Chapter 7: Redox Reactions

Introduction

Overview of Redox Reactions

• Chemistry studies matter and its transformations via chemical reactions.

• Redox reactions involve transfer of electrons, changing one substance into another.

• They occur in physical, biological, industrial, pharmaceutical, metallurgical, and agricultural processes.

• Environmental phenomena like hydrogen fuel use and ozone depletion involve redox processes.

Solved Examples

• Ex 1: Identify if the reaction 2Mg + O2 → 2MgO is redox.
  Step 1: Mg gains oxygen → oxidation.
  Step 2: O2 gains electrons → reduction.
  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. 2Mg + O2 → 2MgO
  C. H2O boiling
  D. NaCl crystallization

• Q2: Redox reactions are important in which of the following fields?
  A. Metallurgy
  B. Textile dyeing
  C. Polymerization
  D. None of these

• Q3: Which environmental issue involves redox reactions?
  A. Acid rain
  B. Ozone hole formation
  C. Soil erosion
  D. Global warming

Answer Key

• Q1: B - It involves simultaneous oxidation and reduction.

• Q2: A - Metallurgy extensively uses redox reactions.

• 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 "Ozone hole" and "Hydrogen fuel" as key examples.

Classical Idea of Redox Reactions-Oxidation and Reduction Reactions

Oxidation

• Addition of oxygen or electronegative element to a substance.

• Removal of hydrogen or electropositive element from a substance.

• Examples:
  2Mg(s) + O2(g) → 2MgO(s)
  2H2S(g) + O2(g) → 2S(s) + 2H2O(l)
  Mg(s) + F2(g) → MgF2(s)
  2K4[Fe(CN)6](aq) + H2O2(aq) → 2K3[Fe(CN)6](aq) + 2KOH(aq)

Reduction

• Removal of oxygen or electronegative element from a substance.

• Addition of hydrogen or electropositive element to a substance.

• Examples:
  

  

  • HgO(s) heated → Hg(l) + O2(g)

  • FeCl3 + H2 → FeCl2 + HCl

  • CH2=CH2 + H2 → C2H6

  • Hg2Cl2 + SnCl2 → Hg2Cl2 + SnCl4

Redox Reactions

• Reactions where oxidation and reduction occur simultaneously.

• Examples:
  

  

  • Fe3O4 + Al → Fe + Al2O3

  • H2S + Cl2 → S + HCl

• 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 gains electrons → reduction.

• Ex 2: In Fe3O4 + 8Al → 4Al2O3 + 3Fe, identify oxidized and reduced species.
  Step 1: Fe3O4 loses oxygen → reduction.
  Step 2: Al gains oxygen → oxidation.

Practice Set

• Q1: Which is oxidation in 2Mg + O2 → 2MgO?
  A. Mg loses oxygen
  B. Mg gains oxygen
  C. O2 loses electrons
  D. Mg loses electrons

• Q2: In the reaction Fe3O4 + Al → Fe + Al2O3, which is reduced?
  A. Fe3O4
  B. Al
  C. Fe
  D. Al2O3

• Q3: Which statement is correct about redox reactions?
  A. Only oxidation occurs
  B. Only reduction occurs
  C. Oxidation and reduction occur simultaneously
  D. Neither oxidation nor reduction occurs

Answer Key

• Q1: B - Mg gains oxygen, so it is oxidized.

• Q2: C - Fe is reduced by gaining electrons.

• Q3: C - Redox reactions involve simultaneous oxidation and reduction.

Memory Aids

• Oxidation: Add oxygen or remove hydrogen.

• Reduction: Remove oxygen or add hydrogen.

• Remember "OIL RIG" for electron transfer: Oxidation Is Loss, Reduction Is Gain.

• Redox reactions always have both oxidation and reduction steps.

Redox Reactions Electron Transfer

Key Concepts

• Oxidation: Loss of electrons in a half reaction.

• Reduction: Gain of electrons in a half reaction.

• Oxidising agent: Substance that gains electrons and gets reduced; causes oxidation of other substances.

• Reducing agent: Substance that loses electrons and gets oxidised; causes reduction of other substances.

