Redox Reactions and Electrochemistry

1. Oxidation Numbers, n-factor and Redox Titrations

• Unusual oxidation states: Fe₃O₄ has Fe = +8/3 (one Fe²⁺, two Fe³⁺); S₂O₃²⁻ has average S = +2 (the two S are −1 and +5 by structure, and 0 and +4 by the average rule); S₄O₆²⁻ has +2.5; H₂O₂ has O = −1; KO₂ has −1/2; OF₂ has O = +2. CrO₅ has Cr = +6 because it has two peroxide groups. In NH₄NO₃ the two N are −3 and +5. In BH₄⁻ hydrogen is −1 and B is +3.
• Equivalents: equivalents = moles × n-factor. At the end point, equivalents of oxidant = equivalents of reductant. For a mixed solution, divide the equivalents by the total volume.
• Disproportionation: Cl₂ in cold dilute alkali gives Cl⁻ + ClO⁻; in hot alkali gives Cl⁻ + ClO₃⁻. Worked pattern 1: 1 mol Cl₂ in 2 L of cold 2 M KOH (4 mol OH⁻): Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O uses 2 mol OH⁻, leaving [Cl⁻] = 0.5 M, [ClO⁻] = 0.5 M and [OH⁻] = 1 M. Also H₂O₂, P₄, Cu⁺, MnO₄²⁻ and HNO₂ disproportionate.
• Worked pattern 2 (Mohr's salt and dichromate): 750 cc of 0.6 M Mohr's salt has 0.45 mol Fe²⁺. Cr₂O₇²⁻ takes 6 e⁻, so it needs 0.45/6 = 0.075 mol, and in 200 cc that is 0.375 M = 375 × 10⁻³ M.
• Worked pattern 3 (iodometry in base): 500 mL of 1.2 M KI (0.6 mol) with 500 mL of 0.2 M KMnO₄ (0.1 mol) in basic medium: MnO₄⁻ → MnO₂ has n = 3, so 0.3 eq of I₂ is liberated (KMnO₄ is limiting). Thiosulphate has n = 1: 0.3 = 0.1 × V, so V = 3 L.
• Worked pattern 4 (balancing for n): BH₄⁻ → H₂BO₃⁻: four H go from −1 to +1, so 8 e⁻ per BH₄⁻. ClO₃⁻ → Cl⁻ takes 6 e⁻. The lowest common multiple is 24, so 3BH₄⁻ + 4ClO₃⁻ → 4Cl⁻ + 3H₂BO₃⁻ + 3H₂O and n = 24.
• Balancing (ion–electron): balance atoms other than H and O, then O with H₂O, then H with H⁺ (acid) or with OH⁻ and H₂O (base), then charge with e⁻. Multiply so the electrons cancel.
• Primary standards should be pure, stable in air, soluble in water, of high molar mass and react stoichiometrically. NaOH and KMnO₄ are secondary standards. Oxalic acid and Mohr's salt are common primary standards for KMnO₄ titrations.

2. Cells, Electrode Potentials and the Series

• Cell potential: E°_cell = E°_cathode − E°_anode using reduction potentials. A cell works if E > 0, i.e. ΔG < 0.
• Intensive versus extensive: E is intensive, so multiplying a half-reaction by 2 leaves E unchanged. ΔG is extensive. To find an unknown potential from two known ones, add ΔG: n₃E°₃ = n₁E°₁ + n₂E°₂.
• Series: the more negative E°, the stronger the reducing agent and the more easily the species is oxidised. Worked: Al (E°_red = −1.66), Cr (−0.74), Fe²⁺ (Fe³⁺/Fe²⁺ = +0.77) and Co²⁺ (Co³⁺/Co²⁺ = +1.81) rank as reducing agents Al > Cr > Fe²⁺ > Co²⁺.
• Disproportionation (Latimer): a species disproportionates if the potential on its right is greater than on its left. Fe²⁺ (+0.77 left of it, −0.44 right) does not.
• Salt bridge: completes the circuit, keeps each half-cell neutral and removes the liquid-junction potential. Its ions need nearly equal mobility (KCl, KNO₃, NH₄NO₃). KCl cannot be used with Ag⁺, Pb²⁺, Hg₂²⁺ or Tl⁺ because chlorides precipitate.
• Notation: anode || cathode (Zn | Zn²⁺ || Cu²⁺ | Cu). For gas electrodes write Pt | H₂ | H⁺. E°(SHE) = 0 at all temperatures. Saturated calomel electrode = +0.28 V.

