Question

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

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

Answer 1

Answer: (D)

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