NCERT Plus One (Class 11) Chemistry • Chapter 7

Redox Reactions

Classical & Electronic Redox, Oxidation Number Rules & Paradoxes, Disproportionation, Equation Balancing & Electrode Potentials

📄 Official NCERT Textbook PDF

Topic 7.1: Classical Idea of Redox Reactions — Oxidation & Reduction

Slice a fresh apple and leave it on the kitchen counter, or watch an iron gate turn reddish-brown during Kerala's monsoon rains, or ignite a magnesium ribbon in the lab until it flares dazzling white! Every time oxygen latches onto a metal or steals hydrogen away from a compound, something equally dramatic happens in reverse to its partner—like two kids on a seesaw. Whenever one goes up (Oxidation), the other MUST come down (Reduction)!

Redox reactions power burning fuels, extraction of metals, caustic soda manufacture, batteries, corrosion, the hydrogen economy, and even atmospheric ozone chemistry. Historically, chemists broadened the meanings of oxidation and reduction step by step:

ProcessClassical DefinitionNCERT Examples
OxidationAddition of oxygen or any electronegative element; OR removal of hydrogen or any electropositive element2Mg + O2 → 2MgO
Mg + F2 → MgF2; Mg + Cl2 → MgCl2
2H2S + O2 → 2S + 2H2O
2K4[Fe(CN)6] + H2O2 → 2K3[Fe(CN)6] + 2KOH
ReductionRemoval of oxygen or any electronegative element; OR addition of hydrogen or any electropositive element2HgO(s) → 2Hg(l) + O2(g)
2FeCl3(aq) + H2(g) → 2FeCl2 + 2HCl
CH2=CH2 + H2 → CH3–CH3
2HgCl2 + SnCl2 → Hg2Cl2 + SnCl4

Notice reaction (7.11): 2HgCl2(aq) + SnCl2(aq) → Hg2Cl2(s) + SnCl4(aq). Electropositive mercury is added to HgCl2 (reduction), while electronegative chlorine is simultaneously added to SnCl2 (oxidation). Because oxidation and reduction always occur together, chemists coined the word "redox"!

NCERT Problem 7.1: Identify the species undergoing oxidation and reduction:
(i) H2S(g) + Cl2(g) → 2HCl(g) + S(s): H2S is oxidised (hydrogen removed / more electronegative Cl added); Cl2 is reduced (hydrogen added).
(ii) 3Fe3O4(s) + 8Al(s) → 9Fe(s) + 4Al2O3(s): Al is oxidised (oxygen added); Fe3O4 is reduced (oxygen removed).
(iii) 2Na(s) + H2(g) → 2NaH(s): Hydrogen is more electronegative than sodium! Hence Na is oxidised and H2 is reduced.
Watch out for 2Na(s) + H2(g) → 2NaH(s)! Many students blindly say "hydrogen is added to sodium, so sodium is reduced." WRONG! Since hydrogen is MORE electronegative than sodium (forming Na+H−), adding H to Na is addition of an electronegative element — so sodium is oxidised and hydrogen is reduced!

Topic 7.2 – 7.2.1: Electronic Concept (OIL RIG) & Competitive Electron Transfer

Think of electrons like rupees moving between two bank accounts: if you transfer ₹500 to your friend, your account drops by ₹500 while theirs rises by ₹500—money never vanishes in mid-air! Similarly, dip a shiny zinc strip into bright blue copper nitrate solution: zinc atoms quietly hand over electrons to blue Cu2+ ions. Within an hour, the blue colour vanishes and reddish copper coats the strip!

