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

Quizplus · Accepted Answer

The solubility product constant for the dissolution of AgCN can be calculated using the Nernst equation:

E°cell = E°cathode - E°anode
E°cell = 0.80 V - (-0.01 V)
E°cell = 0.81 V

n = number of electrons transferred = 1
F = Faraday's constant = 96485 C/mol
R = gas constant = 8.314 J/(mol K)
T = temperature = 298 K

Ksp = exp(nF(E°cell)/(RT))
Ksp = exp((1)(96485 C/mol)(0.81 V)/(8.314 J/(mol K)(298 K)))
Ksp = 2.1 x 10^-14

Therefore, the solubility product of AgCN is 2.1 x 10^-14 at 25°C. The correct choice is (C).

Given the Following Standard Reduction Potentials, Ag+(aq)+ E- →\rarr→ Ag(s)E° = 0

Given the Following Standard Reduction Potentials, Ag⁺(aq)+ E^- $\rarr$ Ag(s)E° = 0