QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 7.4NCERT Class 11 · Chemistry · Chapter 7

Redox Titrations and Electrochemical Cells

Redox principles manifest in two profoundly impactful technological domains: quantitative analytical chemistry through Redox Titrimetry, and energy conversion through Electrochemical (Galvanic) Cells. In this module, we examine titrimetric indicators, permanganometry, dichromatometry, iodometry, the construction of the Daniell cell, the mechanics of salt bridges, and the predictive power of the Electrochemical Series.

1. Redox Titrations: Analytical Principles & Indicators

A redox titration is a volumetric analytical method used to determine the unknown concentration of a reducing or oxidizing analyte by reacting it with a standard titrant of known concentration based on stoichiometric electron equivalence.

Equivalents of Oxidant = Equivalents of Reductant  ⇒  N₁ V₁ = N₂ V₂  ⇒  (M₁ × n₁) V₁ = (M₂ × n₂) V₂

Three Primary Classes of Redox Indicators:

  1. Self-Indicators (e.g., KMnO₄):

    Potassium permanganate acts as its own indicator. In acidic solution, dark purple MnO₄− (Mn in +7) is reduced to practically colorless Mn²+ ions:

    MnO₄− + 8H+ + 5e− → Mn²+ + 4H₂O  (n-factor = 5)

    As long as the reducing agent (e.g., Fe²+, C₂O₄²−) is present in the flask, every drop of KMnO₄ is instantly reduced and decolored. At the exact equivalence point, all reductant is consumed; the very next drop of KMnO₄ remains unreduced, imparting a permanent faint pink color to the solution.

  2. Internal Redox Indicators (e.g., Diphenylamine with K₂Cr₂O₇):

    Potassium dichromate (K₂Cr₂O₇) is an excellent primary standard, but its reduction product Cr³+ is deep green while Cr₂O₇²− is orange. The transition from orange to green is gradual, making visual endpoint detection impossible. An internal indicator, diphenylamine (in the presence of phosphoric acid H₃PO₄), is added. At the equivalence point, the first excess drop of dichromate oxidizes diphenylamine to an intensely colored blue-violet complex.

  3. Starch Indicator in Iodometric & Iodimetric Titrations:

    Elemental iodine (I₂) forms an intensely colored, deep-blue adsorption complex with starch amylose. When titrated against sodium thiosulphate (Na₂S₂O₃ / "hypo"), the solution remains dark blue until the last trace of I₂ is reduced to colorless iodide (I−), producing a razor-sharp blue-to-colorless endpoint.

    I₂ + 2S₂O₃²− → 2I− + S₄O₆²−  (Tetrathionate ion)

2. Electrochemical Cells: The Daniell Cell Architecture

If a redox reaction is carried out in a single beaker, electron transfer occurs directly between colliding particles, releasing energy entirely as dissipated heat. However, if the oxidation and reduction half-reactions are physically segregated into two distinct half-cells connected by an external electronic conductor and an internal ionic conductor, the chemical free energy is directly converted into useful electrical energy.

+-----------------------------------------------------------------------------------+ | ARCHITECTURE OF THE DANIELL (GALVANIC) CELL | +-----------------------------------------------------------------------------------+ Electrons (e-) Flow: Anode to Cathode -------> Conventional Current (I): Cathode to Anode <--- [ Voltmeter: E_cell = +1.10 V ] / Zinc Rod (-) Copper Rod (+) [ ANODE ] [ CATHODE ] | | +-----+-----+ +-------------------+ +-----+-----+ | | | SALT BRIDGE | | | | 1 M ZnSO4 |=======| (KCl in Agar) |===| 1 M CuSO4 | | | +-------------------+ | | +-----------+ +-----------+ OXIDATION HALF-CELL REDUCTION HALF-CELL Zn(s) → Zn2+ + 2e- Cu2+ + 2e- → Cu(s) Anode dissolves Copper deposits CELL NOTATION: Zn(s) | Zn2+(aq, 1 M) || Cu2+(aq, 1 M) | Cu(s) +-----------------------------------------------------------------------------------+
Figure 7.5: The classic Daniell cell featuring physical separation of half-reactions and salt bridge connection.

Role and Mechanism of the Salt Bridge:

A salt bridge is an inverted U-tube filled with a hot agar-agar or gelatin gel impregnated with a concentrated solution of an inert electrolyte such as KCl, KNO₃, or NH₄NO₃.