In ionic reactions like 2Na(s) + Cl2(g) → 2Na+Cl−(s) or 2Na(s) + H2(g) → 2Na+H−(s) (Problem 7.2), we split the reaction into two half-reactions showing explicit electron movement:

OIL RIG Memory Rule:
• Oxidation: Loss of electron(s) → 2Na(s) → 2Na+ + 2e−  (Na = Reducing agent / Electron donor)
• Reduction: Gain of electron(s) → Cl2(g) + 2e− → 2Cl−  (Cl2 = Oxidising agent / Electron acceptor)

7.2.1 Competitive Electron Transfer Reactions

When different metals compete for electrons in aqueous salt solutions, clear winners emerge:

  1. Zinc in Cu(NO3)2 / CuSO4 (Fig. 7.1): Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). The blue colour of Cu2+ disappears and reddish metallic Cu coats the Zn strip. Passing H2S gas in alkaline NH3 medium gives a white ZnS precipitate, confirming Zn2+ ions! Placing Cu in ZnSO4 gives no reaction (even the ultra-sensitive black CuS test with H2S fails) → equilibrium strongly favours products.
  2. Copper in AgNO3 (Fig. 7.2): Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). The colourless solution turns blue due to Cu2+ formation, and silvery Ag deposits on the Cu rod.
  3. Cobalt in NiSO4: Co(s) + Ni2+(aq) ⇌ Co2+(aq) + Ni(s). At equilibrium, moderate concentrations of both Ni2+ and Co2+ exist—neither reactants nor products dominate completely.
Electron-Releasing Tendency Order:   Zn > Cu > Ag   (Basis of the Metal Activity Series & Galvanic Cells!)
Do NOT confuse "substance oxidised" with "oxidising agent"! The species that gets oxidised (loses e−) gives electrons to the other partner, so it IS the Reducing Agent (Reductant)! Conversely, the species that gets reduced (gains e−) IS the Oxidising Agent (Oxidant).

Topic 7.3: Oxidation Number Rules, Special Exceptions (OF₂, KO₂, H₂O₂, NaH) & Stock Notation

Electron transfer is crystal clear in ionic Na+Cl−, yet what about covalent H2O or CH4 + 4Cl2 → CCl4 + 4HCl? Nobody hands over full electrons there—atoms share them unequally! So chemists invented an ingenious bookkeeping score called Oxidation Number: pretend every shared pair belongs 100% to the more electronegative atom, and count the resulting imaginary charge!

Oxidation Number (Oxidation State) denotes the charge an atom appears to have in a compound/ion when shared electron pairs in covalent bonds are assigned entirely to the more electronegative atom.

6 Rules for Assigning Oxidation Number

  1. Free / Uncombined Elements: Always 0 (H2, O2, O3, Cl2, P4, S8, Na, Mg, Al).
  2. Monoatomic Ions: Equal to the ion charge (Na+ = +1, Mg2+ = +2, Fe3+ = +3, Cl− = −1). In compounds, Group 1 alkali metals = +1, Group 2 alkaline earth metals = +2, and Aluminium = +3.
  3. Oxygen (Usually −2) — 3 Crucial Exceptions:
    • Peroxides (H2O2, Na2O2, CaO2): −1
    • Superoxides (KO2, RbO2): −½
    • Oxygen fluorides: OF2 → +2 and O2F2 → +1 (positive because F is more electronegative than O!)
  4. Hydrogen (Usually +1): Becomes −1 in binary metal hydrides like LiH, NaH, CaH2 (and complex hydrides like NaBH4).
  5. Halogens: Fluorine (F) is ALWAYS −1 in every compound. Other halogens (Cl, Br, I) are −1 as halides, but positive in oxoacids/oxoanions (e.g., Cl = +7 in ClO4−).
  6. Algebraic Sum Rule: Sum of oxidation numbers of all atoms = 0 in a neutral compound, and equals the net charge in a polyatomic ion (e.g., in CO32−, C + 3(−2) = −2 ⇒ C = +4).

Stock Notation (Alfred Stock)

Write the metal's oxidation number as a Roman numeral inside parentheses right after its symbol in the formula:

