Govt Exams
When E°cell is negative, ΔG° = -nFE°cell will be positive, indicating a non-spontaneous reaction under standard conditions.
In a galvanic cell, oxidation occurs at the anode (negative electrode), while reduction occurs at the cathode (positive electrode).
For Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 1 mole of Cu. Therefore, 2 Faradays deposit 2 moles of Cu = 2 × 64 = 128 g.
The electrode with higher reduction potential acts as cathode. Since E°(Hg₂Cl₂/Hg) = 0.27 V > E°(PbSO₄/Pb) = -0.36 V, the calomel electrode (Hg-Hg₂Cl₂) is the cathode.
Molar conductivity = (κ × 1000)/C = (2.0 × 10⁻⁴ × 1.0 × 1000)/0.01 = 20 S·cm²/mol. where κ is conductivity and C is molarity.
Using Nernst: E = E° - (0.059/n)log Q. For this cell, n = 1, Q = [H⁺]/[Ag⁺] = 0.1/0.1 = 1, log Q = 0. At different concentrations: Q = [H⁺]²/[Ag⁺] = 0.01/0.1 = 0.1, so E = 0.80 - (0.059)(-1) = 0.859 ≈ 0.82 V
ΔG° = -RT ln K and ΔG° = -nFE°cell are both valid relationships. They can be combined as: -nFE° = -RT ln K or nFE° = RT ln K.
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.