Electrochemistry

Λm = κ / c

Step 1: Molar conductivity is defined as the conductance of the volume of solution that contains one mole of electrolyte. Step 2: The formula is Λm = κ/c, where κ is conductivity and c is the molar concentration. Step 3: If κ is in S m⁻¹ and c is in mol m⁻³, then Λm is in S m² mol⁻¹. Step 4: Alternatively, Λm (S cm² mol⁻¹) = [κ (S cm⁻¹) × 1000 cm³/L] / [c (mol/L)]. Step 5: This formula shows that even though κ decreases with dilution, Λm increases because the volume containing 1 mole of electrolyte increases faster than κ decreases.

It increases slowly and linearly

Step 1: Strong electrolytes are completely dissociated in solution at all concentrations. So the number of ions doesn't increase significantly on dilution. Step 2: As concentration decreases, interionic attractions between oppositely charged ions decrease. This allows ions to move more freely, slightly increasing Λm. Step 3: For strong electrolytes, Λm follows Debye-Hückel-Onsager equation: Λm = Λ°m − A√c. This shows a slow, linear increase with decrease in √c. Step 4: The increase is slow because strong electrolytes are already fully dissociated. Step 5: Weak electrolytes (like acetic acid) s

The sum of the limiting molar conductivities of individual ions (cation and anion)

Step 1: Kohlrausch observed that at infinite dilution, each ion contributes independently to the total molar conductivity of the electrolyte. Step 2: The law states: Λ°m = ν₊λ°₊ + ν₋λ°₋, where ν₊ and ν₋ are the number of cations and anions respectively. Step 3: For NaCl: Λ°m(NaCl) = λ°(Na⁺) + λ°(Cl⁻) = 50.1 + 76.3 = 126.4 S cm² mol⁻¹. Step 4: This law is especially useful for weak electrolytes (like CH₃COOH) where Λ°m cannot be found by direct extrapolation. Step 5: Using Kohlrausch's law: Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl).

Quantity of electricity passed through the electrolyte

Step 1: Michael Faraday studied electrolysis extensively in 1833-34 and published his famous laws. Step 2: Faraday's First Law states: The mass of substance deposited (or dissolved) at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed. Step 3: Mathematically: m ∝ Q, where Q = I × t (current × time) in Coulombs. Step 4: One Faraday (96500 C) deposits one mole of a monovalent ion (e.g., 108 g of Ag from Ag⁺). Step 5: Temperature, solvent, and electrode size do not directly determine the amount deposited — the charge (Q) is the key factor.