• Example reaction with oxidising agent (nitrate ion):
  Cu(s) + 4H+(aq) + 2NO3−(aq) → Cu2+(aq) + 2H2O(l) + 2NO2(g)

• Example reaction with reducing agent (tin(II)):
  Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)

Solved Examples

• Ex 1: Identify oxidising and reducing agents in Na + Cl2 → NaCl.
  - Na loses electrons (oxidised) → reducing agent.
  - Cl2 gains electrons (reduced) → oxidising agent.

Practice Set

• Q1: Which species is oxidised in Cu + 2Ag+ → Cu2+ + 2Ag?
  A. Cu
  B. Ag+
  C. Cu2+
  D. Ag

• Q2: In the reaction Zn + Cu2+ → Zn2+ + Cu, what is the reducing agent?
  A. Zn
  B. Cu2+
  C. Zn2+
  D. Cu

• Q3: Which ion acts as oxidising agent in Cu + 4H+ + 2NO3− → Cu2+ + 2H2O + 2NO2?
  A. Cu
  B. H+
  C. NO3−
  D. NO2

Answer Key

• Q1: A. Cu is oxidised (loses electrons) forming Cu2+.

• Q2: A. Zn is reducing agent as it loses electrons and oxidises.

• Q3: C. NO3− is oxidising agent as it gains electrons and gets 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

Key Concepts

• Metal with higher electron releasing tendency displaces metal ion with lower tendency from solution.

• Zn + Cu2+ → Zn2+ + Cu: Zinc oxidised, copper ion reduced.

• Cu + 2Ag+ → Cu2+ + 2Ag: Copper oxidised, silver ion reduced.

• Electron releasing tendency order: Zn > Cu > Ag.

• Basis for galvanic cells: chemical reactions producing electrical energy via electron transfer.

Solved Examples

• Ex 1: Predict if Zn will displace Ag+ from solution.
  - Zn > Ag in electron releasing tendency → Zn displaces Ag+.

• Ex 2: Write half reactions for Cu + 2Ag+ → Cu2+ + 2Ag.
  - Oxidation: Cu → Cu2+ + 2e−
  - Reduction: 2Ag+ + 2e− → 2Ag

Practice Set

• Q1: Which metal will displace Cu2+ from solution?
  A. Ag
  B. Zn
  C. Cu
  D. Au

• Q2: In Cu + 2Ag+ → Cu2+ + 2Ag, which is oxidised?
  A. Cu
  B. Ag+
  C. Cu2+
  D. Ag

• Q3: Arrange metals in increasing order of electron releasing tendency:
  A. Ag < Cu < Zn
  B. Zn < Cu < Ag
  C. Cu < Zn < Ag
  D. Ag < Zn < Cu

Answer Key

• Q1: B. Zn displaces Cu2+ due to higher electron releasing tendency.

• Q2: A. Cu is oxidised, losing electrons to Ag+.

• Q3: A. Ag < Cu < Zn is correct order of electron releasing tendency.

Memory Aids

• Remember: Zn > Cu > Ag for electron releasing tendency.

• Higher metal in series displaces lower metal ions from solution.

• Galvanic cells use these displacement reactions for electricity.

Oxidation Number

Rules for Assigning Oxidation Numbers

• Oxidation number (ON) is the charge an atom appears to have in a compound.

• ON of elements in free state = 0.

• ON of monoatomic ion = charge on the ion.

• Oxygen ON = -2 usually; exceptions: peroxides (-1), superoxides (-1/2), OF2 (+2), O2F2 (+1).

• Hydrogen ON = +1; in metal hydrides = -1.

• Fluorine ON = -1; exception: HOF (+1).

• Sum of ONs in neutral compound = 0; in polyatomic ion = ion charge.

• Fractional ON is average when atoms have different ONs in a compound.

• Stock notation uses Roman numerals to indicate metal ON, e.g., FeSO4 = iron (II) sulfate.

• Oxidation: increase in ON; Reduction: decrease in ON.

• Oxidising agent: increases ON of element (oxidant).

• Reducing agent: decreases ON of element (reductant).

• Redox reactions involve change in ON of species.

Solved Examples

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

Let Fe = x; H2O = 0, NO = +1, SO4 = -2

x + 5(0) + 1 - 2 = 0 ⇒ x - 1 = 0 ⇒ x = +1

Ex 2: ON of Fe in Fe3O4.

