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17.2 Redox Titrations: Equivalence Potential, Indicators, Iodometry

★★★ 2021–2025: 20 parts, 190 marks (Geo-Scientist Paper I 19 parts / 185 m · Paper II 1 / 5 m). Paper I gives one full question (Q7 or Q8) to redox titrations every year.

Syllabus line. Geo-Scientist Paper I (redox titrations): “Standard redox potentials, Nernst equation. Influence of complex formation, precipitation and change of pH on redox potentials, Normal Hydrogen Electrode (NHE). Feasibility of a redox titration, redox potential at the equivalence point, redox indicators. Redox potentials and their applications. Principles behind iodometry, permanganometry, dichrometry, difference between iodometry and iodimetry. Principles of estimation of iron, copper, manganese, chromium by redox titration.” Electrochemical cells and the Nernst equation in full are in 8.2; acid–base titrations in 17.1.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

17.2.1 Electrode potentials, Nernst equation, feasibility ★★★

(17.2.1)
(17.2.2)
  • H⁺ in the half-reaction: every H⁺ goes into Q. MnO₄⁻/Mn²⁺ at [MnO₄⁻] 0.1 M, [Mn²⁺] 0.010 M, pH 3: Q = 0.010/[0.1 × (10⁻³)⁸] = 10²³; E = 1.51 − (0.05916/5)(23) = 1.238 V. Each pH unit lowers it by (8/5)(0.05916) = 0.095 V Geo 2022 Q7(a) 10m.SN
  • Complexing or precipitating the oxidised form lowers E, the reduced form raises it. That is why H₃PO₄ (complexes Fe³⁺) lowers the Fe³⁺/Fe²⁺ potential, and why Cu²⁺ oxidises I⁻ although E°(Cu²⁺/Cu⁺) 0.15 V < E°(I₂/I⁻) 0.54 V: CuI precipitates, so Cu²⁺/CuI is 0.86 V. Formal potential E°′ = potential in a stated medium (Fe³⁺/Fe²⁺ 0.68 V in 1 M H₂SO₄).SN
  • Feasibility: 99.9 % completion at equivalence needs log K ≥ 6, i.e. the two E° values must differ by ≥ 0.35 V (n = 1) or ≥ 0.18 V (n = 2). Cl₂ + 2Fe²⁺: E°cell = 1.36 − 0.77 = 0.59 V, log K = 2 × 0.59/0.05916 = 19.9, K ≈ 9 × 10¹⁹: Fe²⁺ is oxidised completely Geo 2025 Q7(c) 10m.SN
  • : (i) is written as an oxidation (−0.77 V), so E°cell = −0.77 + 1.77 = +1.00 V, ΔG° = −2 × 96485 × 1.00 = −193 kJ mol⁻¹: spontaneous. Never multiply E by coefficients Geo 2024 Q1(e) 5m.
  • Reducing power rises as E° falls: Ag < Mg < Na < Ca < Li (+1.98, −2.37, −2.71, −2.87, −3.04 V) Geo 2024 Q8(b) 5m.

17.2.2 Titration curve and equivalence-point potential ★★★

diphenylamine 0.73–0.79 Vferroin 1.00–1.12 V02550751000.40.60.81.01.21.41.6volume of 0.100 M Ce4+ added / mLE / V vs SHEhalf-way: E = E°′(Fe) = 0.68equivalence 1.06 V200 %: E = E°′(Ce) = 1.44dashed: with H3PO4 (E°′ Fe ≈ 0.61)
Fig. 17.2.1 Computed curve: 50 mL of 0.100 M Fe²⁺ with 0.100 M Ce⁴⁺ in 1 M H₂SO₄.
Draw it in 60 s: E 0.4–1.6 V vs mL; slow rise through 0.68 V at 25 mL, vertical jump at 50 mL through 1.06 V, flat near 1.44 V; shade the two indicator bands.
  1. Before equivalence E is set by the analyte couple: E = E°′Fe + 0.05916 log([Fe³⁺]/[Fe²⁺]); at half-way E = E°′Fe.
  2. After equivalence by the titrant couple: E = E°′Ce + 0.05916 log([Ce⁴⁺]/[Ce³⁺]). The curve is independent of dilution.SN
  3. At equivalence add the two Nernst equations (weighted by n):
(17.2.3)

Solved PYQ Geo 2025 Q6(b) 15m

25 mL of 0.1 N FeSO₄ titrated with 0.1 N Cr₂O₇²⁻ at pH 1.0; E°(Fe³⁺/Fe²⁺) 0.77 V, E°(Cr₂O₇²⁻/Cr³⁺) 1.36 V. E at equivalence?

