Given the data at 25 °C Ag+I−→AgI+e− E∘=0.152V Ag→Ag++e− E∘=−0.800VWhat is the value of log Ksp∘for Agl? (2.303 RT/IF = 0.059 V)

Given the data at 25 °C $Ag + I^{-} \to AgI + e^{-} E^{\circ} = 0.152 V Ag \to {Ag}^{+} + e^{-} E^{\circ} = - 0.800 V$
What is the value of log $K_{sp}^{\circ}$ for Agl? (2.303 RT/IF = 0.059 V)

A
-16.13
B
-8.12
C
+8.612
D
-37.83

Solution:

$E °$ for the reaction $AgI \to {Ag}^{+} + I^{-} is E_{cell}^{0} = (- 0.800 - 0.152) V = - 0.952 V$

$\log K_{sp}^{\circ} = \frac{E_{cell}^{\circ}}{R T / F} = \frac{- 0.952 V}{0.052 V} = - 16.13$

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