NCERT Problem 7.3 (Stock Notation):
HAuCl4 → HAu(III)Cl4  |  Tl2O → Tl2(I)O  |  FeO → Fe(II)O  |  Fe2O3 → Fe2(III)O3
CuI → Cu(I)I  |  CuO → Cu(II)O  |  MnO → Mn(II)O  |  MnO2 → Mn(IV)O2
Also: Au(I)Cl / Au(III)Cl3, Sn(II)Cl2 / Sn(IV)Cl4, Hg2(I)Cl2 / Hg(II)Cl2.
Oxidation-Number Definition:
• Oxidation: Increase in ON (Reducing agent)  |  Reduction: Decrease in ON (Oxidising agent)
• Problem 7.4: 2Cu2O(+1) + Cu2S(+1, −2) → 6Cu(0) + SO2(+4) ⇒ Cu(+1→0) reduced (Cu2O oxidant); S(−2→+4) oxidised (Cu2S reductant).
Three classic NCERT Exercise 7.1 traps: (1) In NaBH4, H is a hydride (−1), so Na(+1) + B + 4(−1) = 0 gives B = +3! (2) In CaO2, Ca is fixed at +2, making it calcium peroxide where O = −1! (3) In OF2, oxygen has a rare positive state (+2) because fluorine ALWAYS takes −1!

Topic 7.3 (Paradoxes): Paradox of Fractional Oxidation Numbers & Special Structures (CrO₅, H₂SO₅, Fe₃O₄)

Hold on—can an atom ever lose one-third or half an electron? No! Yet if you apply textbook algebra blindly to magnetite (Fe3O4), iron comes out as +8/3! Worse still, calculate chromium in deep-blue CrO5 using O = −2 and you get +10—even though Chromium only has 6 valence electrons! Why? Because algebra returns an average and ignores real chemical bonds—only molecular structure reveals the truth!

Whenever naive algebra yields a fractional oxidation number OR an oxidation number exceeding the element's maximum group state, look immediately at the compound's actual chemical structure:

Compound / IonNaive Algebra (Fallacy / Average)True Structure ExplanationReal Oxidation States
CrO5
(Chromium peroxide)		x + 5(−2) = 0 ⇒ +10 ❌
(Impossible! Cr max is +6)		Butterfly structure: 1 oxo (=O at −2) + two peroxo rings (–O–O–) (4 O at −1)Cr = +6
1 O = −2; 4 O = −1
H2SO5
(Caro's acid, Ex 7.5)		2 + x + 5(−2) = 0 ⇒ +8 ❌
(Impossible! S max is +6)		HO–SO2–O–OH has 1 peroxo (–O–O–) linkage (2 O at −1) + 3 normal O (−2)S = +6
3 O = −2; 2 O = −1
Fe3O4
(Magnetite, Ex 7.2)		3x + 4(−2) = 0 ⇒ +8/3Mixed oxide FeO · Fe2O3 (one Fe2+ + two Fe3+ ions)+2, +3, +3
(Average = +8/3)
C3O2
(Carbon suboxide)		3x + 2(−2) = 0 ⇒ +4/3Linear O=C=C*=C=O; central C* bonds only to identical C atoms+2, 0, +2
(Average = +4/3)
Br3O8
(Tribromooctaoxide)		3x + 8(−2) = 0 ⇒ +16/3Two terminal Br atoms hold 3 oxo O (+6 each); central Br* holds 2 O (+4)+6, +4, +6
(Average = +16/3)
S4O62−
(Tetrathionate ion)		4x + 6(−2) = −2 ⇒ +2.5 (+5/2)−O3S–S*–S*–SO3−; middle S*–S* atoms bond only to S+5, 0, 0, +5
(Average = +2.5)
KI3 (Ex 7.2a)+1 + 3x = 0 ⇒ −1/3K+[I2 · I−]: one I2 molecule coordinated to one I− ion0, 0, −1
(Average = −1/3)
NCERT Problem 7.7 (Why Pb3O4 reacts differently with HCl vs HNO3):
Red lead Pb3O4 is actually 2PbO · PbO2 (two Pb(+2) + one Pb(+4)).
• With HCl: 2PbO neutralises HCl (acid–base), while Pb(+4)O2 oxidises Cl− to Cl2(g) (redox): Pb3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O.
• With HNO3: Since HNO3 is already a strong oxidant, it only reacts acid–base with basic PbO, leaving PbO2 untouched: Pb3O4 + 4HNO3 → 2Pb(NO3)2 + PbO2↓ + 2H2O!
Exams LOVE pairing H2SO5 (+6, one peroxo bond) and H2S2O8 (Marshall's acid, +6, one peroxo bridge) against H2S2O7 (Oleum / Pyrosulphuric acid, Ex 7.1g)! In H2S2O7, simple algebra gives 2(+1) + 2x + 7(−2) = 0 ⇒ S = +6 directly—so H2S2O7 has ZERO peroxo bonds (just an S–O–S oxide bridge)!

