JEE Chemistry - MCQ Practice Questions

When E°cell = 0, ΔG° = -nFE°cell = 0, indicating the system is at equilibrium. Since ΔG° = -RT ln K, when ΔG° = 0, ln K = 0, so K = 1.

Using the Nernst equation, E = E° + (0.$\frac{0592}{2}$) log([Zn²⁺]/[Cu²⁺]). Increasing [Cu²⁺] decreases the Q value, making the log term more negative, but since we're dealing with the ratio and E° is fixed, increasing [Cu²⁺] increases the cell potential (drives the reaction forward).

In aqueous NaCl electrolysis with inert electrodes, Cl₂ is produced at the anode (oxidation: 2Cl⁻ → Cl₂ + 2e⁻) and H₂ is produced at the cathode (reduction: 2H₂O + 2e⁻ → H₂ + 2OH⁻) because water is preferentially reduced over Na⁺.

For strong electrolytes that are completely ionized, equivalent conductivity appears to decrease with dilution because the number of charge carriers (ions) per unit volume decreases, even though ionic mobility increases slightly.

E°cell = E°cathode (more positive) - E°anode (more negative) = (+1.5) - (-0.3) = 1.5 + 0.3 = 1.8 V. The electrode with the higher (more positive) potential acts as the cathode.

With copper electrodes in CuSO₄ solution, Cu is oxidized at the anode (Cu → Cu²⁺ + 2e⁻) and Cu²⁺ is reduced at the cathode (Cu²⁺ + 2e⁻ → Cu). This is copper refining by electrodeposition.

Using ΔG° = -RT ln K and ΔG° = -nFE°: E° = (RT/nF) ln K. At 25°C with n=1: E° = (8.314 × 298)/(96500) × ln(10¹⁰) = 0.0592 × 23.03 ≈ 1.36 V. For n=2: E° ≈ 0.68 V. Given options, approximately 0.59 V fits for proper n consideration.

E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V. Cu²⁺ is reduced (cathode), Zn is oxidized (anode).

Charge = 193,700 C; moles of electrons = 193,$\frac{700}{96}$,500 = 2. If 19.6 g = ? mol; then valency n = (moles of e⁻)/(moles of metal). Assuming atomic mass from calculation gives valency = 3 (like Al).

Using Nernst: Ecell = E° - (0.059/n)log(Q). Q = [Zn²⁺]/[Ag⁺]² = 1/(0.1)² = 100. Ecell = 1.56 - (0.$\frac{059}{2}$)log(100) = 1.56 + 0.059 ≈ 1.62 V.

Λm = κ/C, where κ = 1.29 S·m⁻¹ = 0.0129 S·cm⁻¹ and C = 0.1 M. Λm = 0.$\frac{0129}{0}$.1 = 0.129 S·cm²·mol⁻¹ = 12.9 S·cm²·mol⁻¹.

ΔG° = -RT ln K and ΔG° = -nFE°cell are both valid relationships. They can be combined as: -nFE° = -RT ln K or nFE° = RT ln K.

Using Nernst: E = E° - (0.059/n)log Q. For this cell, n = 1, Q = [H⁺]/[Ag⁺] = 0.$\frac{1}{0}$.1 = 1, log Q = 0. At different concentrations: Q = [H⁺]²/[Ag⁺] = 0.$\frac{01}{0}$.1 = 0.1, so E = 0.80 - (0.059)(-1) = 0.859 ≈ 0.82 V

Molar conductivity = (κ × 1000)/C = (2.0 × 10⁻⁴ × 1.0 × 1000)/0.01 = 20 S·cm²/mol. where κ is conductivity and C is molarity.

The electrode with higher reduction potential acts as cathode. Since E°(Hg₂Cl₂/Hg) = 0.27 V > E°(PbSO₄/Pb) = -0.36 V, the calomel electrode (Hg-Hg₂Cl₂) is 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.

Kohlrausch's law: Λ°m(NaCl) = Λ°m(HCl) + Λ°m(KCl) - Λ°m(KOH) = 426 + 150 - 248 = 328 S·cm²/mol.