Question

Given the data at ${25}^{0}C$,
$Ag+I^{-}\to AgI+e^{-};E^0=0.152V$
$Ag\to A{g}^{+}+e^{-};E^0=-0.800V$

What is the value of log $K_{sp}$ for $AgI$?
$(Given:2.303RT/F=0.059V)$

• A. $-16.13$
• B. $-8.12$
• C. $+8.612$
• D. $-37.83$

Answer 1

Answer: (A)

$-16.13$

At cathode: $A{g}^{+}+e^{-}⟶Ag$ $E_{red}^o=0.800V$

At anode: $Ag+I^{-}⟶AgI+e^{-}$ $E_{oxd}^o=0.152V$

______________________

Cell reaction $A{g}^{+}+I^{-}⟶AgI$ $E_{cell}^o=0.8+0.152V=0.952V$

Note: The reaction will take place at cathode whose reduction potential is higher.

Using Nernst Equatiuon,

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{1}{\left[A{g}^{+}\right]\left[I^{-}\right]}$

At $Eq{b}^m,\Delta G=0$, as $\Delta G=-nF{E}_{cell}$

so, $E_{cell}=0$

$\left[A{g}^{+}\right]\left[I^{-}\right]=K_{sp}$

$\Rightarrow 0=E_{cell}^o-0.0591\log \frac{1}{K_{sp}}$

$\Rightarrow 0=0.952V+0.0591\log K_{sp}$

$\Rightarrow \log K_{sp}=\frac{-0.952V}{0.0591}=-16.13V$

$\Rightarrow \log K_{sp}=-16.13V$