The solubility product of silver iodide is 8.3 x 10^–17 and the standard potential (reduction) of Ag, Ag⁺ electrode is +0.800 volts at 25°C. The standard potential of Ag, \(AgI|I^–\) electrode (reduction) from these data is

–0.15 V

The correct answer is option 4. –0.15 V.

\(E_{I^−/AgI/ Ag} = E^o_{Ag^+/Ag} + \frac{0.0591}{1}log[Ag^+] = E^o_{Ag^+/Ag}\)

and \(E^o_{I^−/AgI/ Ag} = E^o_{Ag^+/Ag} + \frac{0.0591}{1}logK_{sp}AgI\)

\(E^o_{I^−/AgI/ Ag} = 0.80 + 0.0591 log[8.3 × 10^{−17}] + 0.80 − 0.95 = −0.15 V\)