Given the data at $25^{\circ} \mathrm{C}$, $ \begin{aligned} & \mathrm{Ag}+\mathrm{I}^{-} \longrightarrow…

Given the data at $25^{\circ} \mathrm{C}$, $ \begin{aligned} & \mathrm{Ag}+\mathrm{I}^{-} \longrightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\circ}=0.152 \mathrm{~V} \\ & \mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \quad \mathrm{E}^{\circ}=-0.800 \mathrm{~V} \end{aligned} $ What is the value of $\log \mathrm{K}_{\mathrm{sp}}$ for $\mathrm{AgI}$ ? $ \left(2.303 \frac{R T}{F}=0.059 \mathrm{~V}\right) $
  1. $-8.12$
  2. $+8.612$
  3. $-37.83$
  4. $-16.13$

Solution

$ \begin{array}{ll} \mathrm{AgI}(\mathrm{s})+\mathrm{e}^{-} \rightleftharpoons \mathrm{Ag}(\mathrm{s})+\mathrm{I}^{-} ; & \mathrm{E}^{\circ}=-0.152 \\ \mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; & \mathrm{E}^{\circ}=-0.8 \\ \hline \mathrm{AgI}(\mathrm{s}) \longrightarrow \mathrm{Ag}^{+}+\mathrm{I}^{-} ; & \mathrm{E}^{\circ}=-0.952 \\ \mathrm{E}_{\mathrm{cell}}^{\circ}=\frac{0.059}{\mathrm{n}} \operatorname{logK} & \\ -0.952=\frac{0.059}{1} \log \mathrm{K}_{\mathrm{sp}} \\ \log \mathrm{K}_{\mathrm{sp}}=-\frac{0.952}{0.059}=-16.135 \end{array} $

Asked in: JEE Main 2006

Practice more Electrochemistry questions on Aicharya