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)

E° for the reactionAgI→$$Ag++I$$− is Ecell $$0=($$−$$0.800$$−$$0.152)V=$$−$$0.952V$$ log⁡Ksp∘$$=Ecell$$∘$$RT/F=$$−$$0.952V0.052V=$$−$$16.13$$

Questions

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)

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detailed solution

Correct option is A

$E °$ 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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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)

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Ready to Test Your Skills?

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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)