3. Nernst Equation and Thermodynamics

• Nernst: E = E° − (0.059/n) log Q at 298 K, where Q is products over reactants and solids and pure liquids are left out. At equilibrium E = 0 and Q = K, so log K = nE°/0.059, and ΔG° = −nFE° = −2.303RT log K.
• Direction of change: E rises when Q falls. Worked pattern 5: for Ag | AgCl | Fe²⁺, Fe³⁺ | Pt the reaction is Ag + Cl⁻ + Fe³⁺ → AgCl + Fe²⁺, with Q = [Fe²⁺]/([Fe³⁺][Cl⁻]). E increases if [Fe²⁺] falls, [Fe³⁺] rises or [Cl⁻] rises.
• Worked pattern 6 (pH in a cell): the O₂/H₂O couple has E = 1.23 − 0.059 pH at 1 bar. For a cell with E°(M²⁺/M) = 0.994 V as cathode and O₂ evolution at the anode, oxygen starts to evolve when E_cell > 0, i.e. 0.994 > 1.23 − 0.059 pH, so pH above 4.
• Worked pattern 7 (Q given): if E°_cell = 0.46 V, n = 6 and Q = 10⁶, then E = 0.46 − (0.059/6)(6) = 0.401 V, that is 4.01 × 10⁻¹ V. The value of n comes from the lowest-integer balanced equation.
• Temperature dependence: ΔS = nF(∂E/∂T)_P, ΔH = −nFE + nFT(∂E/∂T)_P. Use joules with F = 96500.

4. Concentration Cells and Special Electrodes

• Electrolyte concentration cell M | Mⁿ⁺(c₁) || Mⁿ⁺(c₂) | M has E° = 0 and E = (0.059/n) log(c_cathode/c_anode). Worked pattern 8: E is positive only when the concentration at the cathode is larger than at the anode. If c₁ is at the cathode, we need c₁ > c₂.
• Gas concentration cell Pt | H₂(P₁) | H⁺ || H⁺ | H₂(P₂) | Pt: E = (0.059/2) log(P₁/P₂), which is positive if P₁ > P₂ (the anode side has the higher pressure).
• Metal–insoluble salt electrode: E°(X⁻|MX|M) = E°(M⁺|M) + 0.059 log K_sp. Worked pattern 9: E°(M⁺|M) = 0.79 V and K_sp = 10⁻¹⁰ gives 0.79 + 0.059(−10) = 0.20 V = 200 mV. The electrode is a weaker oxidising agent than the free ion because M⁺ is held in the solid.
• K_sp and K from cells: measure E° of a cell whose net reaction is the dissolution, then log K_sp = nE°/0.059.

5. Electrolysis and Faraday's Laws

• Selective discharge: at the cathode the ion with the higher reduction potential is discharged first; at the anode the one with the lower reduction potential is. Overvoltage stops O₂ from forming at the anode in brine, so Cl₂ comes off.
• Faraday's laws: W = (M/n)(It/96500), with t in seconds. Same charge through different electrolytes in series: W₁/W₂ = E₁/E₂. Current efficiency = (actual mass/theoretical mass) × 100. 1 F = 96500 C liberates 1 equivalent.
• Worked pattern 10 (two stages): 300 mg of Cu (63.54 g/mol) is deposited from acidic Cu²⁺ using 2 × 0.3/63.54 = 9.44 × 10⁻³ eq, giving O₂ = 9.44 × 10⁻³/4 = 2.36 × 10⁻³ mol at the anode. Then 600 mA for 28 min = 1008 C = 0.01045 F of water electrolysis gives O₂ = 0.01045/4 = 2.61 × 10⁻³ mol. Total O₂ = 4.97 × 10⁻³ mol × 22 400 mL = 111 mL at STP.
• Electrolysis of CuSO₄ with inert electrodes makes the solution acidic (H⁺ forms at the anode); with copper electrodes the anode dissolves and the cathode grows by the same amount (refining).
• Lead accumulator: discharge consumes 2 mol H₂SO₄ per 2 F, i.e. 1 mol H₂SO₄ per mole of electrons.