Topic 7.3.1: Four Types of Redox Reactions & Disproportionation (H₂O₂, P₄, Cl₂)

Every redox reaction fits neatly into one of four story archetypes: two atoms teaming up (Combination), one molecule splitting apart (Decomposition), a stronger player stealing a spot (Displacement), or the wildest twist of all—Disproportionation, where the exact same element plays BOTH winner and loser at the same time, just like making household liquid bleach from chlorine gas and alkali!

1. Combination Reactions (A + B → C)

At least one reactant (A or B) MUST be in the elemental state: C(s) + O2(g) → CO2(g), 3Mg(s) + N2(g) → Mg3N2(s), and all combustion reactions like CH4(g) + 2O2(g) → CO2(g) + 2H2O(l).

2. Decomposition Reactions

Breakdown of a compound into two or more components, at least one in the elemental state: 2H2O → 2H2 + O2, 2NaH → 2Na + H2, 2KClO3 → 2KCl + 3O2, 2Pb(NO3)2 → 2PbO + 4NO2 + O2.

3. Displacement Reactions (X + YZ → XZ + Y)

  • (a) Metal Displacement: Used widely in metallurgy: CuSO4 + Zn → Cu + ZnSO4, V2O5 + 5Ca → 2V + 5CaO, TiCl4 + 2Mg → Ti + 2MgCl2, Cr2O3 + 2Al → Al2O3 + 2Cr.
  • (b) Non-Metal Displacement (Hydrogen & Halogens):
    • Cold water (Na, K, Ca, Sr, Ba): 2Na + 2H2O → 2NaOH + H2(g).
    • Steam (Mg, Fe): Mg + 2H2O → Mg(OH)2 + H2; 2Fe + 3H2O → Fe2O3 + 3H2.
    • Acids (Mg > Zn > Fe, plus Cd, Sn; NOT Ag, Au): Zn + 2HCl → ZnCl2 + H2(g).
    • Halogen Oxidising Power: F2 > Cl2 > Br2 > I2. Cl2 displaces Br− and I− in the Layer Test (giving orange-brown Br2 or violet I2 in CCl4). Note: F2 attacks water directly (2H2O + 2F2 → 4HF + O2), and F− can only be oxidised back to F2 electrolytically!

4. Disproportionation Reactions

An element in an intermediate oxidation state is simultaneously oxidised AND reduced (requiring at least 3 accessible oxidation states):

• H2O2: 2H2O2(−1) → 2H2O(−2) + O2(0)
• White Phosphorus in alkali: P4(0) + 3OH− + 3H2O → PH3(−3) + 3H2PO2−(+1)
• Sulphur in alkali: S8(0) + 12OH− → 4S2−(−2) + 2S2O32−(+2) + 6H2O
• Chlorine in cold alkali (Bleach): Cl2(0) + 2OH− → ClO−(+1) + Cl−(−1) + H2O
• Problem 7.6(d): 2NO2(+4) + 2OH− → NO2−(+3) + NO3−(+5) + H2O
NCERT Problem 7.5: Which of ClO−, ClO2−, ClO3−, ClO4− does NOT disproportionate?
ClO4− cannot disproportionate because Cl is already at its maximum (+7) oxidation state! The others do:
3ClO−(+1) → 2Cl−(−1) + ClO3−(+5)  |  6ClO2−(+3) → 4ClO3−(+5) + 2Cl−(−1)  |  4ClO3−(+5) → Cl−(−1) + 3ClO4−(+7).
Two Top Board Traps: (1) CaCO3(s) → CaO(s) + CO2(g) is a decomposition reaction, NOT a redox reaction (Ca +2, C +4, O −2 never change)! (2) Unlike Cl2, Br2, and I2, Fluorine (F2) NEVER disproportionates (2F2 + 2OH− → 2F− + OF2 + H2O) because F cannot exhibit any positive oxidation state!

