Given the data at 25°C, Ag + I- → AgI + e-; E°= 0.152 V Ag → Ag+ + e- ; E° = - 0.800 V What is the value of log Ksp for Agl ? (2.303 (RT/F) = 0.059 V)

Question

Given the data at 25°C, Ag + I- → AgI + e-; E°= 0.152 V Ag → Ag+ + e- ; E° = - 0.800 V What is the value of log Ksp for Agl ? (2.303 (RT/F) = 0.059 V) (A) -8.12 (B) 8.612 (C) -37.83 (D) -16.13

Tardigrade Admin · Accepted Answer

AgI(s) + e- leftharpoons Ag(s) + I-; E° = - 0.152+Ag (s)→ Ag+ + e- E° = - 0.8 Agl(s) → Ag+ + I- E° = - 0.952 E°cell = (0.059/n) log Ksp - 0.952 = (0.059/1) log Ksp log Ksp = (-0.952/0.059) = -16.135

Q. Given the data at 25∘C, Ag+I−→AgI+e−;E∘=0.152V Ag→Ag++e−;E∘=−0.800V What is the value of log Ksp​ for Agl ? (2.303FRT​=0.059V)

-8.12

8.612

-37.83

-16.13

AgI(s)+e−⇌Ag(s)+I−;E∘=−0.152+Ag(s)→Ag++e−E∘=−0.8 Agl(s)→Ag++I−E∘=−0.952 Ecell∘​=n0.059​logKsp​ −0.952=10.059​logKsp​ logKsp​=0.059−0.952​=−16.135

Q. Given the data at 25 $^{\circ}$ C,
$A g + I^{-} \to A g I + e^{-}; E^{\circ} = 0.152 V$
$A g \to A g^{+} + e^{-}; E^{\circ} = - 0.800 V$
What is the value of log $K_{s p}$ for Agl ?
$(2.303 \frac{RT}{F} = 0.059 V)$