6. Conductance and Kohlrausch's Law

• Definitions: G = 1/R, κ = G(l/A) with cell constant l/A. Λ_m = 1000κ/C in S cm² mol⁻¹ (κ in S cm⁻¹, C in mol/L). Λ_m = Λ_eq × n-factor. 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.
• Dilution: κ falls (fewer ions per cm³); Λ_m rises. Strong electrolytes follow Λ_m = Λ°_m − A√c. Worked pattern 11: Λ_m is 96.1 at c = 0.04 and 94.9 at c = 0.25, so slope = (94.9 − 96.1)/(0.5 − 0.2) = −4 and A = 4 S cm² mol⁻¹/(mol L⁻¹)^1/2.
• Kohlrausch's law: Λ°_m(A_xB_y) = xλ°(A^y+) + yλ°(B^x−). Use it for weak electrolytes: α = Λ_m/Λ°_m and K_a = Cα²/(1 − α) = CΛ_m²/[Λ°_m(Λ°_m − Λ_m)].
• Worked pattern 12 (find Λ°_m): a weak acid HX has pH 5 and conductance 4 × 10⁻⁵ S in a cell with l/A = 15 cm⁻¹. κ = 4 × 10⁻⁵ × 15 = 6 × 10⁻⁴ S cm⁻¹. Cα = [H⁺] = 10⁻⁵ M, so Λ°_m = 1000κ/(Cα) = 6 × 10⁴ S cm² mol⁻¹ = 6 S m² mol⁻¹.
• Worked pattern 13 (compare two weak acids): HQ at 0.18 M has Λ_m equal to 1/30 of HZ at 0.02 M, with the same Λ°_m. So α_Q = α_Z/30, and K_a = Cα² gives K_Z/K_Q = (0.02 × 900)/0.18 = 100. pK_a(HQ) − pK_a(HZ) = log 100 = 2.
• Solubility from conductivity: S = 1000κ_salt/Λ°_m (subtract the κ of water), then K_sp = S² for a 1:1 salt.
• Ionic mobility: in water Li⁺ is the most hydrated, so λ(Li⁺) < λ(Na⁺) < λ(K⁺) < λ(Rb⁺) < λ(Cs⁺). H⁺ and OH⁻ conduct exceptionally well (Grotthuss hopping).

7. Batteries, Fuel Cells and Corrosion

• Lead storage cell: discharge Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O (the acid becomes dilute and its density falls); charging reverses it.
• Fuel cell: H₂ is oxidised and O₂ reduced at porous carbon electrodes with Pt or Pd; the product is water and there is no pollution.
• Corrosion: anode Fe → Fe²⁺ + 2e⁻; cathode O₂ + 4H⁺ + 4e⁻ → 2H₂O; Fe²⁺ is then oxidised to hydrated Fe₂O₃·xH₂O (rust). Galvanisation (zinc coat) works even when scratched because zinc is the more active metal; a sacrificial Mg or Zn anode protects buried pipes.

8. How JEE Frames Questions

• Numerical: n-factor from a balanced redox equation, titration volume with equivalents, E of a modified cell, mass or volume from Faraday's law, Λ°_m or A from conductivity data, pK_a difference from conductivity.
• Statements: which change raises or lowers E; which cell is a valid concentration cell; ranking reducing agents from E° values.
• Data questions: take two points and find a slope; reconstruct the cell reaction, then apply Nernst.

9. Quick Sheet and Checklist

Before the exam, check you can:

• Find oxidation numbers of unusual species and the n-factor of KMnO₄, K₂Cr₂O₇, Fe²⁺, C₂O₄²⁻ and S₂O₃²⁻.
• Write the cell reaction and Q from a cell diagram and predict whether E rises or falls.
• Handle a concentration cell and a metal–insoluble salt electrode.
• Compute the mass or volume from Faraday's law in a multi-stage electrolysis.
• Use Kohlrausch's law for α, K_a, Λ°_m and solubility, and keep the units straight.