1
Introduction — Oxidation & Reduction
REDOX = REDuction + OXidation
- Oxidation = Loss of electron (e⁻) | Increase in Oxidation Number | Increase in non-metallic character
- Reduction = Gain of electron (e⁻) | Decrease in Oxidation Number | Increase in metallic character
- In a Redox reaction: simultaneous oxidation and reduction take place
Examples of Oxidation (loss of e⁻)
- Na → Na⁺ + e⁻
- Mg → Mg²⁺ + 2e⁻
- Al → Al³⁺ + 3e⁻
- O + 2e⁻ → O²⁻ (reduction)
- N + e⁻ → N⁻ (reduction)
Oxidising Agent (O.A.) & Reducing Agent (R.A.)
- Oxidising Agent (O.A.) = Species which gets reduced (gains e⁻) during the reaction | Causes oxidation of other species
- Reducing Agent (R.A.) = Species which gets oxidised (loses e⁻) during the reaction | Causes reduction of other species
- e.g. In: Mg + Cl₂ → MgCl₂ — Mg is R.A. (gets oxidised), Cl₂ is O.A. (gets reduced)
Practice Questions
Q: Identify Oxidising Agent and Reducing Agent in: 2K + 2H₂O → 2KOH + H₂
K: 0 → +1 (oxidised → R.A.); H in H₂O: +1 → 0 in H₂ (reduced → O.A.)
Ans: K = R.A., H₂O = O.A. ✓
Q: Identify O.A. & R.A. in: 2CuI₂ → 2CuI + I₂
Cu: +2 → +1 (reduced → O.A.); I: −1 → 0 in I₂ (oxidised → R.A.); This is disproportionation
Ans: CuI₂ acts as both O.A. (Cu part) and R.A. (I part) ✓
2
Oxidation Number (O.N.) / Oxidation State
What is Oxidation Number?
- It is real or imaginary charge on an element when it is in combined state
- Real Charge → Ionic Bond | Imaginary Charge → Covalent Bond
- To find O.N. in covalent bond: We assume that more electronegative atom completely takes away the bonding pair of electrons
- In Covalent Bond: XA < XB → XA is +ve, XB is −ve (where X = electronegativity)
⚠️ Rules for Calculating Oxidation Number
- Oxidation no. of element in free state = 0 (e.g. Na, Fe, O₂, S₈, P₄ all have O.N. = 0)
- Sum of O.N. of all elements in a compound = charge on compound
- For neutral compound: Σ O.N. = 0
- For ion: Σ O.N. = charge on ion
- Alkali Metals (Li, Na, K, Rb, Cs): Show +1 in all their compounds
- Alkaline Earth Metals (Be, Mg, Ca, Sr, Ba): Show +2 in all compounds; Al → +3 in all compounds
- Fluorine (F): Always −1 in all compounds (most electronegative; can never have +ve O.N.)
- Oxygen (O):
- In most compounds: −2
- In peroxides [−O−O−]: −1 (e.g. Na₂O₂, H₂O₂, BaO₂)
- In superoxides: −1/2 (e.g. NaO₂, KO₂, RbO₂)
- In compounds with F (OF₂): +2
- In O₂F₂: +1
- Hydrogen (H):
- In most compounds: +1
- In metal hydrides (NaH, CaH₂, AlH₃): −1
- In HOF: +1
- Halogens (Cl, Br, I):
- −1 in all compounds (generally)
- Can show +1, +3, +5, +7 with more electronegative elements (like O)
- In HClO: Cl = +1 | HClO₂: Cl = +3 | HClO₃: Cl = +5 | HClO₄: Cl = +7
⚡ Special Cases for O.N.
- O.N. of element in metal complex: Consider ligand charges: NH₃=0, H₂O=0, CN⁻=−1, SO₄²⁻=−2, C₂O₄²⁻=−2, CO=0
- For finding avg O.N.: Avg O.N. = Total charge on element / Number of atoms of that element
- Carbon suboxide (C₃O₂): x+4/3 → x = 4/3 for each C
Solved Examples — Finding O.N.
