Question

Given the data at 25°C

\[\ce{Ag + I- -> AgI + e-}\]; E° = – 0.152 V

\[\ce{Ag -> Ag+ + e-}\]; E° = – 0.800 V

The value of log K_sp for AgI is ______.

Options

• – 8.12

• +8.162

• – 37.83

• – 16.13

Answer 1

The value of log K_sp for AgI is –16.13.

Explanation:

\[\ce{Ag + I- -> AgI + e-}\]; E° = – 0.152 V

\[\ce{Ag -> Ag+ + e-}\]; E° = – 0.800 V

\[\ce{AgI -> Ag+ + I-}\] ; E° = – 0.952 V

\[\ce{E{^{\circ}_{cell}} = \frac{0.0591}{1} log K}\]

= \[\ce{\frac{0.0591}{1} log K_{sp}}\]

\[\ce{log K_{sp} = \frac{E{^{\circ}_{cell}}}{0.0591}}\]

= \[\ce{\frac{-0.952 × 1}{0.0591}}\]

= –16.13