JEE Chemistry - MCQ Practice Questions

In a galvanic cell, the anode is the negative electrode where oxidation occurs. The cathode is positive where reduction occurs.

Faraday's first law states that the mass of substance deposited/dissolved during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.

The Nernst equation: E = E° - (RT/nF)ln(Q) calculates the cell potential when concentrations are not at standard state (1 M).

A dry cell or Leclanche cell is a primary cell that cannot be recharged. Lead-acid and Ni-Cd are secondary cells; Li-ion is rechargeable.

In electroplating, the object to be plated is made the cathode (negative terminal) where reduction occurs, causing metal deposition.

The fundamental relationship is ΔG° = -nFE°cell, where n is number of electrons, F is Faraday constant (96485 C/mol), and E°cell is standard cell potential.

When E°cell > 0, the cell reaction is spontaneous. ΔG° = -nFE°cell is negative, indicating spontaneity.

Upon dilution, the degree of ionization increases due to reduced ion-ion interactions (interionic forces), leading to more free ions and higher molar conductivity.

In molten NaCl electrolysis, Na⁺ is reduced at cathode to Na metal. Cl⁻ is oxidized to Cl₂ at anode. Water is absent, so no oxygen evolution.

HCl is a strong electrolyte with complete ionization. NaCl is also strong but HCl has higher molar conductivity. CH₃COOH and NH₃ are weak electrolytes with low ionization.

Λ∞ for HCl ≈ 426 S·cm²·mol⁻¹ (sum of Λ∞H⁺ ≈ 350 and Λ∞Cl⁻ ≈ 76). This is higher than monovalent salts due to high mobility of H⁺.

In a concentration cell, E°cell = 0, so EMF = (RT/nF)ln(C₁/C₂), depending only on the concentration ratio, temperature, and number of electrons transferred.

Cu²⁺ + 2e⁻ → Cu. Moles of Cu = 3.$\frac{2}{64}$ = 0.05 mol. Charge = 0.05 × 2 = 0.1 F (since 1 F = 1 mole of electrons).

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.

In dilute H₂SO₄, H⁺ is preferentially reduced over water at the cathode (less negative reduction potential), producing H₂ gas.

In a galvanic cell, oxidation occurs at the anode. The anode is the negative electrode where electrons are released and the species loses electrons.

Faraday constant (F) = 96485 Coulombs per mole of electrons. This represents the charge carried by one mole of electrons.