A concentration cell with a cell potential of +0.4759 V at 298 K is made using silver electrodes and aqueous silver ion solutions at 298 K.In one compartment, the Ag⁺ concentration is 1.000 M.In the other compartment, the Ag⁺ concentration is due to the dissolution of slightly soluble AgI.Calculate the value of K_sp for AgI.

Question

Quizplus · Accepted Answer

The cell potential (E) is related to the concentrations of the ions in the two compartments by the Nernst equation. For a concentration cell involving Ag+ ions, the equation simplifies to E = (0.0592/n) log(Q), where Q is the reaction quotient, n is the number of electrons transferred (which is 1 for Ag+/Ag), and E is given as 0.4759 V. Since one compartment has Ag+ at 1.000 M and the other has Ag+ at a concentration due to the dissolution of AgI, we can express Q as [Ag+] from AgI / 1.000 M. The solubility product (Ksp) for AgI is the product of the concentrations of its ions in a saturated solution, which can be represented as [Ag+][I-] since AgI dissociates into Ag+ and I-. Given the cell potential and the known concentration of Ag+ in one compartment, we can solve for the concentration of Ag+ in the other compartment and thus determine Ksp for AgI. The correct choice involves calculating Ksp based on the given cell potential and the relationship between cell potential, ion concentrations, and Ksp, leading to the conclusion that the Ksp value corresponds to the choice that involves a calculation based on these principles.

A Concentration Cell with a Cell Potential of +0