Q42 P - Chemistry: Molecular Nature Of Matter And Change

Question

Use the following data to determine if Cr²⁺(aq) can be prepared by the reaction of  with Cr²⁺(aq) :

$\begin{array}{rcl}\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathit{C}{\mathit{r}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}{\mathit{E}}^{\mathbf{0}}& \mathbf{=}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{41}\mathit{V}\\ \mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{e}}^{\mathbf{-}}\mathit{C}\mathit{r}\mathbf{\left(}\mathit{s}\mathbf{\right)}{\mathit{E}}^{\mathbf{0}}& \mathbf{=}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{74}\mathit{V}\\ \mathit{C}{\mathit{r}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathit{C}\mathit{r}\mathbf{\left(}\mathit{s}\mathbf{\right)}{\mathit{E}}^{\mathbf{0}}& \mathbf{=}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{91}\mathit{V}\end{array}$

Step-by-Step Solution

Answer

$C{r}_{\left(aq\right)}^{2+}$ can be formed from Cr_(a) and width="41" height="30" style="max-width: none; vertical-align: -11px;" $C{r}_{\left(aq\right)}^{3+}$ because the resulting $E_{enll}^o$ is positive.

Step 1: Definition of Coordination Compound

A coordination complex consists of a central atom or ion, which is typically metallic and is termed the coordination centre, and a surrounding array of bound molecules or ions, that are successively referred to as ligands or complexing agents.

Step 2: Solve the given statements

Cr_(o) Is oxidized to $C{r}_{\left(an\right)}^{3+}$, therefore it is the anode

$$\begin{gathered} E_{hal f-out}^o=-0.74. \\ C{r}_{\left(a\right)}\to C{r}_{\left(aq\right)}^{3+}+3e^{-} \end{gathered}$$

Cr³⁺ is then reduced to Cr²⁺, therefore it is at the cathode.

$E_{half}^o-an=-0.41 C{r}_{\left(aq\right)}^{3+}+e^{-}\to C{r}_{\left(a\right)}^{2+}$

Overall reaction: $C{r}_{\left(s\right)}+2C{r}_{\left(a\right)}^{3+}\to 3C{r}_{\left(a(q\right)}^{2+}$

$$\begin{gathered} E_{oul}^0=E_{reduction}^0-E_{ox idation}^0 \\ {E}_{cell}^0=E_{cathode}^0-E_{anode}^0 \\ {E}_{cell}^0=-0.41-\left(-0.74\right) \\ {E}_{cell}^a=0.35V \end{gathered}$$

The E_cell is positive, therefore the reaction can proceed spontaneously.