Cr₂O₇²⁻
2x+7(−2)=−2
Cr = +6
Na₂S₂O₃
2(+1)+2x+3(−2)=0
S = +2
Na₂S₄O₆
2(+1)+4x+6(−2)=0
S = +2.5
H₂S₂O₈
2+2x+8(−2)=0
S = +7 → each S=+6 (peroxide O−O bond; 2 O's are −1)
Fe₃O₄
Mixed: Fe₂O₃+FeO
2×(+3)+1×(+2)=+8/3 avg; Fe = +8/3
Pb₃O₄
2PbO + PbO₂
Pb = +8/3 avg (+2 and +4 both)
O.N. in Covalent Bonds — Structural Examples
H₂S₂O₈ (Peroxodisulphuric acid) — O.N. of each S = +6
Structure has O−O bond (peroxide linkage). 2 oxygen atoms are −1 (peroxide), remaining 6 are −2
2(+1) + 2x + 6(−2) + 2(−1) = 0 → 2+2x−12−2=0 → x = +6
O.N. in Coordination Compounds
- [V(CO)₆]²⁻: x + 5(0) = −2 → V = +2... wait x+6(0)=−2 → x=−2 (but 6 CO) actually: x = −2
- [Fe(CN)₆]³⁻: x + 6(−1) = −3 → x = +3
- [PtCl₆]²⁻: x + 6(−1) = −2 → x = +4
- [Co(NH₃)₅Cl]²⁺: x + 5(0) + (−1) = +2 → x = +3
Practice Questions — O.N.
Q: O.N. of underlined element in each: (i) HCl (ii) H₂SO₄ (iii) H₃PO₄ (iv) NH₃ (v) HNO₃ (vi) Cr₂O₇²⁻ (vii) MnO₄⁻
Using Σ O.N. = charge on species formula
Ans: Cl=−1, S=+6, P=+5, H=−3(N), N=+5, Cr=+6, Mn=+7 ✓
Q: O.N. of S in Na₂S₄O₆ (Tetrathionate)?
2(+1)+4x+6(−2)=0; 2+4x−12=0; x=+2.5; Avg O.N. of S = +2.5
Ans: +2.5 ✓
Q: O.N. of Fe in Fe₃O₄?
Fe₃O₄ = FeO·Fe₂O₃; Fe in FeO=+2, Fe in Fe₂O₃=+3; Avg = (2+3+3)/3 = 8/3
Ans: +8/3 ✓
Q: O.N. of C in carbon suboxide C₃O₂?
3x+2(−2)=0; x=+4/3
Ans: +4/3 ✓
Q: O.N. of S in H₂S₂O₈?
Has peroxide (O−O) linkage; 2H: +2; 2S: 2x; 6O(regular): −12; 2O(peroxide): −2; total=0; 2+2x−12−2=0; x=+6 for each S
Ans: S = +6 ✓ (not +7)
Q: O.N. of S in Na₂S₂O₃?
2(+1)+2x+3(−2)=0; 2+2x−6=0; x=+2; Avg O.N. of each S = +2. But structure: one S is −2, one S is +6; avg = (+6+(−2))/2 = +2
Ans: +2 avg ✓
Q: O.N. of N in HN₃?
+1+3x=0; x=−1/3
Ans: N = −1/3 ✓
3
Identifying Oxidation & Reduction in Reactions
Method
- Oxidation: Oxidation Number increases (↑)
- Reduction: Oxidation Number decreases (↓)
- Find O.N. of each element before and after reaction
- Element whose O.N. increases → gets oxidised → is the Reducing Agent
- Element whose O.N. decreases → gets reduced → is the Oxidising Agent
Example: 2KMnO₄ + 5Fe²⁺ + 8H⁺ → 2Mn²⁺ + 5Fe³⁺ + K₂SO₄ + H₂O
- Mn: +7 → +2 (decreases, reduction) → KMnO₄ is O.A.
- Fe: +2 → +3 (increases, oxidation) → Fe²⁺ is R.A.
Practice Questions
Q: Identify which reactions are redox and which are not: (a) 2KClO₃ → 2KCl + 3O₂ (b) CaO + H₂O → Ca(OH)₂ (c) Zn + H₂SO₄ → ZnSO₄ + H₂ (d) AgNO₃ + NaCl → AgCl + NaNO₃
(a) O: −2→0 in O₂; Cl: +5→−1; Both change → Redox ✓ (b) O.N. don't change → NOT redox (c) Zn: 0→+2; H: +1→0 → Redox ✓ (d) Only ionic exchange, O.N. don't change → NOT redox
Ans: (a) & (c) are Redox ✓
Q: In rxn: MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O. Identify O.A. and R.A.
Mn: +7→+2 (reduction); Fe: +2→+3 (oxidation)
Ans: MnO₄⁻ = O.A. (gets reduced) | Fe²⁺ = R.A. (gets oxidised) ✓
Q: In: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O. Identify O.A. and R.A.
