JEE Chemistry - MCQ Practice Questions

In a concentration cell: E°cell = 0 (identical electrodes), electrodes are chemically identical but in different concentrations, and cell potential depends on concentration difference via Nernst equation.

Molar conductivity (Λm) decreases with dilution for both strong and weak electrolytes, but the decrease is greater for weak electrolytes due to increased ionization upon dilution.

A salt bridge completes the internal circuit and allows ion migration to maintain electrical neutrality as electrons flow through the external circuit. This prevents charge accumulation.

The equation explicitly shows that 5 electrons (5e⁻) are transferred in this reduction half-reaction where Mn goes from +7 to +2 oxidation state.

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).

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.

For Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 1 mole of Cu. Therefore, 2 Faradays deposit 2 moles of Cu = 2 × 64 = 128 g.

As dilution increases, ions are farther apart, reducing interionic attractions and increasing ionic mobility, thus increasing molar conductivity.

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.

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.

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

At cathode with copper electrodes in CuSO₄: Cu²⁺ ions are preferentially reduced as their reduction potential (+0.34 V) is higher than H⁺ (-0.83 V).