JEE Chemistry - MCQ Practice Questions

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).

The standard free energy change ΔG° = -nFE°cell, where n is moles of electrons, F is Faraday constant, and E° is standard cell potential.

In electrorefining, impure copper acts as anode and undergoes oxidation. Pure copper deposits at cathode. More reactive impurities go into solution.

At cathode: 2H⁺ + 2e⁻ → H₂ (1 mol H₂). At anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ (0.5 mol O₂). With 2 mol e⁻: 1 mol H₂ (11.2 L) and 0.5 mol O₂ (5.6 L).

As the reaction proceeds, product concentrations increase while reactant concentrations decrease, reducing the driving force according to Nernst equation: E = E° - (0.059/n)log(Q).

The species with highest reduction potential (+1.36 V) is Cl₂, making it the strongest oxidizing agent. Higher E° values indicate greater tendency to accept electrons.

At cathode: Na⁺ + e⁻ → Na; moles of Na = 2.$\frac{3}{23}$ = 0.1 mol. At anode: 2Cl⁻ → Cl₂ + 2e⁻; for 0.1 mol Na, electrons = 0.1 mol, so Cl₂ moles = 0.$\frac{1}{2}$ = 0.05 mol. Volume at STP = 0.05 × 22.4 = 1.12 L.

Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.$\frac{059}{2}$)log(10) = -0.0295 ≈ -0.0296 V.

Using ln([A]₀/[A]ₜ) = kt, ln(0.$\frac{5}{0}$.25) = k × 30, ln(2) = k × 30, k = 0.$\frac{693}{30}$ = 0.0231 s⁻¹

The overall reaction is obtained by adding all steps and canceling intermediates: A + B + D → E + F. Rate law is determined by the slow step: rate = k[A][B]

For a first-order reaction, ln[A] = ln[A]₀ - kt, so a plot of ln[A] vs t gives a straight line with slope -k.

A catalyst provides an alternative reaction pathway with lower activation energy, thus increasing the rate constant k without affecting A or T.