Calculate the solubility product of AgI in water at 25°C [\(E^o_{(Ag^+, Ag)} = +0.799 volt\)] :- given \(AgI + e^– \rightarrow Ag + I^-\) \(E^o = –0.151\)
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 7.91 x 10–17
\(AgI + e^– \rightarrow Ag + I^-\) \(E^o = –0.151\)
\(Ag^+ + e^- \rightarrow Ag, E^o = 0.799\)
\(AgI\) ⇌ \(Ag^+I^-\)
\(E^o = –0.151 – 0.799\)
\(E^o = – 0.95\)
\(E = E^o – \frac{0.00.0591}{1}logK_{sp}\)
\(–16.074 = log_{10}K_{sp}\)
\(K_{sp} = antilog(–16.074)\)
\(K_{sp} = 7.91 × 10^{–17}\)