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

First, let's write the half-reactions and the overall reaction for this concentration cell:

Cathode: Ag+ (1 M) + e- → Ag (s)
Anode: Ag (s) → Ag+ (x M) + e-
Overall: Ag+ (1 M) ⇌ Ag+ (x M)

The standard reduction potential for the cathode half-reaction is +0.80 V, and since the overall cell potential is positive, we know that the anode half-reaction has a higher potential. Therefore, we can use the Nernst equation to relate the two concentrations:

Ecell = Ecathode - Eanode = +0.4759 V

Ecell = (0.0592 V/n) log(Q)
where n is the number of electrons transferred in the cell reaction, and Q is the reaction quotient, which is the ratio of product concentrations to reactant concentrations raised to their stoichiometric coefficients.

At equilibrium, Q = Ksp:

Ksp = [Ag+]x [I-]y

Since the cathode compartment is at 1 M and the anode compartment is at x M, the reaction quotient is:

Q = [Ag+]cathode/[Ag+]anode = 1/x

Substituting into the Nernst equation:

Ecell = (0.0592 V/n) log(1/x) = (0.0592 V/1) log(x)

0.4759 V = 0.0592 V log(x)
x = 10^(0.4759 V/0.0592 V) = 10^8.0428 M

Now we can calculate the solubility product constant:

Ksp = [Ag+][I-] = (1 M)(10^(-8.0428 M)) = 8.362 × 10^-17

Therefore, the answer is (A).

A Concentration Cell with a Cell Potential of +0 ×\times× 10-17 B)7

A Concentration Cell with a Cell Potential of +0 $\times$ 10^-17
B)7