Topic 7.3.2: Balancing Redox Equations: Oxidation Number & Half-Reaction (Acid/Base) Methods

Try balancing Cr2O72− + Fe2+ + H+ → Cr3+ + Fe3+ + H2O by trial and error—you could spend ten minutes guessing coefficients! Yet with NCERT's two systematic toolkits, you simply match total electrons lost to electrons gained, balance oxygen with water, and every single atom and ionic charge clicks shut on the very first try!

Method 1: Oxidation Number Method (5 Steps)

  1. Write skeletal formulas for reactants and products.
  2. Assign oxidation numbers to identify atoms whose ON changes.
  3. Calculate ON increase/decrease per formula unit and cross-multiply so Total Increase = Total Decrease.
  4. Balance total ionic charge: add H+ in acidic medium or OH− in basic medium.
  5. Balance H and O atoms by adding H2O to the appropriate side; verify O atoms match!
NCERT Problem 7.8 (Acidic — ON Method): Balance Cr2O72− + SO32− → Cr3+ + SO42−.
• Cr (+6 → +3, ↓3 per Cr = ↓6 for Cr2O72−); S (+4 → +6, ↑2). Multiply SO32− by 3 & Cr3+ by 2: Cr2O72− + 3SO32− → 2Cr3+ + 3SO42−.
• LHS charge = −8, RHS charge = 0 ⇒ add 8H+ on LHS; then add 4H2O on RHS:
Cr2O72−(aq) + 3SO32−(aq) + 8H+(aq) → 2Cr3+(aq) + 3SO42−(aq) + 4H2O(l).

NCERT Problem 7.9 (Basic — ON Method): MnO4− + Br− → MnO2 + BrO3−
• Mn (+7 → +4, ↓3); Br (−1 → +5, ↑6) ⇒ 2MnO4− + Br− → 2MnO2 + BrO3−.
• LHS charge = −3, RHS charge = −1 ⇒ add 2OH− on RHS; add 1H2O on LHS:
2MnO4−(aq) + Br−(aq) + H2O(l) → 2MnO2(s) + BrO3−(aq) + 2OH−(aq).

Method 2: Half-Reaction (Ion-Electron) Method (7 Steps)

1. Separate into Oxidation & Reduction halves → 2. Balance atoms other than O & H → 3. Balance O using H2O and H using H+ → 4. (Basic medium only) Add one OH− to BOTH sides for each H+ (combining H++OH− → H2O) → 5. Balance charge using e− → 6. Equalise e− & add → 7. Verify atoms & net charge!
Acidic Classic (Eq. 7.58): Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O & 6(Fe2+ → Fe3+ + e−) ⇒ Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.
Basic Classic (Problem 7.10): MnO4− + I− → MnO2 + I2 ⇒ 2MnO4− + 6I− + 4H2O → 2MnO2 + 3I2 + 8OH−.
In basic medium, never try guessing where OH− and H2O go directly in Half-Reaction Method! First balance O and H cleanly in acidic style using H2O and H+, then add equal OH− ions to BOTH sides for every H+ (turning H+ + OH− into H2O) and cancel duplicate waters. Always verify total charge on LHS = RHS at the end!

Topic 7.3.3 – 7.3.4: Redox Titrations (KMnO₄, K₂Cr₂O₇, I₂–Thiosulphate) & Limitations of ON

Step into a chemistry lab during volumetric practicals and look at a purple burette dripping into a clear flask: the moment the final Fe2+ ion is oxidised, a single extra drop of KMnO4 paints the entire solution a delicate pale pink! Unlike acid–base titrations, some redox reagents shine as their own dramatic endpoint signal!

Just like acid–base titrations track pH equivalence, redox titrations determine the strength of an unknown oxidant or reductant using three classic indicator strategies:

Redox SystemIndicator TypeReaction & Endpoint Colour Change
1. KMnO4 vs Fe2+ / C2O42−
(Acidified with dil. H2SO4)		Self-Indicator
(MnO4− itself!)		MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Deep purple MnO4− reduces to colourless Mn2+ until first lasting pale pink tinge (~10−6 M MnO4−) appears!
2. K2Cr2O7 vs Fe2+External Redox Indicator: DiphenylamineCr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Right after equivalence, excess Cr2O72− oxidises diphenylamine to an intense blue colour!
3. Cu2+ / I2 vs S2O32−
(Iodometric Titration)		Starch
(Deep blue with I2)		2Cu2+(aq) + 4I−(aq) → Cu2I2(s) + I2(aq) (kept soluble as KI3)
I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq)
Deep blue I2–starch colour disappears right at endpoint!