Cl₂: 0→−1 in NaCl (reduction) and 0→+1 in NaOCl (oxidation); Same element oxidised and reduced → Disproportionation; Cl₂ is both O.A. and R.A.
Ans: Cl₂ is both O.A. and R.A. (Disproportionation) ✓
4
Disproportionation & Comproportionation
① Disproportionation Reaction
- Same element simultaneously undergoes both oxidation and reduction
- i.e. the intermediate O.N. state converts to both higher and lower O.N. states
- e.g. Cl₂ + 2NaOH → NaCl + NaOCl + H₂O (Cl: 0 → −1 and +1)
- e.g. 3Cl₂ + 6NaOH (hot) → 5NaCl + NaClO₃ + 3H₂O (Cl: 0 → −1 and +5)
- e.g. 2H₂O₂ → 2H₂O + O₂ (O: −1 → −2 and 0)
- e.g. 3HNO₂ → HNO₃ + 2NO + H₂O (N: +3 → +5 and +2)
- e.g. P₄ + NaOH + H₂O → PH₃ + NaH₂PO₂ (P: 0 → −3 and +1)
② Comproportionation Reaction (Conproportionation)
- Two different O.N. states of same element combine to give intermediate O.N. state
- Reverse of disproportionation
- e.g. Cu⁰ + Cu²⁺ → 2Cu⁺ (0 and +2 → +1)
- e.g. S²⁻ + S → S₂²⁻ (−2 and 0 → −1)
- e.g. NH₄NO₃ → N₂O + 2H₂O (N: −3 and +5 → +1)
⚡ Ritrick — Identifying Disproportionation
If same element shows two different O.N. on product side when it had single O.N. on reactant side → Disproportionation
Condition for disproportionation: The element must have at least one O.N. state BELOW and one ABOVE its current state
Elements that cannot disproportionate: F (always −1), O in OF₂ (always +2)
Practice Questions
Q: Which of the following is a disproportionation reaction? (a)2KMnO₄→K₂MnO₄+MnO₂+O₂ (b)CuO+H₂→Cu+H₂O (c)2Cu²⁺+4I⁻→2CuI+I₂ (d)2SO₂+O₂→2SO₃
(a) Mn: +7 → +6 and +4 → two different O.N. → Disproportionation; (c) Cu: +2→+1; I: −1→0 → simple redox not disproportionation
Ans: (a) ✓ — 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
Q: Identify the disproportionation reaction from: (a)2H₂O₂→2H₂O+O₂ (b)Fe+CuSO₄→FeSO₄+Cu (c)2KMnO₄→MnO₂+K₂MnO₄+O₂ (d)Both (a)&(c)
(a) O in H₂O₂: −1 → −2(H₂O) and 0(O₂) → Disproportionation; (c) Mn: +7 → +4 and +6 → Disproportionation
Ans: (d) Both (a) & (c) ✓
Q: Which element cannot undergo disproportionation? (a)Cl₂ (b)H₂O₂ (c)MnO₄²⁻ (d)F₂
F always shows −1 O.N., it can't be further oxidised (no higher O.N. possible for F). F can only be reduced (0→−1).
Ans: (d) F₂ ✓
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Balancing Redox Reactions — Oxidation Number Method
Steps for O.N. Method
- Write the skeleton equation
- Find O.N. of all elements; identify which elements change O.N.
