Govt. Exams
Entrance Exams
The fundamental relationship is ΔG° = -nFE°cell, where n is number of electrons, F is Faraday constant (96485 C/mol), and E°cell is standard cell potential.
Upon dilution, the number of ions per unit volume decreases (concentration effect) and ionic mobility also increases slightly due to reduced ion-ion interactions, but overall conductivity decreases because the decrease in ion concentration dominates.
In electroplating iron with copper, iron (the object to be plated) acts as the cathode where Cu²⁺ ions are reduced and deposit as copper metal. The anode is made of copper.
The Gibbs free energy change is related to cell potential by ΔG° = -nFE°cell, where n is the number of electrons transferred, F is Faraday's constant, and E° is the standard cell potential.
Fluorine has the highest reduction potential among the halogens, approximately +2.87 V. Reduction potentials decrease down the halogen group: F₂ > Cl₂ > Br₂ > I₂.
Cu²⁺ + 2e⁻ → Cu. For 2 faradays (2 moles of electrons), moles of Cu deposited = 2/2 = 1 mol. Mass = 1 × 64 = 64 g. Wait, let me recalculate: 2 faradays means 2 moles of e⁻. Cu²⁺ requires 2e⁻, so 2 moles of e⁻ deposits 1 mole of Cu = 64 g. The answer should be B, but checking: if 2 faradays, then 2×64/2 = 64g. Correction: Answer is 64g (option B). However, reviewing: 2F of charge produces 2 mol e⁻, which deposits 64g Cu. The listed answer C (128g) would require reconsideration - using the formula: mass = (charge × molar mass)/(n × F) = (2F × 64)/(2 × 96500) is incorrect reasoning. Correct: 2 faradays deposit 1 mole of Cu = 64g.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Cu²⁺ is reduced (cathode) and Zn is oxidized (anode).
A negative E° (E°cell < 0) indicates a non-spontaneous reaction under standard conditions. ΔG° = -nFE°, so negative E° gives positive ΔG°.
The metal that is more easily oxidized (more reactive) acts as the anode. Metal X, being more reactive, loses electrons and acts as the anode (negative electrode).
