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Calculate the solubility product of silver iodide at 25°C given the following data: -0.15 +0.54 +0.80 Agl(s) + e-→ Ag(s) +「 12(s) + 2e-→ 2「 Ag+ + e-→ Ag(s) A) 2.9 x 10 B) 1.9 x 104 C) 2.1 x 10-12 D) 9.0×10-17 E) 2.4 x 10-24Step-by-Step approach please?

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Answer #1

Ans : D) 9.0 x 10-17

AgI disolves as :

AgI = Ag+ + I-

ksp = [Ag+] [I-]

The first and third reaction can be added to get the desired reaction for the dissolution of AgI

AgI + e- = Ag + I-   Eo = -0.15 V

Ag = Ag+ + e-   Eo = -0.80 V

So Eo for the reaction will be -0.15 + -0.80 = -0.95 V

Now using the Nernst equation :

Ecell = Eocell - (0.059/n) log K

0 = -0.95 - (0.059/1) log K

Log K = 0.95 / -0.059

K = 9.0 x 10-17

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