3x + 4(-2) = 0 ⇒ 3x - 8 = 0 ⇒ x = +8/3 (fractional ON)

Practice Set

• Q1: ON of oxygen in H2O2 is:

  • A) -2

  • B) -1

  • C) -1/2

  • D) 0

• Q2: In FeSO4, the ON of Fe is:

  • A) +1

  • B) +2

  • C) +3

  • D) +6

• Q3: ON of Cr in CrO5 is:

  • A) +4

  • B) +5

  • C) +6

  • D) +7

Answer Key

• Q1: B) -1; oxygen in peroxide has ON -1.

• Q2: B) +2; Fe in FeSO4 is iron (II) sulfate.

• Q3: C) +6; CrO5 has two peroxo bonds, Cr ON = +6.

Memory Aids

• Oxygen exceptions: peroxides (-1), superoxides (-1/2), OF2 (+2).

• Hydrogen exceptions: metal hydrides (-1).

• Fluorine always -1 except in HOF (+1).

• Stock notation: Roman numerals indicate metal ON.

Oxidation Number of Important Compounds

Key Examples

• C in C3O2: Two carbons have +2 ON, middle carbon 0.

• 

• Br in Br3O8: Two Br atoms +6 ON, one Br +4 ON.

• 

• S in S4O62−: Two S atoms +5 ON, two S atoms 0.

• 

• Cl in Ca(OCl)Cl: Cl in OCl− is +1; overall Cl ON in compound is -1.

• Cr in CrO5: Cr ON is +6 due to peroxo bonds.

• 

Solved Examples

Ex 1: Determine ON of C in C3O2.

Two end carbons bonded to O have +2 ON; middle carbon is 0.

Ex 2: ON of Br in Br3O8.

Two Br atoms have +6 ON; one Br atom has +4 ON.

Practice Set

• Q1: ON of S in S4O62− for two S atoms is:

  • A) 0

  • B) +2

  • C) +5

  • D) +6

• Q2: ON of Cl in OCl− is:

  • A) -1

  • B) 0

  • C) +1

  • D) +2

• Q3: ON of Fe in [Fe(H2O)5(NO)]SO4 is:

  • A) 0

  • B) +1

  • C) +2

  • D) +3

Answer Key

• Q1: C) +5; two S atoms have +5 ON in tetrathionate ion.

• Q2: C) +1; Cl in hypochlorite ion OCl− is +1.

• Q3: B) +1; Fe ON calculated as +1 in complex.

Memory Aids

• In polyatomic ions, sum of ON = ion charge.

• Peroxo oxygen ON = -1; normal oxygen ON = -2.

• Average ON used when multiple oxidation states present.

Types of Redox Reactions

Classification and Characteristics

• Combination reactions: Two or more reactants combine; at least one reactant in elemental form.

  • Example: H2 + Cl2 → 2HCl (H: 0→+1, Cl: 0→-1)

  • 

• Decomposition reactions: Compound breaks into two or more products; one product must be elemental for redox.

  • Example: 2H2O → 2H2 + O2 (H: +1→0, O: -2→0)

  • 

• Displacement reactions: An atom/ion replaces another in a compound.

  • Metal displacement: TiCl4 + 2Mg → Ti + 2MgCl2

  • 

  • Non-metal displacement: Cl2 + 2KBr → 2KCl + Br2

  • 

• Disproportionation reactions: Single species is simultaneously oxidized and reduced.

  • Example: 3ClO2− → 2ClO3− + Cl−

  • 

• Comproportionation reactions: Reverse of disproportionation; reactants with same element in different ON form product with same ON.

  • Example: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O (lead storage battery)

  • 

Solved Examples

Ex 1: Identify type of reaction: Zn + H2SO4 → ZnSO4 + H2.

Non-metal displacement; Zn oxidized, H reduced.

Ex 2: Reaction in lead-acid battery.

Comproportionation; Pb and PbO2 react to form PbSO4.

Practice Set

• Q1: Which is a disproportionation reaction?

  • A) 2H2O → 2H2 + O2

  • B) 3ClO2− → 2ClO3− + Cl−

  • C) Zn + H2SO4 → ZnSO4 + H2

  • D) TiCl4 + 2Mg → Ti + 2MgCl2

• Q2: In which reaction does a metal displace another metal?

