Electrochemistry: Kohlrausch's and Ostwald Dilution Law | IIT JAM Chemistry

Electrochemistry for  IIT JAM: Electrochemistry studies the relationship between electrical and chemical phenomena, focusing on processes involving electron transfer. This field is crucial for understanding batteries, corrosion, and biological systems, and forms a significant part of competitive chemistry curricula. 

This discussion will cover essential concepts such as molar conductivity, ionic activity, solubility products, and transport numbers, demonstrated through problem-solving.

Introduction to Electrochemistry

Electrochemistry is a core area of physical chemistry that explains the relationship between electrical energy and chemical reactions. It deals with concepts such as electrolytes, conductivity, ionic movement, and electrochemical laws. 

A clear understanding of electrochemistry is essential for competitive exams like IIT JAM, as it forms the base for solving numerical and conceptual problems.

Molar Conductivity in Complex Reaction Mixtures

Calculating molar conductivity (λm) for different species in solutions undergoing chemical reactions requires careful consideration of concentration changes. For a solution with conductivity (κ) in S cm⁻¹ and concentration (M) in mol/L, the molar conductivity is given by:

λm = (κ * 1000) / M

When 0.1 M NaOH reacts with an equal volume of 0.1 M HCl, neutralization occurs: NaOH + HCl → NaCl + H₂O. The resulting NaCl solution will have a new concentration. If equal volumes are mixed, the total volume doubles, halving the concentration of the product salt (e.g., from 0.1 M to 0.05 M NaCl if starting with 0.1 mol). The molar conductivity of the formed NaCl can then be calculated using its new concentration and the solution's conductivity.

In mixed electrolyte solutions, the total conductivity of the solution is the sum of contributions from each ionic species. This can be expressed as:

κ_solution = Σ (cᵢ * λᵢ)

Here, cᵢ is the concentration and λᵢ is the molar conductivity of each component. This principle allows for determining the conductivity of an individual component in a complex mixture by isolating its contribution from the total measured conductivity.

Dissociation of a Weak Acid: Kohlrausch's Law and Ostwald Dilution Law

To characterise a weak acid, its degree of dissociation (α) and dissociation constant (Ka) are essential. These can be determined using conductivity data.

1. Molar Conductivity at Infinite Dilution (λm⁰)

For a weak acid (HA), its molar conductivity at infinite dilution (λm⁰) cannot be directly measured. Instead, Kohlrausch's Law of Independent Migration of Ions is applied. This law states that at infinite dilution, each ion contributes independently to the total molar conductivity of the electrolyte.

For a weak acid, HCr (Crotonic Acid), λm⁰ (HCr) = λm⁰ (H⁺) + λm⁰ (Cr⁻).

Using strong electrolytes, this can be determined:

λm⁰ (HCr) = λm⁰ (HCl) + λm⁰ (NaCr) - λm⁰ (NaCl)

For example, given λm⁰ (HCl) = 426, λm⁰ (NaCr) = 83, and λm⁰ (NaCl) = 126 S cm² mol⁻¹,

λm⁰ (HCr) = 426 + 83 - 126 = 383 S cm² mol⁻¹.

2. Molar Conductivity at Given Concentration (λmᶜ)

This is calculated using the weak acid's measured conductivity (κ) at a specific concentration (c):

λmᶜ = (κ * 1000) / c

For instance, for 0.01 M Crotonic Acid with κ = 3.83 × 10⁻⁵ S cm⁻¹:

λmᶜ = (3.83 × 10⁻⁵ * 1000) / 0.01 = 38.3 S cm² mol⁻¹.

3. Degree of Dissociation (α)

The degree of dissociation for a weak electrolyte is the ratio of its molar conductivity at a given concentration to its molar conductivity at infinite dilution:

α = λmᶜ / λm⁰

Using the values above, α = 38.3 / 383 = 0.1.

4. Dissociation Constant (Ka)

The dissociation constant is determined using the Ostwald Dilution Law:

Ka = cα² / (1 - α)

Given c = 0.01 M and α = 0.1:

Ka = (0.01 * (0.1)²) / (1 - 0.1) = (0.01 * 0.01) / 0.9 ≈ 1.1 × 10⁻⁵.

It is important to note that the (1 - α) term should not be approximated to 1 unless α is very small (typically α ≤ 0.05).

Molar Conductivity Formula Variants and Unit Management

The correct formula for molar conductivity (λm) depends on the units of conductivity (κ) and concentration (c). The factor of 1000 acts as a unit conversion between volume units (e.g., cm³ and L).

Solubility and Solubility Product (Ksp) from Conductivity Measurements

For sparingly soluble salts, conductivity measurements can determine their solubility (S) and solubility product (Ksp). In a saturated solution of a sparingly soluble salt, the concentration is very low, allowing its molar conductivity (Λm) to be approximated as the molar conductivity at infinite dilution (Λm⁰).

1. Molar Conductivity at Infinite Dilution (Λm⁰)

Using Kohlrausch's Law and ionic molar conductivities (λ⁰), calculate Λm⁰ for the salt.

