Question

For $AgI$, the solubility product $K_{sp}=1\times {10}^{-16}$. What will be the electrode potential of $Ag$ electrode at $25$${}^{o}C$ in a saturated solution of AgI in water (write answer as nearest integer after multiplying it by $10$)? $E_{A{g}^{+}/Ag}^o=0.80$

Answer 1

$A{g}^{+}+e^{-}\to$ $E^{0}A{g}^{+}|Ag=0.80V$
According to Nernst equation
$E=E^{0}A{g}^{+}|Ag-\frac{0.059}{n}log\frac{1}{\left[A{g}^{+}\right]}$
In a saturated solution of AgI,
$AgI\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I^{-}\left(aq\right)$
$K_{sp}\left(AgI\right)=\left[A{g}^{+}\right]\left[I^{-}\right]=1\times {10}^{-16}$
But $\left[A{g}^{+}\right]=\left[I^{-}\right]$
Hence, $\left[A{g}^{+}\right]=\sqrt{1\times {10}^{-16}}=1\times {10}^{-8}$.
Substitute the values in the Nernst equation.
$E=0.80-\frac{0.059}{1}log\frac{1}{1\times {10}^{-8}}=0.328V$.