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

role="math" localid="1663934413634" $$\begin{gathered} {\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}{\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{V} \\ }^{\mathbf{+}} \\ \mathbf{Ag}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{,}\mathbf{}{\mathbf{E}}^{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{37}\mathbf{V} \end{gathered}$$

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.

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$1.88\times {10}^7$.