Question

Calculate the solubility product constant of $AgI$ from the following values of standard electrode potentials.
$E_{A{g}^{+}/Ag}^{\circ }=0.80volt$ and $E_{I/AgI/Ag}^{\circ }=-0.15volt$ at ${25}^{\circ }C$.

Answer 1

Solubility product of $AgI=\left[A{g}^{+}\right]\left[I^{-}\right]$
Two half reactions for the cell are:
$Ag\to A{g}^{+}+e^{-}$ Anode (Oxidation)
$AgI+e^{-}\to Ag+I^{-}$
Cell reaction $\overline{AgI\to A{g}^{+}+I^{-}}$
Applying Nernst equation,
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{1}\log \frac{\left[A{g}^{+}\right]\left[I^{-}\right]}{\left[AgI\right]}$
At equilibrium, $E_{cell}=0$ and $\left[AgI\right]=1$
So, $\log \left[A{g}^{+}\right]\left[I^{-}\right]=\frac{E^{\circ }}{0.0591}$
$E_{cell}^{\circ }=-0.80-0.15=-0.95volt$
$\log \left[A{g}^{+}\right]\left[I^{-}\right]=-\frac{0.95}{0.0591}=-16.0774$
Solubility product of $AgI=8.4\times {10}^{-17}$.