Given the data at 25, ^oC:Ag + I^{-} rightarr | vedclass

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(B) The cell reactions are:$(I) \ Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$$(II) \ Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$Subtractin

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$-16.13$

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(B) The cell reactions are:$(I) \ Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$$(II) \ Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$Subtracting $(I)$ from $(II)$ gives the reaction for the solubility product:$AgI \rightarrow Ag^{+} + I^{-}$$E^o_{cell} = E^o_{(II)} - E^o_{(I)} = 0.152 - (-0.800) = -0.952\, V$Using the relation $E^o_{cell} = \frac{0.059}{n} \log K_{sp}$ at $25\, ^oC$:$-0.952 = \frac{0.059}{1} \log K_{sp}$$\log K_{sp} = -\frac{0.952}{0.059} = -16.135 \approx -16.13$

List – $I$	List – $II$
$(a)$ Henry’s law	$(i)$ $p_1 = x_1 p_1^o$
$(b)$ Raoult’s law	(ii) $p = K_H x$
$(c)$ First law of thermodynamics	(iii) $\Lambda_m^o = \nu_+ \lambda_+^o + \nu_- \lambda_-^o$
$(d)$ Kohlrausch’s law	(iv) $\Delta U = q + w$

Given the data at $25\, ^oC$:
$Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$
$Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$
What is the value of $\log K_{sp}$ for $AgI$? $(2.303\, RT/F = 0.059\, V)$

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Given the data at $25\, ^oC$:$Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$$Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$What is the value of $\log K_{sp}$ for $AgI$? $(2.303\, RT/F = 0.059\, V)$