  • Electrical Circuit Completion: It permits the migration of ions between the two half-cells without allowing the bulk solutions to mix mechanically.
  • Maintenance of Electrical Neutrality: During cell discharge, Zn²+ ions accumulate in the anode compartment, creating an excess positive charge. Simultaneously, Cu²+ ions deposit at the cathode, leaving an excess of SO₄²− negative ions. The salt bridge feeds anions (Cl−) into the anode beaker and cations (K+) into the cathode beaker, preserving neutrality and eliminating the liquid junction potential.
  • Mobility Requirement: The ionic mobility (transport velocity) of the cation and anion must be nearly identical (e.g., K+ and Cl−) to prevent unequal diffusion potentials.

3. Standard Electrode Potentials and the Electrochemical Series

When a metal electrode is dipped into a solution of its own ions, an electrical potential difference develops across the metal-solution interface, known as the Electrode Potential.

By international IUPAC agreement, standard electrode potentials are defined strictly as Standard Reduction Potentials (E°) measured at 298 K, 1 bar pressure, and 1.0 M solute concentration against the Standard Hydrogen Electrode (SHE), assigned an arbitrary reference potential of exactly 0.00 V.

E°_cell = E°_cathode (Reduction) − E°_anode (Oxidation)
+-----------------------------------------------------------------------------------+ | ELECTROCHEMICAL SERIES: THERMODYNAMIC TRENDS | +-----------------------------------------------------------------------------------+ Standard Reduction Potential E° (V) Redox Couple ^ -3.05 V (Most Negative) Li+ + e- <===> Li(s) | -2.87 V Ca2+ + 2e- <===> Ca(s) | -2.71 V Na+ + e- <===> Na(s) | -2.36 V Mg2+ + 2e- <===> Mg(s) | -0.76 V Zn2+ + 2e- <===> Zn(s) | -0.44 V Fe2+ + 2e- <===> Fe(s) | 0.00 V (Arbitrary Standard) 2H+ + 2e- <===> H2(g) [SHE] | +0.34 V Cu2+ + 2e- <===> Cu(s) | +0.80 V Ag+ + e- <===> Ag(s) | +1.33 V Cr2O7^2- + 14H+ + 6e- <===> 2Cr3+ | +1.51 V MnO4- + 8H+ + 5e- <===> Mn2+ v +2.87 V (Most Positive) F2(g) + 2e- <===> 2F- TREND DIRECTION: • Moving UP (Negative E°): Reducing power increases (Li is strongest reductant). • Moving DOWN (Positive E°): Oxidizing power increases (F2 is strongest oxidant). +-----------------------------------------------------------------------------------+
Figure 7.6: The Electrochemical Series showing relative oxidizing and reducing capabilities.
Predicting Redox Spontaneity:

The standard Gibbs free energy change of a redox reaction is connected to its standard cell potential by:

ΔG° = − n F E°_cell

A redox reaction is thermodynamically spontaneous under standard conditions if and only if:

E°_cell > 0  ⇒  ΔG° < 0

4. Solved Examples & Numerical Applications

Example 1: Calculate the standard cell potential (E°_cell) and determine the spontaneous direction of the reaction for a galvanic cell composed of a nickel electrode dipped in 1 M Ni²+ and an aluminium electrode dipped in 1 M Al³+. Given: E°(Ni²+/Ni) = -0.25 V and E°(Al³+/Al) = -1.66 V.
Solution:

Step 1: Identify anode and cathode:

The electrode with the more negative reduction potential has a stronger tendency to undergo oxidation (lose electrons) and acts as the Anode.

E°(Al³+/Al) = -1.66 V < E°(Ni²+/Ni) = -0.25 V ⇒ Aluminium is the Anode; Nickel is the Cathode.

Step 2: Write half-reactions and cell notation:

  • Anode (Oxidation): 2Al(s) → 2Al³+(aq) + 6e−
  • Cathode (Reduction): 3Ni²+(aq) + 6e− → 3Ni(s)
  • Cell Representation: Al(s) | Al³+(aq) || Ni²+(aq) | Ni(s)

Step 3: Calculate standard cell potential:

E°_cell = E°_cathode − E°_anode = (-0.25 V) − (-1.66 V) = +1.41 V.

Since E°_cell > 0, the reaction is spontaneous in the forward direction: 2Al(s) + 3Ni²+(aq) → 2Al³+(aq) + 3Ni(s).

Example 2: How many mL of 0.02 M KMnO₄ solution are required to completely oxidize 25.0 mL of 0.10 M FeSO₄ solution in the presence of dilute sulfuric acid?
Solution:

Step 1: Determine n-factor of reactants:

  • In acidic medium: MnO₄− + 8H+ + 5e− → Mn²+ + 4H₂O ⇒ n-factor of KMnO₄ = 5.
  • Fe²+ → Fe³+ + e− ⇒ n-factor of FeSO₄ = 1.

