Govt Exams
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.
At the cathode (reduction occurs), Cu²⁺ + 2e⁻ → Cu. This is why copper electrorefining works - pure copper deposits on the cathode.
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.
ΔG° = -nFE° = -2 × 96485 × 0.50 ≈ -96485 J/mol ≈ -96.5 kJ/mol. Using F ≈ 96500 C/mol simplifies to -96.5 kJ/mol.