At 25^circ text{C},given the following data: | vedclass

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(D) The cell reaction for the dissolution of $AgI$ is: $AgI_{(s)} \rightarrow Ag^+_{(aq)} + I^-_{(aq)}$.This can be obtained by subtracting the first

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$-16.13$

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(D) The cell reaction for the dissolution of $AgI$ is: $AgI_{(s)} \rightarrow Ag^+_{(aq)} + I^-_{(aq)}$.This can be obtained by subtracting the first reaction from the second reaction:$(Ag_{(s)}$ $\rightarrow Ag^+_{(aq)} + e^-) - (Ag_{(s)} + I^-_{(aq)}$ $\rightarrow AgI_{(s)} + e^-)$ $\Rightarrow AgI_{(s)}$ $\rightarrow Ag^+_{(aq)} + I^-_{(aq)}$.The standard cell potential $E^o_{cell} = E^o_{ox} + E^o_{red} = (-0.800 \ V) - (0.152 \ V) = -0.952 \ V$.Using the Nernst equation at equilibrium: $E^o_{cell} = \frac{0.0591}{n} \log K_{sp}$.Here $n = 1$,so $-0.952 = 0.0591 \log K_{sp}$.$\log K_{sp} = \frac{-0.952}{0.0591} \approx -16.11 \approx -16.13$ (considering standard approximation $0.059$).

At $25^\circ \text{C}$,given the following data: $Ag_{(s)} + I^-_{(aq)} \rightarrow AgI_{(s)} + e^-$,$E^o = 0.152 \ V$; $Ag_{(s)} \rightarrow Ag^+_{(aq)} + e^-$,$E^o = -0.800 \ V$. What is the value of $\log \ K_{sp}$ for $AgI$? (where $K_{sp} = \text{solubility product}$)

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