7.3.4 Limitations of Oxidation Number Concept

Oxidation number is a helpful formal bookkeeping tool, not always a physical charge. In modern chemistry, oxidation is more fundamentally visualised as a decrease in electron density and reduction as an increase in electron density around the reacting atom(s).

Lab Viva Trap: Why is KMnO4 NEVER acidified with HCl or HNO3? Because MnO4− oxidises HCl into Cl2 gas (spoiling the titration reading!), while HNO3 is itself a strong oxidant! Always acidify with dilute H2SO4. Also note NCERT Ex 7.14: mild oxidant I2 oxidises thiosulphate (S2O32−) only to tetrathionate (S4O62−, +2.5), whereas stronger Br2 oxidises it all the way to sulphate (SO42−, +6)!

Topic 7.4: Electrode Processes, Daniell Cell & Standard Electrode Potentials (E°)

Dip a zinc strip straight into blue copper sulphate and all that chemical energy simply heats up the beaker. Now separate them into two half-cell beakers, link the zinc and copper rods with an external wire, and bridge the solutions with an upside-down jelly U-tube (Salt Bridge)—suddenly electrons sprint through the wire to generate electricity! You just built a Daniell Cell, the ancestor of every modern battery!

1. Redox Couple & The Daniell Cell

A redox couple pairs the oxidised and reduced forms of a species separated by a slash: Oxidised / Reduced (Zn2+/Zn and Cu2+/Cu).

  • Anode (−, Oxidation): Zn(s) → Zn2+(aq) + 2e−
  • Cathode (+, Reduction): Cu2+(aq) + 2e− → Cu(s)
  • External Wire: Electrons flow from Zn → Cu; conventional electric current flows in the opposite direction (Cu → Zn)!
  • Salt Bridge: A U-tube filled with KCl or NH4NO3 solidified in agar-agar jelly. It allows ion migration to maintain electrical neutrality and complete the internal circuit without letting the two solutions mix.

2. Standard Electrode Potential (E°) — NCERT Table 7.1

Measured at 298 K, 1 M concentration (and 1 atm / 1 bar for gases) against the Standard Hydrogen Electrode (SHE), assigned E°(H+/H2) = 0.00 V by convention.

Reduction Half-Reaction (Oxidised + ne− → Reduced)E° / VChemical Meaning
F2(g) + 2e− → 2F−+2.87Strongest Oxidising Agent (easiest to reduce)
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O+1.51Very strong oxidant
Cl2 + 2e− → 2Cl−  |  Cr2O72− → 2Cr3++1.36 / +1.33Strong oxidants (weaker than F2)
Ag+ + e− → Ag  |  Fe3+ + e− → Fe2+  |  I2 → 2I−+0.80 / +0.77 / +0.54Positive E° → weaker reductants than H2
Cu2+ + 2e− → Cu(s)+0.34Cannot displace H2 from dilute acids
2H+ + 2e− → H2(g)0.00Reference Standard Hydrogen Electrode (SHE)
Fe2+ → Fe  |  Zn2+ + 2e− → Zn(s)−0.44 / −0.76Negative E° → stronger reductant than H2
Mg2+ → Mg  |  Na+ → Na  |  K+ → K−2.36 / −2.71 / −2.93Powerful alkali/alkaline-earth reducing metals
Li+ + e− → Li(s)−3.05Strongest Reducing Agent (most negative E°)
Predicting Cell EMF & Feasibility:
E°cell = E°cathode − E°anode  →  Reaction is feasible / spontaneous if E°cell > 0.
Daniell Cell: E°cell = (+0.34 V) − (−0.76 V) = +1.10 V!
Always check the sign! Table 7.1 lists Standard REDUCTION Potentials. Therefore: (1) The species with the most POSITIVE E° (F2, +2.87 V) is the strongest OXIDISING agent (NOT Li!). (2) The species with the most NEGATIVE E° (Li, −3.05 V) is the strongest REDUCING agent! (3) Use mnemonic LOAN for Galvanic cells: Left = Oxidation = Anode = Negative!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 1 Marks): Using Stock notation, represent the following compounds: (a) HAuCl4, (b) Fe2O3, (c) MnO2, (d) CuI.