Step 2: Apply the law of chemical equivalence:

Equivalents of KMnO₄ = Equivalents of FeSO₄

(M₁ × n₁ × V₁) = (M₂ × n₂ × V₂)

(0.02 M × 5 × V₁) = (0.10 M × 1 × 25.0 mL)

0.10 × V₁ = 2.50

V₁ = 2.50 / 0.10 = 25.0 mL.

Example 3: A sample of 0.20 g of impure copper ore was dissolved and treated with excess KI. The liberated iodine required 20.0 mL of 0.05 M Na₂S₂O₃ solution for complete titration. Calculate the percentage of copper in the ore. (Atomic mass of Cu = 63.5 g/mol).
Solution:

Step 1: Write chemical stoichiometry:

2Cu²+ + 4I− → Cu₂I₂(s) + I₂

I₂ + 2S₂O₃²− → 2I− + S₄O₆²−

From stoichiometry: 2 moles of Cu²+ ≡ 1 mole of I₂ ≡ 2 moles of S₂O₃²−.

Therefore, Moles of Cu = Moles of S₂O₃²− consumed.

Step 2: Calculate moles and mass of Cu:

Moles of S₂O₃²− = M × V(L) = 0.05 × (20.0 / 1000) = 1.0 × 10&sup-3; mol.

Moles of Cu = 1.0 × 10&sup-3; mol.

Mass of Cu = 1.0 × 10&sup-3; mol × 63.5 g/mol = 0.0635 g.

Step 3: Compute percentage purity:

% Cu = (Mass of Cu / Mass of ore) × 100 = (0.0635 g / 0.20 g) × 100 = 31.75%.

Example 4: Given the standard potentials: E°(Fe³+/Fe²+) = +0.77 V and E°(I₂/I−) = +0.54 V, state whether Fe³+ ions will spontaneously oxidize I− ions to I₂ in aqueous solution under standard conditions.
Solution:

Step 1: Write the proposed redox reaction:

2Fe³+(aq) + 2I−(aq) → 2Fe²+(aq) + I₂(s)

Step 2: Identify half-reactions:

  • Cathode (Reduction): Fe³+ + e− → Fe²+  (E° = +0.77 V)
  • Anode (Oxidation): 2I− → I₂ + 2e−  (E° = +0.54 V)

Step 3: Calculate E°_cell:

E°_cell = E°_cathode − E°_anode = +0.77 V − (+0.54 V) = +0.23 V.

Since E°_cell > 0, ΔG° is negative, and Fe³+ will indeed spontaneously oxidize I− to I₂.

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Why can a salt bridge not be made using KCl for a cell containing silver (Ag⁺) or lead (Pb²⁺) ions?
If a KCl salt bridge is used with Ag⁺ or Pb²⁺ solutions, chloride ions (Cl⁻) migrating from the salt bridge will react with the metal ions to form insoluble precipitates (AgCl or PbCl₂). Precipitation clogs the junction pores, changes ionic concentrations unpredictably, and alters electrode potentials. In such cases, KNO₃ or NH₄NO₃ salt bridges must be used.
Q2. What is the equivalent mass of KMnO₄ in acidic, neutral, and strongly alkaline media?
Equivalent mass = Molar Mass (M) / n-factor: (1) In acidic medium: MnO₄⁻ → Mn²⁺ (change of 5e⁻, n = 5) ⇒ Equivalent Mass = M / 5 = 158 / 5 = 31.6 g/eq. (2) In neutral / faintly alkaline medium: MnO₄⁻ → MnO₂ (change of 3e⁻, n = 3) ⇒ Equivalent Mass = M / 3 = 158 / 3 = 52.67 g/eq. (3) In strongly alkaline medium: MnO₄⁻ → MnO₄²⁻ (change of 1e⁻, n = 1) ⇒ Equivalent Mass = M / 1 = 158 g/eq.
Q3. Why does copper not react with dilute HCl to produce hydrogen gas, but reacts vigorously with concentrated HNO₃?
The standard reduction potential of copper (E° = +0.34 V) is more positive than hydrogen (E° = 0.00 V); hence Cu cannot reduce non-oxidizing H⁺ ions from dilute HCl. In contrast, concentrated HNO₃ is a potent oxidizing agent whose nitrate ion (NO₃⁻) has a high reduction potential (+0.96 V), readily oxidizing copper to Cu²⁺ while reducing nitrogen to brown NO₂ gas: Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O.
Q4. What is the difference between an electrolytic cell and a galvanic cell?
A galvanic (voltaic) cell converts chemical energy from a spontaneous redox reaction (ΔG < 0, E_cell > 0) into electrical energy (anode is negative, cathode is positive). An electrolytic cell uses external electrical energy from a power supply to drive a non-spontaneous chemical redox reaction (ΔG > 0, anode is positive, cathode is negative).
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