Chemistry - Electrochemistry Question with Solution | TestHub

A) As ${\mathrm{Ag}}^{+}$ has more positive reduction potential than $[\mathrm{Ag}({\mathrm{NH}}_3{)}_2{]}^{+}$, it is a better reducing agent.

B) The reaction between $[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}$ and sulfide ion is spontaneous.

$$\begin{array}{rl}& [\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}+{\mathrm{e}}^{-}⟶\mathrm{Ag}+2{\mathrm{CN}}^{-};{\mathrm{E}}^{\circ }=-0.31\mathrm{V}\\ & 2\mathrm{Ag}+{\mathrm{S}}^{2-}⟶{\mathrm{Ag}}_2\mathrm{S}+2{\mathrm{e}}^{-};{\mathrm{E}}_{\mathrm{ox}}^{\circ }=+0.69\mathrm{V}\\ & \\ & 2[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}+{\mathrm{S}}^{2-}⟶{\mathrm{Ag}}_2\mathrm{S}+4{\mathrm{CN}}^{-};{\mathrm{E}}^{\circ }=+0.38\mathrm{V}\end{array}$$

C) Formation of $[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}$ is more spontaneous, and hence this complex is more stable.

$\begin{array}{rl}& {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Ag};{\mathrm{E}}^{\circ }=+0.7991\mathrm{V}\\ & \mathrm{Ag}+2{\mathrm{CN}}^{-}⟶[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}+{\mathrm{e}}^{-};{\mathrm{E}}_{\mathrm{ox}}=+0.31\mathrm{V}\\ & \\ & 2{\left[\mathrm{Ag}{\left(CN\right)}_2^{}\right]}^{-}+S_{}^{2-}⟶{}^{}A{g}_{2}S+4C{N}^{-};{\mathrm{E}}^{\circ }=+1.1091\mathrm{V}\end{array}$

$${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}Ag;{\mathrm{E}}^{\circ }=+0.7991\mathrm{V}$$

$$\mathrm{Ag}+2{\mathrm{NH}}_3⟶{\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}+{\mathrm{e}}^{-};{\mathrm{E}}_{\mathrm{ox}}=-0.3731\mathrm{V}$$

$$2{\left[\mathrm{Ag}{\left(CN\right)}_2^{}\right]}^{-}+S_{}^{2-}⟶{}^{}A{g}_{2}S+4C{N}^{-};{\mathrm{E}}^{\circ }=+0.426\mathrm{V}$$

D)

$$\begin{array}{rl}& [\mathrm{Ag}({\mathrm{NH}}_3{)}_2{]}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Ag}+2{\mathrm{NH}}_3;{\mathrm{E}}^{\circ }=+0.373\mathrm{V}\\ & \mathrm{Ag}+{\mathrm{I}}^{-}⟶\mathrm{AgI}+{\mathrm{e}}^{-};{\mathrm{E}}_{\mathrm{ox}}^{\circ }=+0.151\mathrm{V}\\ & \\ & [\mathrm{Ag}({\mathrm{NH}}_3{)}_2{]}^{+}+{\mathrm{I}}^{-}⟶\mathrm{AgI}+2{\mathrm{NH}}_3;{\mathrm{E}}^{\circ }=+0.524\mathrm{V}\end{array}$$