Electrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small. Calculate the value of K_sp for CdS from the following data.
CdS(s) + 2e^- $\rightarrow$ Cd(s) + S²^-(aq)
E° = -1.21 V
Cd²⁺(aq) + 2e^- $\rightarrow$ Cd(s)
E° = -0.40 V

Question

Electrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small. Calculate the value of K_sp for CdS from the following data.
CdS(s) + 2e^-

\rightarrow

Cd(s) + S²^-(aq)
E° = -1.21 V
Cd²⁺(aq) + 2e^-

\rightarrow

Cd(s)
E° = -0.40 V

Quizplus · Accepted Answer

First, we need to write the balanced equation for the reaction:
CdS(s) + Cd(s) → 2CdS1U1P12
We can obtain the cell potential for this reaction by adding the cell potentials given for the two half-reactions:
E°cell = E°reduction (CdS1U1P12+S1S1P0(aq) + 2eS1U1P1-S1S1P0 F1F1F1S1F1F1F10 Cd(s)) - E°oxidation (CdS(s) + 2eS1U1P1-S1S1P0S1S1P0S1U1P1)
E°cell = (-0.40 V) - (-1.21 V) = 0.81 V

The equilibrium constant is related to the standard cell potential by the Nernst equation:
Ecell = E°cell - (RT/nF)ln(K)
where R is the gas constant, T is the temperature (in Kelvin), F is the Faraday constant, n is the number of electrons transferred in the reaction, and ln(K) is the natural logarithm of the equilibrium constant.

At room temperature (25°C or 298 K), we have:
Ecell = 0.81 V
R = 8.314 J/(mol·K)
F = 96,485 C/mol
n = 2

Plugging in the values, we get:
0.81 V = -0.0257 V ln(K)
ln(K) = -31.52

Solving for K, we get:
K = 3.57 x 10^-14

The equation for the dissolution of CdS is:
CdS(s) ⇌ Cd2+(aq) + S2-(aq)

The solubility product constant, Ksp, is given by:

Ksp = [Cd2+][S2-]

Assuming that the dissolution of CdS is complete (i.e., all of the CdS has dissolved), then [Cd2+] = [S2-] = x, where x is the molar concentration of Cd2+ and S2-. Therefore:

Ksp = x^2

Substituting the value of Ksp and solving for x, we get:
Ksp = 3.57 x 10^-14 = x^2
x = 1.89 x 10^-7 M

Since the concentration of CdS is equal to its solubility (assuming complete dissolution), we can calculate the mass of CdS that dissolves in a given volume of water:
mass = molar mass × moles = molar mass × concentration × volume

The molar mass of CdS is 144.46 g/mol.

Substituting the values, we get:
mass = 144.46 g/mol × (1.89 x 10^-7 mol/L) × 1000 mL = 2.73 x 10^-5 g

Therefore, the best choice is B, 4 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P128S1S1P0, since it gives the correct value of Ksp.

Electrochemical Cell Potentials Can Be Used to Determine Equilibrium Constants →\rightarrow→

Electrochemical Cell Potentials Can Be Used to Determine Equilibrium Constants $\rightarrow$