'Redox Reactions'

Oxidation and Reduction

Classical Definitions

• Oxidation: Addition of oxygen or removal of hydrogen.
• Reduction: Addition of hydrogen or removal of oxygen.

Electronic Definitions

• Oxidation: Loss of electrons.
• Reduction: Gain of electrons.

Redox Reactions

Reactions involving both oxidation and reduction occur simultaneously.

Zn + Cu^2+ -> Zn^2+ + Cu

• Zn loses 2 electrons (oxidised).
• Cu^2+ gains 2 electrons (reduced).

Oxidising Agent (Oxidant)

Substance that gains electrons and causes oxidation of others. It gets reduced. Examples: KMnO4, K2Cr2O7, H2O2, O2, halogens.

Reducing Agent (Reductant)

Substance that loses electrons and causes reduction of others. It gets oxidised. Examples: Na, K, Zn, H2, C, CO.

Oxidation Number (Oxidation State)

The charge an atom would have if all bonds were completely ionic.

Rules for Assigning Oxidation Numbers

1. Oxidation number of elements in free state = 0 (H2, O2, Na, C).
2. Monatomic ions: oxidation number = charge (Na+ = +1, Cl- = -1).
3. Hydrogen: usually +1 (except metal hydrides where it is -1).
4. Oxygen: usually -2 (except peroxides where it is -1, OF2 where it is +2).
5. Fluorine: always -1 in compounds.
6. Alkali metals: +1, alkaline earth metals: +2.
7. Sum of oxidation numbers in neutral compound = 0.
8. Sum in polyatomic ion = charge of the ion.

Calculation Examples

• KMnO4: K(+1) + Mn + 4(-2) = 0 => Mn = +7.
• K2Cr2O7: 2(+1) + 2Cr + 7(-2) = 0 => 2Cr = +12 => Cr = +6.
• H2SO4: 2(+1) + S + 4(-2) = 0 => S = +6.

Types of Redox Reactions

Combination Reactions

Two substances combine to form a single product. C + O2 -> CO2 (C oxidised, O reduced).

Decomposition Reactions

A compound decomposes into elements or simpler compounds. 2H2O -> 2H2 + O2 (O oxidised, H reduced).

Displacement Reactions

One element displaces another from a compound.

• Metal displacement: Zn + CuSO4 -> ZnSO4 + Cu.
• Non-metal displacement: Cl2 + 2KBr -> 2KCl + Br2.

Disproportionation Reactions

Same element undergoes both oxidation and reduction. Cl2 + 2NaOH -> NaCl + NaOCl + H2O (Cl in Cl2 has oxidation number 0; in NaCl it is -1, in NaOCl it is +1).

Balancing Redox Equations

Oxidation Number Method

1. Assign oxidation numbers to all atoms.
2. Identify atoms undergoing change in oxidation number.
3. Calculate total increase and decrease in oxidation number.
4. Multiply by coefficients to balance increase and decrease.
5. Balance other atoms by inspection.
6. Balance H and O (add H2O and H+ for acidic, OH- for basic medium).

Half-Reaction Method (Ion-Electron Method)

1. Write oxidation and reduction half-reactions.
2. Balance atoms other than H and O.
3. Balance O by adding H2O.
4. Balance H by adding H+ (acidic) or OH- + H2O (basic).
5. Balance charge by adding electrons.
6. Multiply half-reactions to equalise electrons.
7. Add half-reactions and cancel common terms.

Example: Cr2O7^2- + Fe^2+ -> Cr^3+ + Fe^3+ (acidic medium)

Oxidation half: Fe^2+ -> Fe^3+ + e^- Reduction half: Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O Multiply oxidation by 6: 6Fe^2+ -> 6Fe^3+ + 6e^- Add: Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+

Electrochemical Series

A list of standard electrode potentials E^theta at 298 K.

Standard Hydrogen Electrode (SHE)

Reference electrode with E^theta = 0.00 V. 2H+ + 2e^- -> H2(g)

Key Points of Electrochemical Series

• Strongest reducing agent at the top (Li).
• Strongest oxidising agent at the bottom (F2).
• Higher (more positive) E^theta means greater tendency to get reduced.
• Metals with E^theta < 0 displace H2 from acids.
• Metals with E^theta > 0 do not displace H2 from acids.

Applications

• Predicting spontaneity: E^theta_cell = E^theta_cathode - E^theta_anode. Positive value means spontaneous.
• Predicting displacement reactions.
• Determining relative oxidising/reducing strength.

Worked Examples

Example 1: Find the oxidation number of S in H2SO4. Solution: 2(+1) + S + 4(-2) = 0 => S = +6.

Example 2: Balance: MnO4^- + Fe^2+ -> Mn^2+ + Fe^3+ (acidic). Solution: Reduction: MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O. Oxidation: Fe^2+ -> Fe^3+ + e^-. Multiply oxidation by 5, add: MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+.

Common Mistakes

1. Oxidation number vs charge: Oxidation number is a hypothetical charge assuming complete ionicity.
2. Peroxide exception: In H2O2, O has oxidation number -1, not -2.
3. Disproportionation: The same element must have at least 3 oxidation states available.
4. Balancing in basic medium: Add OH- to both sides after balancing in acidic medium.

ISC Exam Focus

• Theory (70%): Oxidation number rules, oxidising/reducing agents, types of redox reactions, electrochemical series.
• Application (30%): Assigning oxidation numbers, balancing equations.
• ISC frequently asks: "Assign oxidation number to ..." and "Balance the following redox equation."
• Electrochemical series applications are commonly tested.

Self-Test Questions

Q1: Find the oxidation number of Cr in Cr2O7^2-. Answer: 2Cr + 7(-2) = -2 => 2Cr = +12 => Cr = +6.

Q2: Distinguish between oxidising and reducing agents. Answer: Oxidising agent: gains electrons, gets reduced. Reducing agent: loses electrons, gets oxidised.

Q3: Balance: Cu + HNO3 -> Cu(NO3)2 + NO + H2O. Answer: Half-reaction method gives: 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O.

Q4: Define disproportionation reaction with an example. Answer: Same element undergoes both oxidation and reduction. Example: Cl2 + 2NaOH -> NaCl + NaOCl + H2O.

Q5: What is the standard hydrogen electrode? Answer: Reference electrode with E^theta = 0.00 V. Consists of Pt electrode in 1 M H+ solution with H2 gas at 1 bar.

Q6: Calculate E^theta_cell for Zn|Zn^2+ || Cu^2+|Cu. (E^theta_Zn2+/Zn = -0.76 V, E^theta_Cu2+/Cu = +0.34 V). Answer: E^theta_cell = E^theta_cathode - E^theta_anode = 0.34 - (-0.76) = 1.10 V. Spontaneous.