JEE Chemistry - MCQ Practice Questions

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).

For Cu²⁺ + 2e⁻ → Cu, 2 moles of electrons (2 Faradays) deposit 1 mole of Cu. Therefore, 3 Faradays deposit $\frac{3}{2}$ = 1.5 moles of Cu.

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.

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 rate of electrodeposition depends on current density, temperature, and ion concentration. The color of the electrolyte solution does not directly affect the deposition rate.

ΔG° = -nFE° = -2 × 96485 × 0.50 ≈ -96485 J/mol ≈ -96.5 kJ/mol. Using F ≈ 96500 C/mol simplifies to -96.5 kJ/mol.

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.

At the cathode (reduction occurs), Cu²⁺ + 2e⁻ → Cu. This is why copper electrorefining works - pure copper deposits on the cathode.

Current I = E/(R + r). For the same emf E, lower internal resistance r means higher current. The cell with lower internal resistance delivers more current.

At E = 0: 0 = 1.2 - (0.$\frac{059}{2}$) log([Zn²⁺]/[Ag⁺]). Solving: log([Zn²⁺]/[Ag⁺]) = (1.2 × 2)/0.059 ≈ 40.68. So [Zn²⁺]/[Ag⁺] ≈ 10⁴⁰.

Overpotential (η) is the excess potential needed to overcome kinetic barriers and drive the reaction at significant rates. It depends on current density and the nature of the electrode.

Conductivity (κ) depends on multiple factors: concentration of ions, mobility of ions (which depends on electrolyte nature, temperature, and solvent nature). These are the primary factors affecting conductivity measurements.

Negative E°cell indicates non-spontaneous reaction with positive ΔG°. The relationship is ΔG° = -nFE°cell.

Maximum EMF = E°cathode - E°anode = 0.80 - (-0.76) = 1.56 V for Zn-Ag cell.

Conductivity (κ) = concentration × molar conductivity. Upon dilution, concentration decreases faster than molar conductivity increases.

At cathode: Na⁺ + e⁻ → Na (reduction). At anode: 2Cl⁻ → Cl₂ + 2e⁻ (oxidation).

For strong electrolytes: Λm = Λ°m - A√c (Debye-Hückel-Onsager equation), showing decrease with √c.

Charge = I × t = 5 × 1930 = 9650 C. Moles of e⁻ = $\frac{9650}{96500}$ = 0.1 mol.

Anode undergoes oxidation (Zn → Zn²⁺ + 2e⁻), losing mass. Cathode undergoes reduction (Cu²⁺ + 2e⁻ → Cu), gaining mass.

Λm = κ × 1000/c (molar conductance), Λ = κ × 1000/(c/n) (equivalent conductance). Therefore, Λm = Λ/n or Λ = Λm × n.