Question

Question: (a) Based only on the standard reduction potentials for the${\mathbf{\text{Cu}}}^{\mathbf{2}\mathbf{+}}\mathbf{\mid }{\mathbf{\text{Cu}}}^{\mathbf{+}}{\mathbf{\text{andI}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\mid }{\mathbf{\text{I}}}^{\mathbf{-}}$the half-reactions, would you expectto be reduced toby?

(b) The formation of solid plays a role in the interaction between${\mathbf{\text{Cu}}}^{\mathbf{2}\mathbf{+}}{\mathbf{\text{andI}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$.

${\mathbf{\text{Cu}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{I}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{⥪}\mathit{C}\mathit{u}\mathit{l}\mathbf{}\left(s\right)\mathbf{}{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{86}\mathit{V}$

Taking into account this added information, do you expect Cu²⁺to be reduced by iodide ion?

Short answer

Answer

a)$\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$would not be reduced to$\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$by${{\text{I}}^{-}}_{\left(aq\right)}$.

b) Yes, the $\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$will be reduced by iodide ion

Step-by-step solution

The reduction potential

The reduction potential of a chemical is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Greater positive values of reduction potential are indicative of a greater tendency to get reduced.

Reduction Cu+(aq) to Cu+ 

a)

To determine $\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$to be reduced to ${\text{Cu}}^{2+}$by${{\text{I}}^{-}}_{\left(aq\right)}$ ,let's write standard reduction potentials for given compounds:

$$\begin{gathered} E^0\left({\text{Cu}}^{2+}\mid {\text{Cu}}^{+}\right)==0.158\text{V} \\ {E}^0\left(I_2|I^{-}\right)=0.535V \end{gathered}$$

To know that the cell formed by the half cells $E^0\left({\text{Cu}}^{2+}\mid {\text{Cu}}^{+}\right)and{E}^0\left(I_2|I^{-}\right)$and is spontaneous, we have to calculate the of the reaction.

$$\begin{gathered} {\text{Cu}}^{\text{+ 2}}\text{+ e}\to {\text{Cu}}^{\text{+}} {\text{E}}_{\text{cathode}}\text{= 0.158} \text{V} \\ {\text{I}}_2+\text{2e} \to 2{\text{I}}^{-} {\text{E}}_{\text{anode}}\text{= 0.535} \text{V} \end{gathered}$$

$$\begin{gathered} \text{∆E = 0.158V - 0.535V} \\ \text{=} \text{- 0.377V} \end{gathered}$$

Since the is negative , $E_{cell}$the reaction is not spontaneous.

This means that $\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$would not be reduced to $\text{C}{{\text{u}}^{2+}}_{\left(aq\right)}$by ${{\text{I}}^{-}}_{\left(aq\right)}$.

Reduction of Cu2+ by Iodide ion.

b)

$C{u}^{2+}\left(aq\right)+I^{-}\left(aq\right)+e^{-}\rightleftarrows CuI\left(s\right)E_{cathode}=0.86V$

$$\begin{gathered} \text{∆E = 0.86V - 0.535V} \\ \text{=} \text{0.32} \text{V} \end{gathered}$$

Since the E_cell is positive , the reaction is not spontaneous.

This means that $\text{C}{{\text{u}}^{+}}_{\left(aq\right)}$would not be reduced to $\text{C}{{\text{u}}^{+}}_{\left(aq\right)}$by ${{\text{l}}^{-}}_{\left(aq\right)}$.