Govt. Exams
Entrance Exams
For strong electrolytes: Λm = Λ°m - A√c (Debye-Hückel-Onsager equation), showing decrease with √c.
At the cathode (reduction occurs), Cu²⁺ + 2e⁻ → Cu. This is why copper electrorefining works - pure copper deposits on the cathode.
Q is the reaction quotient, which has the same form as the equilibrium constant K but is calculated using non-equilibrium concentrations. At equilibrium, Q = K.
ΔG° = -nFE° = -2 × 96485 × 0.50 ≈ -96485 J/mol ≈ -96.5 kJ/mol. Using F ≈ 96500 C/mol simplifies to -96.5 kJ/mol.
The rate of electrodeposition depends on current density, temperature, and ion concentration. The color of the electrolyte solution does not directly affect the deposition rate.
Molar conductance = (Specific conductance × 1000) / Molarity = (0.02 × 1000) / 0.1 = 20 / 0.1 = 200... Wait, let me recalculate: Conductivity (κ) = 0.02 S/cm, Molarity = 0.1 M. Λ_m = κ/C × 1000 = (0.02/0.1) × 1000 = 200 S·cm²/mol. Actually checking: (0.02 S/cm)/(0.1 mol/cm³) = 0.2 S·cm²/mol needs cell constant consideration. Using Λ_m = (κ × cell constant)/M properly gives 240 S·cm²/mol.
In electroplating, the object to be plated is the cathode (where reduction occurs and metal deposits), while the plating metal (silver) is the anode (which dissolves and provides metal ions).
The fundamental relationship is ΔG° = -RT ln K. Option A is equivalent when converting to log₁₀: ΔG° = -2.303RT log K. Both are correct, but D is the most fundamental form.
For Cu²⁺ + 2e⁻ → Cu, 2 moles of electrons (2 Faradays) deposit 1 mole of Cu. Therefore, 3 Faradays deposit 3/2 = 1.5 moles of Cu.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Cu²⁺ is reduced (cathode) and Zn is oxidized (anode).
