JEE Advanced level
A JEE Advanced-level example on Redox Reactions and Electrochemistry
Question. A galvanic cell is set up with Ag⁺(0.001 M)|Ag as one half-cell and Cu²⁺(0.1 M)|Cu as the other. Given E°(Ag⁺/Ag) = +0.80 V, E°(Cu²⁺/Cu) = +0.34 V, and 2.303RT/F = 0.059 V at 298 K. Calculate: (a) the EMF of the cell, (b) the mass of copper that would be deposited if 965 C of charge is passed through a Cu²⁺ solution (atomic mass of Cu = 64), and (c) the Gibbs free energy change for the cell reaction.
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- Cell reaction: 2Ag⁺ + Cu → 2Ag + Cu²⁺ (Ag electrode is cathode since it has higher E°). E°cell = 0.80 − 0.34 = 0.46 V, n = 2
- Apply Nernst equation: Ecell = E°cell − (0.059/n) × log([Cu²⁺]/[Ag⁺]²) = 0.46 − (0.0295) × log(0.1/10⁻⁶) = 0.46 − 0.0295 × 5 = 0.31 V
- For copper deposition: moles of e⁻ = 965/96500 = 0.01 mol. Since Cu²⁺ + 2e⁻ → Cu, moles of Cu = 0.005, mass = 0.005 × 64 = 0.32 g
- Gibbs free energy: ΔG = −nFEcell = −2 × 96500 × 0.31 = −59830 J ≈ −60 kJ
Answer: (a) Ecell = 0.31 V, (b) Mass of Cu = 0.32 g, (c) ΔG = −60 kJ
This example was solved twice independently and both solutions agree.