Govt Exams
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).
As dilution increases, ions are farther apart, reducing interionic attractions and increasing ionic mobility, thus increasing molar conductivity.
For Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 1 mole of Cu. Therefore, 2 Faradays deposit 2 moles of Cu = 2 × 64 = 128 g.
The anode is made of copper (the plating metal) which gets oxidized and dissolved, while the cathode (iron object) receives Cu²⁺ ions that are reduced to form the copper coating.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Zinc is oxidized (anode) and copper is reduced (cathode).