- Calculate increase in O.N. (oxidation) and decrease in O.N. (reduction)
- Multiply equations by suitable factors so that total increase = total decrease
- Balance the charge by adding H⁺ (acidic medium) or OH⁻ (basic medium)
- Balance H and O by adding H₂O
Solved Example — Acidic Medium
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic medium)
- Mn: +7 → +2 (decrease of 5) | Fe: +2 → +3 (increase of 1)
- Multiply: Fe²⁺ × 5 to balance electrons
- MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺
- Balance charge: +8H⁺ on left side
- Balance O: 4H₂O on right side
- MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O ✓
Solved Example — Basic Medium
Cl₂ + NaOH → NaCl + NaOCl + H₂O (basic medium)
- Cl₂: Cl: 0 → −1 (decrease of 1) and 0 → +1 (increase of 1)
- Already balanced (1 Cl → +1, 1 Cl → −1)
- Balanced: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O ✓
Practice Questions
Q: Balance in acidic medium: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺
Cr: +6→+3 (decrease 3); 2Cr → decrease 6; Fe: +2→+3 (increase 1); Need 6 Fe²⁺; Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Ans: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O ✓
Q: Balance in acidic medium: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂
Mn: +7→+2 (decrease 5); C: +3→+4 (increase 1); 2C per oxalate → increase 2 per C₂O₄²⁻; Need 5 oxalate for 2 MnO₄⁻; 2MnO₄⁻+5C₂O₄²⁻+16H⁺→2Mn²⁺+10CO₂+8H₂O
Ans: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O ✓
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Balancing Redox Reactions — Ion-Electron (Half-Reaction) Method
Steps for Ion-Electron Method
- Split the equation into two half-reactions: Oxidation half and Reduction half
- Balance each half-reaction separately:
- Balance all atoms except H and O
- Balance O by adding H₂O
- Balance H by adding H⁺ (acid) or OH⁻ (base)
- Balance charge by adding electrons (e⁻)
- Multiply both half-reactions so electrons cancel
- Add both half-reactions
Solved Example — Ion-Electron Method (Acidic)
MnO₄⁻ + I⁻ → Mn²⁺ + I₂ (acidic medium)
Reduction half: MnO₄⁻ → Mn²⁺
- Balance Mn: OK | Balance O: +8H₂O right | Balance H: +8H⁺ left | Balance e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation half: 2I⁻ → I₂
Multiply reduction×2, oxidation×5 then add:
2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O ✓
Important Rules
- Acidic medium: Balance O by H₂O, balance H by H⁺
- Basic medium: Balance O by H₂O, balance H by OH⁻; If H⁺ appears, add equal OH⁻ to both sides (H⁺ + OH⁻ → H₂O)
- Neutral medium: Use H₂O to balance both H and O
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Applications of Redox Reactions
① Important Redox Reactions to Remember
- 2F₂ + 2H₂O → 4HF + O₂ (F₂ oxidises water)
- Cl₂ + 2NaOH (cold, dil.) → NaCl + NaOCl + H₂O (Bleaching powder formation)
- 3Cl₂ + 6NaOH (hot, conc.) → 5NaCl + NaClO₃ + 3H₂O
- 2H₂O₂ → 2H₂O + O₂ (decomposition — disproportionation)
- P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ (disproportionation of white P)
- Fe + CuSO₄ → FeSO₄ + Cu (redox — Fe displaces Cu)
- Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag (Cu displaces Ag)
- 2K + 2H₂O → 2KOH + H₂ (K displaces H)
② Bleaching Action of Cl₂
- Cl₂ + H₂O → HCl + HOCl
- HOCl → HCl + [O] (nascent oxygen)
- [O] + coloured substance → colourless substance
- Bleaching by Cl₂ is permanent
③ Bleaching Action of SO₂
- SO₂ + H₂O → H₂SO₃
- H₂SO₃ + coloured substance → colourless compound
- Bleaching by SO₂ is temporary (on heating, colour returns)
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Special Cases & Important Reactions
Carbon Oxidation States in Organic Compounds
- CH₄: C = −4 (most reduced state of C)
- CH₃OH: C = −2
- HCHO (formaldehyde): C = 0
- HCOOH (formic acid): C = +2
- CO₂: C = +4 (most oxidised state of C)
- CH₃COOH: C of CH₃ = −3; C of COOH = +3; Avg = 0
- Series from most reduced → most oxidised: CH₄ → CH₃OH → HCHO → HCOOH → CO₂
Nitrogen Oxidation States
- N₂ = 0 | NH₃ = −3 | N₂H₄ = −2 | N₂H₂ = −1 | N₂O = +1
- NO = +2 | N₂O₃ = +3 (HNO₂) | NO₂ = +4 | N₂O₅ = +5 (HNO₃)
- Range: −3 to +5
Sulphur Oxidation States
- H₂S = −2 | S = 0 | SO₂ = +4 | SO₃ = +6 | H₂SO₄ = +6
- Na₂S₂O₃ = +2 (avg) | Na₂S₄O₆ = +2.5 (avg)
- H₂S₂O₈ (peroxodisulphuric acid) = +6 each S
- H₂SO₅ (peroxomonosulphuric/Caro's acid) = +6
Manganese Oxidation States
- MnO₄⁻ = +7 (permanganate, purple) | MnO₄²⁻ = +6 (manganate, green)
- MnO₂ = +4 (pyrolusite, black) | Mn₂O₃ = +3 | MnO = +2 | Mn = 0
- In acidic medium: KMnO₄ → Mn²⁺ (colourless) → gain 5e⁻
- In neutral/faintly basic: KMnO₄ → MnO₂ (brown) → gain 3e⁻
- In strongly basic: KMnO₄ → MnO₄²⁻ (green) → gain 1e⁻
Important O.A. and R.A.
