which species is reduced?
  A. Zn²⁺
  B. Cu²⁺
  C. Zn
  D. Cu

Answer Key

• Q1: B - Salt bridge maintains electrical neutrality by allowing ion flow.
• Q2: B - Cu²⁺/Cu has higher EB0 than Zn and Li; H⁺/H₂ is zero by convention.
• Q3: B - Cu²⁺ gains electrons and is reduced to Cu metal.

Memory Aids

• Redox couple = oxidised + reduced forms.
• Salt bridge = ion flow, no mixing.
• Electrode potential: tendency to gain/lose electrons.
• Standard hydrogen electrode EB0 = 0.00 V.
• Electrochemical series: negative EB0 = strong reducing agent.

Predicting Reducing and Oxidising Power

Key Concepts

• Lower electrode potential → stronger reducing agent.
• Order of reducing agents: Li > Mg > Al > Zn > Fe > Co > Ni > Sn > Pb > Cu > I^- > Fe²⁺ > Hg > Ag > Br^- > Cl^- > F^-
• Higher electrode potential → stronger oxidising agent.
• Order of oxidising agents: F₂ > Cl₂ > Br₂ > I₂
• Fluorine is the strongest oxidising agent.
• Lithium is the strongest reducing agent due to high hydration energy.
• Salt bridge electrolytes must have cations and anions with similar ionic mobilities.
• Equivalent mass of oxidising/reducing agent = Molar mass / Change in oxidation state per mole.

Solved Examples

• Ex 1: Identify the strongest reducing agent among Li, Cu, and Ag.
  - Li has lowest electrode potential → strongest reducing agent.
• Ex 2: Arrange F₂, Cl₂, Br₂ in order of oxidising power.
  - F₂ > Cl₂ > Br₂

Practice Set

• Q1: Which is the strongest oxidising agent?
  A. Cl₂
  B. F₂
  C. Br₂
  D. I₂
• Q2: Which element is the strongest reducing agent?
  A. Cu
  B. Li
  C. Ag
  D. Hg
• Q3: What is the equivalent mass of an oxidising agent with molar mass 96 and oxidation state change 3?
  A. 32
  B. 48
  C. 96
  D. 288

Answer Key

• Q1: B - Fluorine (F₂) is the strongest oxidising agent.
• Q2: B - Lithium (Li) is the strongest reducing agent due to low electrode potential and high hydration energy.
• Q3: A - Equivalent mass = 96 / 3 = 32.

Memory Aids

• Reducing power ↑ as electrode potential ↓.
• Oxidising power ↑ as electrode potential ↑.
• Fluorine = strongest oxidiser.
• Lithium = strongest reducer.
• Equivalent mass = molar mass / change in oxidation state.