ElectrochemistryHard

Question

Select the correct option if it is known that Ksp (AgCl) > Ksp (AgBr) > Ksp (AgI)

Options

A.$E_{I^{-}|AgI|Ag}^{o} > E_{Br^{-}|AgBr|Ag}^{o} > E_{Cl^{-}|AgCl|Ag}^{o}$
B.$E_{I^{-}|AgI|Ag}^{o} < E_{Br^{-}|AgBr|Ag}^{o} < E_{Cl^{-}|AgCl|Ag}^{o}$
C.$E_{I^{-}|AgI|Ag}^{o} < E_{Cl^{-}|AgCl|Ag}^{o} < E_{Br^{-}|AgBr|Ag}^{o}$
D.$E_{I^{-}|AgI|Ag}^{o} = E_{Br^{-}|AgBr|Ag}^{o} = E_{Cl^{-}|AgCl|Ag}^{o}$

Solution

$E_{X^{-}|AgX|Ag}^{o} = E_{Ag^{+}|Ag}^{o} - \frac{RT}{F}.\ln\frac{1}{K_{sp}}$

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