For Co₂(Fe(CN)₆): Co₂(Fe(CN)₆) (s) ⇌ 2 Co²⁺ (aq) + [Fe(CN)₆]⁴⁻ (aq)

Λm⁰ (salt) = 2 * λ⁰(Co²⁺) + 1 * λ⁰([Fe(CN)₆]⁴⁻)

For example, with λ⁰(Co²⁺) = 86 and λ⁰([Fe(CN)₆]⁴⁻) = 444 S cm² mol⁻¹,

Λm⁰ (salt) = 2 * 86 + 1 * 444 = 172 + 444 = 616 S cm² mol⁻¹.

2. Conductivity of the Salt (κ_salt)

The measured conductivity of a saturated solution includes contributions from both the dissolved salt and the solvent water. Therefore, the conductivity of the salt alone is:

κ_salt = κ_solution - κ_water

For example, if κ_solution = 2.06 × 10⁻⁶ and κ_water = 0.41 × 10⁻⁶ S cm⁻¹, then κ_salt = 1.65 × 10⁻⁶ S cm⁻¹.

3. Solubility (S)

Rearrange the molar conductivity formula to find the concentration (which is S for a saturated solution):

S = (κ_salt * 1000) / Λm⁰

Using the example values, S ≈ 2.6 × 10⁻⁶ mol/L.

4. Solubility Product (Ksp)

Ksp is calculated from the solubility (S) and the salt's stoichiometry. For a salt AₓBᵧ, Ksp = xˣyʸS⁽ˣ⁺ʸ⁾.

For Co₂(Fe(CN)₆): Ksp = (2S)² * (S)¹ = 4S³

Substituting S = 2.6 × 10⁻⁶ mol/L, Ksp ≈ 7.68 × 10⁻¹⁷.

Activity and Non-Ideal Solutions

For non-ideal solutions, activity (a) replaces concentration to represent the "effective concentration" of a species. Activity is related to molal concentration (m) by the activity coefficient (γ): a = γ * m.

Key definitions:

1. Activity of a Salt (AₓBᵧ): a_salt = (a₊)ˣ * (a₋)ʸ = (a_±)⁽ˣ⁺ʸ⁾
   It can also be expressed as a_salt = (xˣyʸ) * (m⁽ˣ⁺ʸ⁾) * (γ_±)⁽ˣ⁺ʸ⁾, where γ_± is the mean ionic activity coefficient.

2. Mean Ionic Activity (a_±): A geometric mean of individual ion activities: a_± = [ (a₊)ˣ * (a₋)ʸ ] ^ (1 / (x+y))

3. Mean Ionic Activity Coefficient (γ±): This very important formula relates the mean coefficient to individual ionic coefficients:
   γ± = [ (γ₊)ˣ * (γ₋)ʸ ] ^ (1 / (x+y))

The Debye-Hückel Limiting Law (DHLL)

DHLL predicts the mean ionic activity coefficient for very dilute solutions:

log(γ_±) = -A |z₊z₋| √I

Where:

• γ_±: Mean ionic activity coefficient.

• A: A constant depending on solvent and temperature. For water at 25°C, A = 0.509.

• z₊, z₋: Charges of the cation and anion.

• I: Ionic strength of the solution.

Ionic Strength (I)

Ionic strength measures the total concentration of ions:

I = ½ Σ (cᵢzᵢ²)

Where cᵢ is the molar concentration and zᵢ is the charge of ion i.

When calculating ionic strength, do not forget the ½ term and do not forget to square the charge (zᵢ²) of each ion, as these are common sources of error.

Calculating Cell Potential with Activity

The Nernst equation can be modified to incorporate activities for more accurate cell potential calculations in non-ideal solutions. For a cell like H₂(1 bar) | HBr(a± = 0.2) | Hg₂Br₂(s) | Hg(l) at 25°C with E°(Hg₂Br₂/Hg, Br⁻) = 0.138 V:

The overall reaction is Hg₂Br₂(s) + H₂(g) → 2Hg(l) + 2Br⁻(aq) + 2H⁺(aq).

The standard cell potential E°_cell = E°_cathode - E°_anode = 0.138 V - 0 V = 0.138 V.

The Nernst Equation (at 25°C) is:

E_cell = E°_cell - (0.0591 / n) * log(Q)

Here, n = 2 (electrons transferred). The reaction quotient (Q) is expressed in terms of activities:

Q = (a_H⁺)² * (a_Br⁻)² = (a_±)⁴

Substituting the mean ionic activity a_± = 0.2:

E_cell = 0.138 - (0.0591 / 2) * log(0.2)⁴

E_cell = 0.138 - (0.0591 * 2) * log(0.2) ≈ 0.22 V.

The Nernst equation with 0.0591 is only valid at 25°C; otherwise, the general form E_cell = E°_cell - (RT / nF) * ln(Q) must be used. Always define Q as [Products] / [Reactants] based on the balanced reaction to avoid errors.

