Answer to Question #211739 in Physical Chemistry for Reese

Question

A concentration cell consists of two half-cells with identical electrodes but different ion concentrations.

Consider the following concentration cell (pictured below).

M(s) | M⁺(satd MX) || M⁺(0.100 M)| M(s)   E_cell = 0.417 V

At the anode, a metal electrode (M) is placed in a saturated aqueous solution containing ions of the same metal (M) and a halogen (X), MX. At the cathode, a second electrode of the same metal (M) is placed in a solution with [M⁺] = 0.100 M.The two half-cells are connected by a salt bridge, and the measured cell voltage is 0.417 V. The cell reaction occurring in this concentration cell is

Reduction: M⁺(0.100 M) + e^- → M(s)

Oxidation: M(s) → M⁺(satd MX) + e^-

From this information, calculate the K_sp of MX. 

Answer 1

$K_{sp}(mx) =S^2=(8.79×10 ^{-9})^2$

$K_{sp}=7.73×10 ^{-17}$