Govt. Exams
Entrance Exams
As dilution increases, ions are farther apart, reducing interionic attractions and increasing ionic mobility, thus increasing molar conductivity.
For Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 1 mole of Cu. Therefore, 2 Faradays deposit 2 moles of Cu = 2 × 64 = 128 g.
The anode is made of copper (the plating metal) which gets oxidized and dissolved, while the cathode (iron object) receives Cu²⁺ ions that are reduced to form the copper coating.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Zinc is oxidized (anode) and copper is reduced (cathode).
The equation explicitly shows that 5 electrons (5e⁻) are transferred in this reduction half-reaction where Mn goes from +7 to +2 oxidation state.
A salt bridge completes the internal circuit and allows ion migration to maintain electrical neutrality as electrons flow through the external circuit. This prevents charge accumulation.
Molar conductivity (Λm) decreases with dilution for both strong and weak electrolytes, but the decrease is greater for weak electrolytes due to increased ionization upon dilution.
In a concentration cell: E°cell = 0 (identical electrodes), electrodes are chemically identical but in different concentrations, and cell potential depends on concentration difference via Nernst equation.
Sodium has a very negative reduction potential (-2.71 V). Water is preferentially reduced instead. Sodium is obtained by electrolysis of molten NaCl. Copper, silver, and gold can be obtained from aqueous solutions.
ΔG° = -nFE°cell = -2 × 96485 × 0.46 = -88,766 J/mol ≈ -88.7 kJ/mol. The negative value confirms spontaneity.
