Given at 25^ : AgI + e^- → Ag + I^- , E^= -0.152 V Ag^+ + e^- → Ag , E^= 0.800 V What is the value of K_sp for AgI ? ( 2.303RT/F = 0.059 V ) - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live

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Given at 2 5^{\circ} C : AgI + e^{-} \to Ag + I^{-}, E^{\circ} = - 0.152 V Ag^{+} + e^{-} \to Ag, E^{\circ} = 0.800 V What is the value of lo g K_{s p} for AgI ? (\frac{2.303 RT}{F} = 0.059 V)

Solution & Step-by-step Explanation

The reaction for K_{s p} is: AgI (s) ⇌ Ag^{+} (a q) + I^{-} (a q) .This can be obtained by subtracting (2) from (1): E_{cell}^{\circ} = E_{1}^{\circ} - E_{2}^{\circ} = - 0.152 - 0.800 = - 0.952 V .We know Δ G^{\circ} = - n F E_{cell}^{\circ} = - RT ln K_{s p} . E_{cell}^{\circ} = \frac{0.059}{n} lo g K_{s p} - 0.952 = \frac{0.059}{1} lo g K_{s p} lo g K_{s p} = \frac{- 0.952}{0.059} \approx - 16.13

Try it yourself before checking the explanation above.

Given at 2 5^{\circ} C : AgI + e^{-} \to Ag + I^{-}, E^{\circ} = - 0.152 V Ag^{+} + e^{-} \to Ag, E^{\circ} = 0.800 V What is the value of lo g K_{s p} for AgI ? (\frac{2.303 RT}{F} = 0.059 V)

-8.12

+8.612

-37.83

-16.13

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