Question

The standard reduction potential of $A{g}^{+}/Ag$ electrode at $298K$ is $0.799volt$. Given for $Agl,K_{sp}=8.7\times {10}^{-17}$, evaluate the potential of the $A{g}^{+}/Ag$ electrode in a saturated solution of $Agl$. Also calculate the standard reduction potential of the $I^{-}/AgI/Ag$ electrode.

Answer 1

In the saturated solution of $AgI$, the half-cell reactions are:
$AgI+e^{-}\to Ag+I^{-}$ Cathode (Reduction)
$Ag\to A{g}^{+}+e^{-}$
Cell reaction $\overline{AgI\to A{g}^{+}+I^{-}}$
$E_{A{g}^{+}/Ag}=E_{A{g}^{+}/Ag}^{\circ }+0.059\log \left[A{g}^{+}\right]$
$\left[A{g}^{+}\right][I^{-}=K_{sp}(AgI)=[A{g}^{+}{]}^2=[I^{-}{]}^2$
So, $[A{g}^{+}{]}^2=8.7\times {10}^{-17}$
$\left[A{g}^{+}\right]=\sqrt{8.7\times {10}^{-17}}=9.3\times {10}^{-9}$
Substituting the values of $E_{A{g}^{\circ }/Ag}^{\circ }$ and $\left[A{g}^{+}\right]$ in the above equation.
$E_{A{g}^{+}/Ag}=0.799+0.059\log (9.3\times {10}^{-9})$
$=0.324volt$
$E_{cell}^{\circ }=0.0591\log K_{sp}\left(AgI\right)$
$=0.0591\log (8.7\times {10}^{-17})$
$=-0.95volt$
$E_{cell}^{\circ }=$ Oxid. pot. of anode + Red. pot. of cathode
Red. pot. of cathode $E_{I^{-}/AgI/Ag}^{\circ }=-0.95-(-0.799)$
$=-0.95+0.799$
$=-0.151volt$.