Question

The solubility product of silver iodide is $8.3\times 10^{-17}$ and the standard reduction potential of $Ag/A{g}^{+}$ electrode is $+0.8$ volts at $25^{\circ }C.$ The standard reduction potential of $Ag,AgI/I^{-}$ electrode from these data is

• A. $-0.30V$
• B. $+0.16V$
• C. $+0.10V$
• D. $-0.16V$

Answer 1

Answer: (D)

Anode reaction $Ag\rightleftharpoons A{g}^{+}+e^{-}$
Cathode reaction $AgI+e^{-}\rightleftharpoons Ag+I^{-}$
$E_{\text{cell }}^{\circ }={\left(E_{\text{reduction }}^{\circ }\right)}_c-{\left(E_{\text{oxidation }}^{\circ }\right)}_a$
$E_{AgI/I^{-}}^{\circ }-E_{Ag/A{g}^{+}}^{\circ }$
$=x-0.8V$
Cell is in equilibrium, thus $E_{\text{cell }}=O$
$\therefore E_{cell}=E_{cell}^{\circ }-\frac{0.059}{1}\log \left[A{g}^{+}\right]\left[I^{-}\right]$
$0=(x-0.8V)\frac{0.06}{1}\log K_{sp}(0.059\approx 0.06)$
$x-0.8V=0.06\log K_{sp}$
$x-0.8V=0.06\log \left(8.3\times 10^{-17}\right)$
$=0.06[\log 8.3-17\log 10]$
$=0.06[0.9191-17]=0.06\times (-16.1)$
$x-0.8V=-0.966$
$x=(-0.966+0.8)V=-0.16V$