  • A) CuSO4 + Zn → ZnSO4 + Cu

  • B) Cl2 + 2KBr → 2KCl + Br2

  • C) 2H2O → 2H2 + O2

  • D) Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O

• Q3: Which reaction is a combination reaction?

  • A) H2 + Cl2 → 2HCl

  • B) 2H2O → 2H2 + O2

  • C) Zn + H2SO4 → ZnSO4 + H2

  • D) 3ClO2− → 2ClO3− + Cl−

Answer Key

• Q1: B) Disproportionation; ClO2− both oxidized and reduced.

• Q2: A) Metal displacement; Zn displaces Cu.

• Q3: A) Combination; H2 and Cl2 combine.

Memory Aids

• Combination: reactants combine; decomposition: compound breaks down.

• Displacement: one element replaces another.

• Disproportionation: same species oxidized and reduced.

• Comproportionation: species with different ON form same ON product.

Balancing of Redox Reactions

Oxidation Number Method

• Write correct formulas for reactants and products.

• Assign oxidation numbers to all elements.

• Identify atoms with ON changes.

• Calculate total increase and decrease in ON; multiply to equalize.

• Add H+ (acidic) or OH− (basic) to balance charges.

• Add H2O to balance hydrogen and 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.

• Balance O by adding H2O; balance H by adding H+ (acidic) or H2O and OH− (basic).

• Add electrons to balance charge in each half.

• Multiply half reactions to equalize electrons.

• Add half reactions and cancel electrons.

Solved Examples

Ex 1: Balance MnO4− + Fe2+ → Mn2+ + Fe3+ in acidic medium using half reaction method.

Oxidation: Fe2+ → Fe3+ + e−

Reduction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

Multiply oxidation by 5, add and cancel electrons.

Practice Set

• Q1: In acidic medium, balance: Fe2+ + Cr2O72− → Fe3+ + Cr3+.

• Q2: Balance in basic medium: SO32− + I2 → SO42− + I−.

• Q3: Using oxidation number method, balance: H2S + Cl2 → S + HCl.

Answer Key

• Q1: 6Fe2+ + Cr2O72− + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

• Q2: SO32− + I2 + 2OH− → SO42− + 2I− + H2O

• Q3: H2S + Cl2 → S + 2HCl

Memory Aids

• Half reaction method: separate oxidation and reduction, balance atoms and charge, combine.

• Oxidation number method: balance ON changes, then atoms and charges.

• Use H+ in acidic, OH− in basic medium for balancing.

Redox Reactions as the Basis for Titrations

Redox Titrations and Indicators

• 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 or reduced at equivalence point.

• Starch used as indicator for iodine-based titrations; forms intense blue complex with I2.

Solved Examples

Ex 1: In KMnO4 titration, color change from purple to colorless indicates end point.

Ex 2: Diphenylamine indicator turns blue when oxidized by Cr2O72− at equivalence point.

Practice Set

• Q1: Which reagent acts as self-indicator in redox titration?

  • A) KMnO4

  • B) Diphenylamine

  • C) Starch

  • D) Phenolphthalein

• Q2: Starch indicator is used to detect:

  • A) Cu2+

  • B) I2

  • C) S2O32−

  • D) Cr2O72−

• Q3: Diphenylamine indicator changes color due to:

  • A) Reduction

  • B) Oxidation

  • C) Acid-base reaction

  • D) Precipitation

Answer Key

• Q1: A) KMnO4; purple color acts as self-indicator.

• Q2: B) I2; starch forms blue complex with iodine.

• Q3: B) Oxidation; diphenylamine oxidized to show color change.

Memory Aids

• KMnO4 is self-indicator (purple color).

• Starch + I2 = intense blue color.

• Diphenylamine changes color upon oxidation.

Redox Reactions and Electrode Processes

Key Concepts

• Redox couple: Contains oxidised and reduced forms of a substance in a half-reaction, e.g., Zn2+/Zn and Cu2+/Cu in Daniell cell.

• Salt bridge: Connects two solutions electrically without mixing them; maintains electrical neutrality.

• Electrode potential: Potential difference at metal/ion interface; indicates tendency to gain or lose electrons.

• Standard electrode potential (E°): Electrode potential measured at 1 M concentration and 298 K; hydrogen electrode is 0.00 V by convention.

• Electrochemical series: List of reducing agents arranged by decreasing strength (activity series).