Transport Numbers: Fraction of Current Carried by Ions

Transport numbers (tᵢ) quantify the fraction of total current carried by a specific ion in an electrolyte solution.

tᵢ = κᵢ / κ_total = (Cᵢ * λᵢ) / Σ(Cⱼ * λⱼ)

Where Cᵢ and λᵢ are the concentration and molar ionic conductivity of ion i.

Example: For a solution with 0.1 M Na₂SO₄ and 0.01 M H₂SO₄, and given ionic conductivities:

1. Ion Concentrations:

• [Na⁺] = 2 * 0.1 = 0.2 M

• [H⁺] = 2 * 0.01 = 0.02 M

• [SO₄²⁻] = 0.1 + 0.01 = 0.11 M

2. Calculate Denominator (Σ(Cⱼλⱼ)):
   Σ(Cⱼλⱼ) = (0.2 * 50) + (0.02 * 350) + (0.11 * 160)

3. Calculate Individual Transport Numbers:

• t(Na⁺) = (0.2 * 50) / Σ(Cⱼλⱼ) ≈ 0.29

• t(H⁺) = (0.02 * 350) / Σ(Cⱼλⱼ) ≈ 0.20

• t(SO₄²⁻) = (0.11 * 160) / Σ(Cⱼλⱼ) ≈ 0.51
  The sum of transport numbers should always be 1.

Determining Concentration Ratios via Transport Numbers

Transport numbers can be used to determine unknown concentration ratios in mixed electrolyte solutions.

Example: For a solution of HCl and NaCl, if the transport number of H⁺, t(H⁺) = 0.5, what is the ratio [HCl] / [NaCl]?

Let [HCl] = C₁ and [NaCl] = C₂.

Then, [H⁺] = C₁, [Na⁺] = C₂, [Cl⁻] = C₁ + C₂.

Using the transport number formula:

0.5 = (C₁ * λ_H⁺) / (C₁ * λ_H⁺ + C₂ * λ_Na⁺ + (C₁ + C₂) * λ_Cl⁻)

Substituting given λ values and solving algebraically for the ratio C₁/C₂ yields approximately 0.46.

Degree of Dissociation of Pure Water

The degree of dissociation (α) for pure water can be determined from its conductivity (κ) and infinite dilution molar conductivities (λ∞) of H⁺ and OH⁻.

α = Λm / Λm⁰

1. Λm⁰ for Water: Using Kohlrausch's Law for H₂O ⇌ H⁺ + OH⁻:
   Λm⁰(H₂O) = λ∞(H⁺) + λ∞(OH⁻)
   Given λ∞(H⁺) = 350 and λ∞(OH⁻) = 198 S cm² mol⁻¹,
   Λm⁰(H₂O) = 350 + 198 = 548 S cm² mol⁻¹.

2. Λm for Water:
   Λm = (κ * 1000) / C
   The molar concentration (C) of pure water is a standard value: C = 55.5 M (Memory Tip: Remember pure water's concentration is 55.5 M).
   Given κ = 0.554 × 10⁻⁷ S cm⁻¹:
   Λm = (0.554 × 10⁻⁷ * 1000) / 55.5 ≈ 1.0 × 10⁻⁶ S cm² mol⁻¹.

3. Degree of Dissociation (α):
   α = (1.0 × 10⁻⁶) / 548 ≈ 1.8 × 10⁻⁹.

Solubility and Solubility Product (Ksp) of CaF₂

Let's apply the principles of solubility and Ksp to CaF₂ from conductivity data.

Given κ_solution = 4.2 × 10⁻⁵ S cm⁻¹, κ_water = 2.0 × 10⁻⁶ S cm⁻¹, λ∞(Ca²⁺) = 104 S cm² mol⁻¹, and λ∞(F⁻) = 48 S cm² mol⁻¹.

1. Conductivity of the Salt (κ_salt):
   κ_salt = κ_solution - κ_water = (4.2 × 10⁻⁵) - (0.2 × 10⁻⁵) = 4.0 × 10⁻⁵ S cm⁻¹.

2. Λm⁰ for CaF₂:
   For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), do not forget the stoichiometric coefficient '2' for the fluoride ion.
   Λm⁰(CaF₂) = λ∞(Ca²⁺) + 2 × λ∞(F⁻) = 104 + 2 * 48 = 104 + 96 = 200 S cm² mol⁻¹.

3. Solubility (S):
   S = (κ_salt * 1000) / Λm⁰ = (4.0 × 10⁻⁵ * 1000) / 200 = (4.0 × 10⁻²) / 200 = 2.0 × 10⁻⁴ mol L⁻¹.

4. Solubility Product (Ksp):
   Ksp = [Ca²⁺][F⁻]². Since [Ca²⁺] = S and [F⁻] = 2S,
   Ksp = (S)(2S)² = 4S³
   Ksp = 4 × (2.0 × 10⁻⁴)³ = 4 × (8.0 × 10⁻¹²) = 3.2 × 10⁻¹¹ M³.