Given the following standard reduction potentials, $\begin{array}{ll} \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \ \mathrm{AgCN}(\mathrm{s})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq}) & \mathrm{E}^{\circ}=-0.01 \mathrm{~V} \end{array}$ calculate the solubility product of AgCN at 25^$\circ$C.

Question

Quizplus · Accepted Answer

The balanced equation for the dissociation of AgCN is:

AgCN(s) ⇌ Ag+(aq) + CN-(aq)

The half-reactions for the reduction of Ag+ and the oxidation of CN- are:

Ag+(aq) + e- ⇌ Ag(s)       E°red = +0.80 V
CN-(aq) ⇌ CNO-(aq) + e-   E°red = +0.30 V

The overall reaction (sum of the two half-reactions) is:

AgCN(s) + e- ⇌ Ag(s) + CN-(aq)      E°red(overall) = +1.10 V

From the Nernst equation, we have:

Ecell = E°cell - (RT/nF) ln Q

At equilibrium, Ecell = 0 and Q = Ksp. Substituting the values above:

0 = 1.10 - (0.0257/1) ln Ksp
ln Ksp = -42.97
Ksp = e^-42.97

Ksp = 2.1 × 10^-14 M^2

Therefore, the correct option is C, 2.1 × 10^-14.

Given the Following Standard Reduction Potentials Calculate the Solubility Product of AgCN at 25