Q21.147CP - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the ${\mathbf{K}}_{\mathbf{f}}$ of $\mathbf{Ag}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{+}}$ from:

$$\begin{gathered} {\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}{\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{V} \\ }^{\mathbf{+}} \\ \mathbf{Ag}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{,}\mathbf{ }{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{37}\mathbf{V} \end{gathered}$$

Step-by-Step Solution

Answer

The ${\mathrm{K}}_{\mathrm{f}}$ of $\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{{)}_2}^{+}$is obtained as: $1.88\times {10}^7$ .

Step 1: Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate a new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Step 2: Evaluate the value of K f

Silver is oxidised during the production of $\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{{)}_2}^{+}$.

$\begin{array}{c}\mathrm{Oxidation}:{\mathrm{Ag}}_{\left(\mathrm{s}\right)}+2{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\to \mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\\ {\mathrm{Reduction}:\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{\left(\mathrm{s}\right)}\\ {\mathrm{Reaction}:\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+2{\mathrm{NH}}_{3\left(\mathrm{aq}\right)}\to \mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{2\left(\mathrm{aq}\right)}^{+}\end{array}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}= 0.37 \mathrm{V}\\ {\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}= 0.80 \mathrm{V}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}= 0.80 \mathrm{V}-0.37 \mathrm{V}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}= 0.43 \mathrm{V}\end{array}$

$\begin{array}{c}{\mathrm{\Delta G}}^{\mathrm{o}}= {-\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}\\ = -1{\mathrm{e}}^{-}\times 96,500\frac{\mathrm{C}}{{\mathrm{mole}}^{-}}\times 0.43 \mathrm{V}\\ = -41,495\mathrm{J}\end{array}$

${\mathrm{\Delta G}}^{\mathrm{o}} = -\mathrm{RTlnK}$in Action the  for  may be calculated as:

$\begin{array}{c}-41,495 = -\mathrm{RTlnK}\\ \mathrm{lnK} = \frac{41,495\mathrm{J}}{\mathrm{RT}}\\ {\mathrm{K}}_{\mathrm{sp}}= \mathrm{e}\frac{41,495\mathrm{J}}{\mathrm{RT}}\\ {\mathrm{K}}_{\mathrm{sp}}= \mathrm{e}\frac{41,495\mathrm{J}}{298\mathrm{K}\times 8.314\frac{\mathrm{J}}{\mathrm{molK}}}\\ {\mathrm{K}}_{\mathrm{sp}} = 1.88\times {10}^7\end{array}$

Therefore, the value is$1.88\times {10}^7$.