(a) HAuCl4: Au = +3 → HAu(III)Cl4
(b) Fe2O3: Fe = +3 → Fe2(III)O3
(c) MnO2: Mn = +4 → Mn(IV)O2
(d) CuI: Cu = +1 → Cu(I)I

Q2 (Kerala HSE SAY / NCERT Prob 7.5 • 1 Marks): Fluorine does not show disproportionation reaction, and ClO4− also does not disproportionate. Give the reason for each.

1. Fluorine (F2): Being the most electronegative element, F cannot exhibit any positive oxidation state (only 0 and −1), so it cannot be simultaneously oxidised and reduced.
2. Perchlorate ion (ClO4−): Chlorine is already in its maximum (+7) oxidation state and cannot be oxidised further.

Q3 (Kerala HSE Model • 1 Marks): All decomposition reactions are not redox reactions. Justify this statement with an example.

For a decomposition reaction to be redox, at least one element must undergo a change in oxidation number (or one product must be elemental). In CaCO3(s) → CaO(s) + CO2(g), the oxidation numbers remain unchanged (Ca = +2, C = +4, O = −2 on both sides); hence it is NOT a redox reaction.

Q4 (Kerala HSE March (Most Asked) • 2 Marks): Calculate the oxidation number of Cr in CrO5 and S in H2SO5. Account for the fallacy obtained by ordinary rules using their structures.

Fallacy by ordinary rules (O = −2): Cr in CrO5 gives +10, and S in H2SO5 gives +8—both impossible because maximum oxidation states of Cr and S are +6.
Structural Reality:
• CrO5 (Butterfly structure): Has 1 oxo (=O at −2) and two peroxo (–O–O–) linkages (4 peroxo O atoms at −1): x + (−2) + 4(−1) = 0 ⇒ Cr = +6.
• H2SO5 (Caro's acid): HO–SO2–O–OH has one peroxo (–O–O–) linkage (2 O at −1) and 3 normal O (−2): 2(+1) + x + 3(−2) + 2(−1) = 0 ⇒ S = +6.

Q5 (Kerala HSE SAY / NCERT Ex 7.2 • 2 Marks): What are the average oxidation numbers of the central elements in Fe3O4, C3O2, and S4O62−? How do you rationalise these fractional values?

Fractional oxidation states represent only the average oxidation number of atoms of the same element occupying different structural environments:
• Fe3O4 (Avg = +8/3): Mixed oxide FeO · Fe2O3 containing one Fe(II) at +2 and two Fe(III) at +3.
• C3O2 (Avg = +4/3): Linear O=C=C*=C=O → terminal carbons = +2, central carbon C* = 0.
• S4O62− (Avg = +2.5): −O3S–S*–S*–SO3− → two terminal S = +5, two inner S* = 0.

Q6 (Kerala HSE Improvement / NCERT Prob 7.4 • 2 Marks): Justify that 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction. Identify the species oxidised, reduced, oxidant, and reductant.

Assigning oxidation numbers: 2Cu2(+1)O(−2) + Cu2(+1)S(−2) → 6Cu(0) + S(+4)O2(−2).
• Reduced (+1 → 0): Cu(I) → Cu(s).
• Oxidised (−2 → +4): S(−2) in Cu2S → SO2.
• Oxidant: Cu(I) (Cu2O / Cu2S); Reductant: S in Cu2S.

Q7 (Kerala HSE March • 2 Marks): Explain the 'Layer Test' used in the laboratory to identify Br− and I− ions using chlorine water and CCl4, giving balanced ionic equations.

