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The standard reduction potential of $Ag^+/Ag$ electrode at 298K is 0.79V.$K_{sp}$ of Ag$^+$ is $8.7\times 10^{-17}$.Evaluate the standard reduction potential of $I^-/AgI/Ag$ electrode

$\begin{array}{1 1}(a)\;-0.149V\\(b)\;-0.37V\\(c)\;-0.26V\\(d)\;-0.4V\end{array}$

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$Ag^++e^-\rightarrow Ag$
$E_{Ag^+/Ag}=E^0_{Ag^+/Ag}-0.059\log \large\frac{1}{[Ag^+]}$
$\Rightarrow E^0_{Ag ^+/Ag}-0.059\log \large\frac{1}{\sqrt{K_{sp}}}$
$E_{Ag^+/Ag}=0.799-0.059\log \large\frac{1}{\sqrt{8.7\times 10^{-17}}}$
$\Rightarrow 0.325V$
$AgI\leftrightharpoons Ag^++I^-$------(i)
$Ag^++e^-\rightarrow Ag$---------(ii)
$AgI+e^-\rightarrow Ag+I^-$--------(iii)
We have learnt that metal-insoluble salt-anion electrode has same potentials.
Thus,eq(ii) & eq(iii) would have same reduction potential.
$\therefore E_{Ag^+/Ag}=E^0{I^-/AgI/Ag}-0.059\log [I^-]$
$0.325=E^0_{I^-/AgI/Ag}-0.059\log \sqrt{K_{sp}}$
$\Rightarrow E^0_{I^-/AgI/Ag}-0.059\log \sqrt{8.7\times 10^{-17}}$
$\therefore E^0_{I^-/AgI/Ag}=-0.149V$
Hence (a) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

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