Final answer:
The equilibrium constant (K) for the reaction Cu(s) + 2Ag+(aq) = Cu2+(aq) + 2Ag(s) at 25°C with a given E°cell of -0.46 V is approximately 2.8 x 10^-16.
Step-by-step explanation:
To calculate the equilibrium constant for the reaction Cu(s) + 2Ag+(aq) = Cu2+(aq) + 2Ag(s) at 25°C with a given standard cell potential (E°cell) of -0.46 V, you can use the Nernst equation and the relationship between the standard free energy change and the equilibrium constant. The Nernst equation is E = E° - (RT/nF)ln(Q), where E is the cell potential, R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in kelvins, n is the number of electrons transferred in the reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. At equilibrium, E = 0 and Q = K (the equilibrium constant). The relationship between the standard free energy change (ΔG°) and E° is ΔG° = -nFE°. Let's calculate the value of the equilibrium constant (K).
The reaction transfers 2 electrons (n = 2), so we have ΔG° = -2 * F * (-0.46 V) = 2 * 96485 C/mol * 0.46 V = 88702.2 J/mol. Next, we use the relationship between ΔG° and K, which is ΔG° = -RTln(K). Solving for K gives us ln(K) = -ΔG°/RT. Substituting the values, we get ln(K) = -88702.2 J/mol / (8.314 J/(mol·K) * 298 K) = -35.68. Taking the exponential of both sides, we get K = e^{-35.68} ≈ 2.8 x 10^{-16}.