• Negative E° means stronger reducing agent than H+/H2 couple; positive E° means weaker.

• Only electrolytes with cations and anions of similar ionic mobility are used in salt bridges.

• Equivalent mass of oxidising/reducing agent = (Equivalent mass) / (Change in oxidation state per mole).

Setup of Daniell Cell

• Zinc rod (anode) in ZnSO4 solution; copper rod (cathode) in CuSO4 solution.

• Zinc oxidises: Zn → Zn2+ + 2e−.

• Electrons flow from zinc to copper through external circuit.

• Copper ions reduce: Cu2+ + 2e− → Cu.

• Salt bridge (K2SO4) maintains charge balance by ion flow.

• Electron flow produces electric current; voltmeter measures voltage.

• Chemical energy converts to electrical energy.

Solved Examples

Ex 1: Write the half-reactions at anode and cathode in Daniell cell.

• Anode (oxidation): Zn → Zn2+ + 2e−

• Cathode (reduction): Cu2+ + 2e− → Cu

Practice Set

• Q1: What is the role of the salt bridge in an electrochemical cell?
  A) To allow mixing of solutions
  B) To complete the circuit and maintain electrical neutrality
  C) To increase voltage
  D) To act as an electrode

• Q2: In Daniell cell, which electrode acts as the cathode?
  A) Zinc
  B) Copper
  C) Salt bridge
  D) Hydrogen

• Q3: Standard electrode potential of hydrogen is:
  A) +0.34 V
  B) 0.00 V
  C) -0.76 V
  D) +1.23 V

Answer Key

• Q1: B - Salt bridge completes the circuit and maintains electrical neutrality without mixing solutions.

• Q2: B - Copper electrode is cathode where reduction occurs.

• Q3: B - Standard hydrogen electrode potential is defined as 0.00 V.

Memory Aids

• Salt bridge = "Charge balancer" without mixing.

• Daniell cell: Zn (anode, oxidation) → Cu (cathode, reduction).

• Standard hydrogen electrode (SHE) potential = 0.00 V by definition.

Electrochemical Series and Prediction of Redox Power

Key Concepts

• Electrochemical series arranges reducing agents by decreasing strength (activity series).

• Lower (more negative) E° → stronger reducing agent.

• Higher (more positive) E° → stronger oxidising agent.

• Examples of reducing agents (strongest to weaker): Li > Mg > Al > Zn > Fe > Co > Ni > Sn > Pb > Cu > I− > Fe2+ > Hg > Ag > Br− > Cl− > F−

• Examples of oxidising agents (strongest to weaker): F2 > Cl2 > Br2 > I2

• Fluorine is the strongest oxidising agent.

• Lithium is the strongest reducing agent due to high hydration energy.

Standard Reduction Potentials Table

• Lists half-reactions and their E° values at 298 K.

• Negative E°: stronger oxidising agents, less likely to gain electrons.

• Positive E°: stronger reducing agents, more likely to gain electrons.

• Hydrogen half-reaction (2H+ + 2e− → H2) is reference at 0.00 V.

Solved Examples

Ex 1: Predict which is stronger reducing agent: Zn or Cu?

• Zn has lower E° than Cu → Zn is stronger reducing agent.

Practice Set

• Q1: Which element is the strongest oxidising agent?
  A) Cl2
  B) F2
  C) Br2
  D) I2

• Q2: Arrange the following in order of increasing reducing power: Cu, Li, Zn.
  A) Li < Zn < Cu
  B) Cu < Zn < Li
  C) Zn < Cu < Li
  D) Cu < Li < Zn

• Q3: Which statement is correct?
  A) Higher E° means stronger reducing agent
  B) Lower E° means stronger oxidising agent
  C) Negative E° means stronger reducing agent
  D) Positive E° means stronger oxidising agent

Answer Key

• Q1: B - Fluorine (F2) is the strongest oxidising agent.

• Q2: B - Increasing reducing power: Cu < Zn < Li (Li strongest).

• Q3: C - Negative E° indicates stronger reducing agent.

Memory Aids

• Remember: "Li is the strongest reducer" due to hydration energy.

• Oxidising agents order: F2 > Cl2 > Br2 > I2.

• Negative E° → strong reducer; Positive E° → strong oxidiser.