Given the following standard reduction potentials, Ag⁺(aq) + e^- → Ag(s) E° = +0.80 V
AgCN(s) + e^- → Ag(s) + CN^-(aq) E° = -0.01 V
Calculate the solubility product of AgCN at 25°C.

Question

Given the following standard reduction potentials, Ag⁺(aq)+ e^- → Ag(s)E° = +0.80 V AgCN(s)+ e^- → Ag(s)+ CN^-(aq)E° = -0.01 V Calculate the solubility product of AgCN at 25°C.

Quizplus · Accepted Answer

The solubility product of AgCN can be calculated using the Nernst equation:E° = E°cell - (RT/nF)lnQwhere E° is the standard reduction potential, E°cell is the cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.In this case, the cell potential is 0 V, since the reaction is at equilibrium. The number of electrons transferred is 1. The temperature is 25°C, which is 298 K. Substituting these values into the Nernst equation, we get:0 V = -0.01 V - (0.0257 V/K)(1)(lnQ)Solving for Q, we get:Q = 10^-14The solubility product of AgCN is equal to Q, so the solubility product of AgCN is 10^-14.

Given the Following Standard Reduction Potentials, Ag+(aq)+ E- → Ag(s)E°

Given the Following Standard Reduction Potentials, Ag⁺(aq)+ E^- → Ag(s)E°