Question

Given the data at 25°C,

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

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

What is the value of K_sp for AgI?

\[\ce{\left(2.303 \frac{RT}{F} = 0.059 V\right)}\]

Options

• −8.12

• +8.612

• −37.83

• −16.13

Answer 1

−16.13

Explanation:

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

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

Eq. (i) can be written as:

\[\ce{AgI_{(s)} + e- -> Ag+ + I-}\]; E° = – 0.800 + (– 0.152) = – 0.952 V

E_cell = \[\ce{E^{\circ}_{cell} - \frac{2.303RT}{nF} log_10 \frac{[Ag^+][I^-]}{[AgI_{(s)}]}}\]

At equilibrium, E_cell = 0. Moreover, n = 1.

∴ 0 = \[\ce{E^{\circ}_{cell} - \frac{2.303 RT}{F} log_10 K_{sp}}\]

or, 0 = −0.952 − 0.059 log₁₀ K_sp

or, log₁₀ K_sp = \[\ce{-\frac{0.952}{0.059}}\]

= −16.13