Since oxidising power decreases down Group 17 (F2 > Cl2 > Br2 > I2), chlorine water oxidises Br− and I− ions to free Br2 and I2, which dissolve in an immiscible CCl4 layer giving characteristic colours:
• Cl2(g) + 2Br−(aq) → 2Cl−(aq) + Br2(l) → Orange-brown CCl4 layer.
• Cl2(g) + 2I−(aq) → 2Cl−(aq) + I2(s) → Violet CCl4 layer.

Q8 (Kerala HSE March (Repeated) • 2 Marks): What is a disproportionation reaction? Write balanced chemical equations for the disproportionation of (i) white phosphorus (P4) and (ii) chlorine (Cl2) in alkaline medium.

Definition: A redox reaction in which an element in an intermediate oxidation state is simultaneously oxidised and reduced.
(i) P4(s) + 3OH−(aq) + 3H2O(l) → PH3(g) + 3H2PO2−(aq)  (P: 0 → −3 & +1)
(ii) Cl2(g) + 2OH−(aq) → ClO−(aq) + Cl−(aq) + H2O(l)  (Cl: 0 → +1 & −1; household bleach)

Q9 (Kerala HSE Model • 2 Marks): How are end-points detected in redox titrations involving (a) KMnO4, (b) K2Cr2O7, and (c) I2 against Na2S2O3?

(a) KMnO4: Acts as a self-indicator; purple MnO4− turns colourless Mn2+ until the first permanent pale pink tinge appears.
(b) K2Cr2O7: Uses external redox indicator diphenylamine, which turns intense blue right after equivalence.
(c) I2 vs S2O32−: Uses starch indicator, which forms a deep blue complex with I2 that vanishes cleanly when all I2 reduces to I−.

Q10 (Kerala HSE March (Standard 3-Mark) • 3 Marks): Balance the following ionic equation by the Oxidation Number method or Half-Reaction method in acidic medium:
Cr2O72−(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)

Step 1 (Halves): Ox: Fe2+ → Fe3+  |  Red: Cr2O72− → Cr3+
Step 2 (Balance Cr): Cr2O72− → 2Cr3+
Step 3 (Balance O & H): Cr2O72− + 14H+ → 2Cr3+ + 7H2O
Step 4 (Balance Charge with e−): Ox: Fe2+ → Fe3+ + e−  |  Red: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
Step 5 (Multiply Ox × 6 & Add):
Cr2O72−(aq) + 6Fe2+(aq) + 14H+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

Q11 (Kerala HSE March (Most Asked) • 3 Marks): Balance the redox equation in basic medium by the Ion-Electron (Half-Reaction) method:
MnO4−(aq) + I−(aq) → MnO2(s) + I2(s)

1. Oxidation half: 2I− → I2 + 2e−
2. Reduction half (acid style first): MnO4− + 4H+ → MnO2 + 2H2O
3. Convert to basic (add 4OH− to both sides) & add e−:
MnO4− + 2H2O + 3e− → MnO2 + 4OH−
4. Equalise electrons (Ox × 3, Red × 2) & add:
6I− → 3I2 + 6e−
2MnO4− + 4H2O + 6e− → 2MnO2 + 8OH−
Final Balanced Equation: 2MnO4−(aq) + 6I−(aq) + 4H2O(l) → 2MnO2(s) + 3I2(s) + 8OH−(aq)

Q12 (Kerala HSE March / NCERT 7.29–7.30 • 4 Marks): (a) Depict the galvanic cell for Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s), showing the negatively charged electrode, current carriers, and half-cell reactions.
(b) Given E° values: K+/K = −2.93 V, Ag+/Ag = +0.80 V, Hg2+/Hg = +0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V, arrange these metals in increasing order of reducing power.

(a) Cell Representation: Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
• Negatively charged electrode (Anode): Zinc electrode (where oxidation occurs: Zn(s) → Zn2+ + 2e−).
• Cathode (+, Reduction): 2Ag+(aq) + 2e− → 2Ag(s).
• Carriers of current: Electrons in the external metallic wire (flowing Zn → Ag; conventional current Ag → Zn) and ions inside the solutions and salt bridge.
(b) Increasing Order of Reducing Power (Lower/more negative E° = stronger reducing agent):
Ag (+0.80 V) < Hg (+0.79 V) < Cr (−0.74 V) < Mg (−2.37 V) < K (−2.93 V)