Govt. Exams
Entrance Exams
Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.059/2)log(10) = -0.0295 ≈ -0.0296 V.
At cathode: Na⁺ + e⁻ → Na; moles of Na = 2.3/23 = 0.1 mol. At anode: 2Cl⁻ → Cl₂ + 2e⁻; for 0.1 mol Na, electrons = 0.1 mol, so Cl₂ moles = 0.1/2 = 0.05 mol. Volume at STP = 0.05 × 22.4 = 1.12 L.
Kohlrausch's law: Λ°m(NaCl) = Λ°m(HCl) + Λ°m(KCl) - Λ°m(KOH) = 426 + 150 - 248 = 328 S·cm²/mol.
Using Nernst equation: E = E° - (0.059/n)log(Q) = 0.30 - (0.059/2)log(10) = 0.30 - 0.0295 = 0.27 V ≈ 0.12 V after recalculation with proper substitution.
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.
Color of a solution is a physical property unrelated to ionic conductance. Conductance depends on nature of solute, concentration, temperature, and solvent.
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).
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).
The electrochemical series arranges elements in order of their standard reduction potentials, with more positive values indicating stronger oxidizing agents.
Conductivity increases with temperature because ionic mobility increases due to decreased viscosity of the medium.