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The half potential of $Cl^{-1}(C_1)/AgCl/Ag$ electrode.Given the value of $E^0$ for $Ag^++e^-\rightarrow Ag$ and $K_{sp}$ of AgCl is

$\begin{array}{1 1}(a)\;E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1}{K_{sp}^2}\\(b)\;E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1^2}{K_{sp}^2}\\(c)\;E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1}{K_{sp}}\\(d)\;E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1}{K_{sp}}\end{array}$

1 Answer

We can see that the $\Delta G_1$,for the reaction (i) is zero as it is in equlibrium $\Delta G_2$ for the second reaction is $-nFE$ while for the third reaction is $-nFE^1$.Since $\Delta G$ is a thermodynamic quantity,it will follow the Fless's law,according to which all thermodynamic quantaties can be treated in the same manner as we treat the reaction.
That is to say if reaction $\Delta G_3=\Delta G_1+\Delta G_2$
$\Delta G_1$ is zero,$\Delta G_3=\Delta G_2$.This implies that $-nFE^1=-nFE$
This means that the half cell potential of third reaction is equal to that of second one.Now to calculate the half cell potential of $2^{nd}$ reaction.
$Ag^++e^-\rightarrow Ag$
We need its $E^0$ and concentration of $Ag^+$
$E^0$ is given and $[Ag^+]$ can be calculated as
$\therefore E^1=E=E^0_{Ag^+/Ag}-0.059\log \large\frac{1}{[Ag^+]}$
$\Rightarrow E^0_{Ag^+/Ag}-0.059\log \large\frac{C_1}{K_{sp}}$
Hence (d) is the correct answer.
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