Calculate the solubility product of AgI in water at 25°C [\(E^o_{(Ag^+, Ag)} = +0.799 volt\)] :- given \(AgI  + e^– \rightarrow Ag + I^-\) \(E^o = –0.151\)

7.91 x 10^–17

The correct answer is option 2. 7.91 x 10^–17

\(AgI  + e^– \rightarrow Ag + I^-\) \(E^o = –0.151\)

\(Ag^+  +  e^- \rightarrow Ag, E^o = 0.799\)

\(AgI\) ⇌ \(Ag^+I^-\) 

\(E^o = –0.151 – 0.799\)

\(E^o = – 0.95\)

\(E = E^o – \frac{0.00.0591}{1}logK_{sp}\)

\(–16.074 = log_{10}K_{sp}\)

\(K_{sp} = antilog(–16.074)\)

\(K_{sp} = 7.91 × 10^{–17}\)