The question is asking for the solubility product constant (Ksp) of AgCl at 25°C. We can use the Nernst equation to solve this problem. The Nernst equation is:

E = E° - (RT/nF) * lnQ

where:
E is the cell potential,
E° is the standard cell potential,
R is the gas constant,
T is the temperature in Kelvin,
n is the number of moles of electrons transferred in the redox reaction,
F is the Faraday constant, and
Q is the reaction quotient.

Given in the problem, we have E = -0.209 V, E° = -0.799 V, T = 25°C = 298 K, and n = 1 (since the reaction is Ag+ + e- - Ag, only one electron is transferred).

We can plug these values into the Nernst equation:

-0.209 = -0.799 - (8.314 * 298 / (1 * 96485)) * lnQ

Solving for Q gives Q = [Ag+][Cl-] = 10^((E°-E)*nF/RT) = 10^((0.799-(-0.209))*96485/(8.314*298)) = 1.77*10^-10

Since the solution is saturated with AgCl, [Ag+] = [Cl-] = sqrt(Ksp). Therefore, Ksp = Q = 1.77*10^-10.

Question

The question is asking for the solubility product constant (Ksp) of AgCl at 25°C. We can use the Nernst equation to solve this problem. The Nernst equation is:

E = E° - (RT/nF) * lnQ

where:
E is the cell potential,
E° is the standard cell potential,
R is the gas constant,
T is the temperature in Kelvin,
n is the number of moles of electrons transferred in the redox reaction,
F is the Faraday constant, and
Q is the reaction quotient.

Given in the problem, we have E = -0.209 V, E° = -0.799 V, T = 25°C = 298 K, and n = 1 (since the reaction is Ag+ + e- -> Ag, only one electron is transferred).

We can plug these values into the Nernst equation:

-0.209 = -0.799 - (8.314 * 298 / (1 * 96485)) * lnQ

Solving for Q gives Q = [Ag+][Cl-] = 10^((E°-E)*nF/RT) = 10^((0.799-(-0.209))*96485/(8.314*298)) = 1.77*10^-10

Since the solution is saturated with AgCl, [Ag+] = [Cl-] = sqrt(Ksp). Therefore, Ksp = Q = 1.77*10^-10.

Knowee AI · Accepted Answer

The question is asking for the solubility product constant (Ksp) of AgCl at 25°C. We can use the Nernst equation to solve this problem. The Nernst equation is:

E = E° - (RT/nF) * lnQ

where:
E is the cell potential,
E° is the standard cell potential,
R is the gas constant,
T is the temperature in Kelvin,
n is the number of moles of electrons transferred in the redox reaction,
F is the Faraday constant, and
Q is the reaction quotient.

Given in the problem, we have E = -0.209 V, E° = -0.799 V, T = 25°C = 298 K, and n = 1 (since the reaction is Ag+ + e- -> Ag, only one electron is transferred).

We can plug these values into the Nernst equation:

-0.209 = -0.799 - (8.314 * 298 / (1 * 96485)) * lnQ

Solving for Q gives Q = [Ag+][Cl-] = 10^((E°-E)*nF/RT) = 10^((0.799-(-0.209))*96485/(8.314*298)) = 1.77*10^-10

Since the solution is saturated with AgCl, [Ag+] = [Cl-] = sqrt(Ksp). Therefore, Ksp = Q = 1.77*10^-10.

A silver wire dipped in 0.1 M HCl solution saturated with AgCl at 25°C develops oxidation potential of –0.209 V. If = –0.799V, the Ksp of AgCl at 25°C is

Question

A silver wire dipped in 0.1 M HCl solution saturated with AgCl at 25°C develops oxidation potential of –0.209 V. If $E^\circ = –0.799V$ , the Ksp of AgCl at 25°C is

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