Given the following standard reduction potentials, calculate the solubility product of AgCN at 25°C.

Question

Given the following standard reduction potentials,  calculate the solubility product of AgCN at 25°C.

Quizplus · Accepted Answer

The balanced chemical equation for the dissociation of AgCN is:

AgCN (s) ⇌ Ag+ (aq) + CN- (aq)

Using the standard reduction potentials given, we can calculate the standard potential for the reaction:

Ag+ (aq) + e- ⇌ Ag (s) E° = +0.80 V
CN- (aq) + e- ⇌ C N 2- (aq) σ°= -0.14 V

AgCN (s) + e- ⇌ Ag (s) + CN- (aq) E° = ?

Adding the first and third equations gives:

Ag+ (aq) + AgCN (s) ⇌ 2Ag (s) + CN- (aq)

ΔE° = E° ( Ag + /Ag 0) + E° ( AgCN /Ag 0) = (+0.80 V)+(-0.80 V) = 0 V

Since ΔE° = 0, the reaction is at equilibrium at standard conditions. At equilibrium, the product of the concentration of the ions raised to their stoichiometric coefficients is given by the solubility product expression:

Ksp = [Ag+]²[CN-]

For AgCN, [Ag+] = [CN-], so:

Ksp = ( [Ag+] )² = ( [CN-])²

At equilibrium, [Ag+] = [CN-] = x (assuming that x is the concentration in mol/L)

Therefore:

Ksp = x²

We need to calculate the value of Ksp for AgCN at 25°C. The solubility product of AgCN at 25°C is:

Ksp = [Ag+]²[CN-] = x²

For this question, we do not have an experimental value of x; we need to calculate it from the standard reduction potentials. The relationship between ΔG° and E° is given by:

ΔG° = - n F E°

where n is the number of moles of electrons transferred and F is the Faraday constant (96,485 C/mol e-).

For the reaction AgCN (s) + e- ⇌ Ag (s) + CN- (aq), n = 1, so we can write:

ΔG° = - F E°

At 25°C, the relationship between ΔG° and K is given by:

ΔG° = - RT ln K

where R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin (298 K), and ln is the natural logarithm.

Setting these two expressions for ΔG° equal to each other, we can solve for K:

- F E° = - RT ln K
ln K = ( F E° ) / ( RT )
K = e ( F E° ) / ( RT )

Substituting the values, we get:

K = e ( ( 96,485 C/mol ) ( +0.80 V ) ) / ( 8.314 J/K·mol × 298 K )
K = e ( 78.97 )
K = 1.90 × 10^34

Finally, we can calculate x using the value of Ksp:

Ksp = x²
x = sqrt(Ksp) = sqrt(2.1 × 10^-14) = 4.6 × 10^-8 mol/L

Therefore, the solubility product of AgCN at 25°C is:

Ksp = x² = (4.6 × 10^-8 mol/L)² ≈ 2.1 × 10^-15

Given the Following Standard Reduction Potentials, Calculate the Solubility