  1. . At equivalence [Fe³⁺] = 0.100 M, [Cr³⁺] = 0.100/3 = 0.0333 M, and [Fe²⁺] = 6[Cr₂O₇²⁻].
  2. E = 0.77 − 0.05916 log([Fe²⁺]/[Fe³⁺]); 6E = 6(1.36) − 0.05916 log([Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴)).
  3. Add: 7E = 0.77 + 8.16 − 0.05916 log(2[Cr³⁺]) − 14 × 0.05916 pH.
  4. 7E = 8.93 + 0.0695 − 0.828 = 8.171; Eeq = 1.17 V. Not 1.276 V (the weighted mean): 14 H⁺ and the 2:1 Cr³⁺ ratio lower it.SN

17.2.3 Redox indicators ★★★

(17.2.4)
  • Redox indicator: a substance whose oxidised and reduced forms have different colours and which changes at a definite potential (not at a definite pH). Criteria: E°In inside the potential jump; fast, reversible, intense colour change; tiny amount; stable in the medium Geo 2025 Q8(a) 15m Geo 2021 Q7(a) 15m.SN
  • End point: the eye sees the reduced colour while [Inred]/[Inox] ≥ 10. Putting 10 in eq. (17.2.4): E = E°In − 0.05916/n, the first potential at which the oxidised colour begins to show; full change at E°In + 0.05916/n (ratio 1/10) Geo 2024 Q8(c) 10m.SN
  • Diphenylamine, E° 0.756 V, n = 2: interval 0.756 ± 0.030 = 0.726–0.786 V, colourless → violet (diphenylbenzidine violet) Geo 2025 Q1(f) 5m.
  • Ferroin = [Fe(phen)₃]²⁺, three 1,10-phenanthroline ligands chelating Fe(II), low-spin: red; oxidised [Fe(phen)₃]³⁺: pale blue; E° 1.06 V in 1 M H₂SO₄, ideal for Ce⁴⁺ titrations (Fig. 17.2.1).SN
  • Why H₃PO₄ with dichromate and diphenylamine: in 1 M H₂SO₄ (E°′ 0.68 V) the indicator reaches 0.76 V when only 95.7 % of Fe²⁺ is oxidised: a premature end point. H₃PO₄ complexes Fe³⁺ (colourless phosphato complex), lowering E°′ to about 0.61 V, so 0.76 V is reached at 99.7 %, inside the jump; it also removes the yellow Fe³⁺ colour Geo 2021 Q1(i) 5m.SN
2 × diphenylamineoxidationirreversible N,N′-diphenylbenzidine: colourless− 2e−, − 2H+reversible, E° = 0.76 V diphenylbenzidine violet (quinonoid diimine)
Fig. 17.2.2 Diphenylamine: how the violet colour forms.
Draw it in 60 s: two diphenylamine → (oxidation, irreversible) → N,N′-diphenylbenzidine, colourless → (− 2e−, − 2H+, reversible) → diphenylbenzidine violet.

17.2.4 Permanganometry ★★★

acidic (H2SO4)MnO4− + 8H+ + 5e−→ Mn2+ + 4H2Ocolourless; n = 5E° = 1.51 Vneutral / weakly alkalineMnO4− + 2H2O + 3e−→ MnO2(s) + 4OH−brown ppt.; n = 3E° = 0.59 Vstrongly alkalineMnO4− + e−→ MnO42−green manganate; n = 1E° = 0.56 VH2SO4 only: HCl is oxidised to Cl2 and HNO3 is itself an oxidant
Fig. 17.2.3 What permanganate becomes.
Draw it in 60 s: three columns: acid → Mn2+ (5e−), neutral → MnO2 (3e−), strong alkali → MnO42− (1e−).
  • KMnO₄ is its own indicator (first permanent pink). Not a primary standard (MnO₂ impurity, decomposes in light): standardise with sodium oxalate Geo 2023 Q7(c) 5m.SN
  • Oxalate is titrated hot (60–70 °C): 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O is slow at room temperature; Mn²⁺ formed catalyses it (autocatalysis) Geo 2021 Q7(b) 15m.SN
  • Zimmermann–Reinhardt (preventive) solution for Fe²⁺ in chloride media: MnSO₄ (Mn²⁺ lowers the MnO₄⁻/Mn²⁺ potential and stops Cl⁻ → Cl₂), H₂SO₄ (keeps the 5-electron path), H₃PO₄ (complexes Fe³⁺: colourless, lowers E°′ Fe, sharper pink). Used when iron ore is dissolved in HCl or reduced with SnCl₂ Geo 2023 Q7(b) 15m.SN

Solved PYQs Geo 2023 Q7(a) 10m Geo 2022 Q1(f) 5m

  1. , so n(Fe²⁺) = 5n(MnO₄⁻).
  2. Mohr's salt: 15.5 mL × 0.1 M = 1.55 mmol MnO₄⁻ → 7.75 mmol Fe²⁺ in 25 mL = 0.310 M; × 392 = 121.5 g L⁻¹.
  3. Iron ore 0.6128 g, 38.4 mL of 0.0198 M KMnO₄: n(Fe) = 5 × 0.0384 × 0.0198 = 3.80 × 10⁻³ mol; 0.2123 g; 34.6 % Fe. (0.1 M KMnO₄ = 0.5 N here.)SN

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