JEE Chemistry - MCQ Practice Questions

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.