Govt Exams
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.
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.
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).
In electrorefining, impure copper acts as anode and undergoes oxidation. Pure copper deposits at cathode. More reactive impurities go into solution.
The standard free energy change ΔG° = -nFE°cell, where n is moles of electrons, F is Faraday constant, and E° is standard cell potential.
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).
Λm = κ × 1000/c (molar conductance), Λ = κ × 1000/(c/n) (equivalent conductance). Therefore, Λm = Λ/n or Λ = Λm × n.
Charge = I × t = 5 × 1930 = 9650 C. Moles of e⁻ = 9650/96500 = 0.1 mol.
For strong electrolytes: Λm = Λ°m - A√c (Debye-Hückel-Onsager equation), showing decrease with √c.