6. Calculate the solubility product of silver iodide at 25°C given the following

Question

6. Calculate the solubility product of silver iodide at 25°C given the following data: 1. AgI(s) + e– Ag(s) + I– E° = –0.15 2. I2(s) + 2e– 2I– E° = +0.54 3. Ag+ + e– Ag(s) E° = +0.80 HINT: To solve this problem, realize that Equation 1 is the same as Equation 3 under non-standard conditions. Apply the Nernst equation to Equation 3 to find what [Ag+] would reduce the potential from +0.80 v to –0.15 v. This is the [Ag+] when AgI dissolves into a solution which has [I–] = 1 M. In a saturated solution of AgI with [I–] = 1M, [Ag+] = Ksp of AgI. A) 2 10–12 B) 3 10–3 C) 2 10–24 D) 9 10–17 E) 2 10–4

Explanation / Answer

AgI(s) + e- ==> Ag(s) + I-, Eº = -0.15 V . . . (1)
I2(s) + 2e- ==> 2I-, Eº = +0.54 V . . . . . . . . (2)
Ag+ + e- ==> Ag(s), Eº = +0.80 V . . . . . . . (3)
An assumptional cell is from 0.5*(2)-(3):
0.5*I2(s) + Ag(s) ==> Ag+ + I-, EMFº = -0.26V . . . (4)
where, of course, [Ag+] = [I-] = 1.000M for the EMF to be -0.26V. When [Ag+] got drastically reduced, EMF turns to be positive, as is from 0.5*(2)-(1):
0.5*I2(s) + Ag(s) ==> AgI(s), EMFº = 0.69V . . . . . (5)
Therefore, AgI(s) should be correctly understood as the realistic saturated concentration of [Ag+] and [I-] in the solution. From (4)-(5):
AgI(s) ==> Ag+ + I-, EMFº = -0.95V
At equilibrium, Eº = RT/nF Ln Ksp, hence:
Ksp = exp(-0.95*1*96485/(8.314*298.15)) = 8.73x10^-17

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