- Strong Oxidising Agents: F₂, O₃, Cl₂, KMnO₄, K₂Cr₂O₇, HNO₃ (conc.), H₂SO₄ (conc.)
- Strong Reducing Agents: Na, Zn, Fe, H₂, H₂S, SO₂, SnCl₂, FeSO₄
- Amphoteric (can act as both): H₂O₂, SO₂, NO₂⁻, Fe²⁺, Cu⁺
Practice Questions — Mixed
Q: In acidic medium KMnO₄ is reduced to Mn²⁺. How many moles of KMnO₄ are needed to oxidise 1 mol Fe²⁺?
Mn: +7→+2, gain 5e⁻ per Mn; Fe: +2→+3, lose 1e⁻ per Fe; 5 Fe²⁺ required per MnO₄⁻; So 1 mol Fe²⁺ needs 1/5 mol KMnO₄
Ans: 1/5 mol KMnO₄ ✓
Q: Which of the following are oxidising agents? (a)H₂SO₄ (dilute) (b)HNO₃ (conc.) (c)Na (d)H₂S
HNO₃(conc) is a strong O.A. (N: +5→+4,+2); H₂SO₄(dil) is not a O.A. generally; Na and H₂S are R.A.
Ans: (b) HNO₃ (conc.) ✓
Q: H₂O₂ acts as both O.A. and R.A. Give one example of each.
As O.A.: H₂O₂ + 2KI → 2KOH + I₂ (I: −1→0, H₂O₂ reduced); As R.A.: 2KMnO₄+5H₂O₂+3H₂SO₄→2MnSO₄+5O₂+8H₂O (O in H₂O₂: −1→0, oxidised)
Ans: O.A. with KI; R.A. with KMnO₄ ✓
Q: O.N. of S in SO₄²⁻?
x+4(−2)=−2; x=+6
Ans: +6 ✓
Q: Which has highest O.N. of Cl? (a)ClO⁻ (b)ClO₂⁻ (c)ClO₃⁻ (d)ClO₄⁻
x+(−2)=−1→x=+1; x+2(−2)=−1→x=+3; x+3(−2)=−1→x=+5; x+4(−2)=−1→x=+7
Ans: (d) ClO₄⁻ (Cl = +7) ✓
Q: Equivalent weight of KMnO₄ in acidic medium?
Mn: +7→+2; change=5e⁻; Equivalent wt = Molar mass/n = 158/5 = 31.6 g
Ans: 31.6 g ✓
Q: Equivalent weight of K₂Cr₂O₇ in acidic medium?
Cr: +6→+3; 2 Cr atoms; total change = 2×3 = 6e⁻; Eq. wt = 294/6 = 49 g
Ans: 49 g ✓
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Important Balancing Practice
| Reaction (Unbalanced) | Medium | Balanced Equation |
|---|
| MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ | Acidic | MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O |
| Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ | Acidic | Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O |
| MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ | Acidic | 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O |
| MnO₄⁻ + I⁻ → Mn²⁺ + I₂ | Acidic | 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O |
| Cl₂ + NaOH → NaCl + NaOCl | Basic (cold) | Cl₂ + 2NaOH → NaCl + NaOCl + H₂O |
| Cl₂ + NaOH → NaCl + NaClO₃ | Basic (hot) | 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O |
| MnO₄⁻ + Br⁻ → Mn²⁺ + Br₂ | Acidic | 2MnO₄⁻ + 10Br⁻ + 16H⁺ → 2Mn²⁺ + 5Br₂ + 8H₂O |
| NO₃⁻ + Zn → Zn²⁺ + NH₄⁺ | Acidic | NO₃⁻ + 4Zn + 10H⁺ → NH₄⁺ + 4Zn²⁺ + 3H₂O |
Important MCQs — Balancing
Q: The coefficient of H₂O in the balanced equation: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic)?
Ans: 4 ✓ (MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O)
Q: Coefficient of H⁺ in: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (acidic)?
Ans: 14 ✓
Q: In balanced: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O, ratio of MnO₄⁻ to C₂O₄²⁻ is?
Ans: 2:5 ✓
Q: In acidic medium MnO₄⁻ gains how many electrons?
Ans: 5 e⁻ (Mn: +7 → +2) ✓
Q: In neutral medium MnO₄⁻ gains how many electrons?
Ans: 3 e⁻ (Mn: +7 → +4 in MnO₂) ✓
Q: In strongly alkaline medium MnO₄⁻ gains how many electrons?
Ans: 1 e⁻ (Mn: +7 → +6 in MnO₄²⁻) ✓