Question

For the following half-reaction, E° = -2.07 V:
$\mathit{A}\mathit{l}{{\mathbf{F}}_{\mathbf{6}}}^{\mathbf{3}\mathbf{-}}\mathbf{+}\mathbf{3}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{A}\mathit{l}\mathbf{+}\mathbf{6}{\mathit{F}}^{\mathbf{-}}$
Using data from Table 11.1, calculate the equilibrium constant at 25°C for the reaction
$\mathit{A}{\mathit{l}}^{\mathbf{3}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{6}{\mathit{F}}^{\mathbf{-}}\left(aq\right)\mathbf{\rightleftharpoons }\mathit{A}\mathit{l}{{\mathbf{F}}_{\mathbf{6}}}^{\mathbf{3}\mathbf{-}}\left(aq\right)$

Short answer

Theequilibrium constant can be found as $\mathbf{6}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{.}$

Step-by-step solution

Overall and half-reaction

The half-reaction can be denoted as:

$$\begin{gathered} Al{F_6}^{3-}+3e^{-}\to Al+6F^{-} E^o=-2.07 V \\ Al\to A{l}^{3+}+3e^{-} E^o=1.66 V \end{gathered}$$

The overall reaction can be written as:

$$\begin{gathered} Al{F_6}^{3-}\left(aq\right)\to A{l}^{3+}\left(aq\right)+6F^{-}\left(aq\right) \\ {E^o}_{cell}=-2.07 V+1.66V \\ =-0.41 V \end{gathered}$$

Calculating Equilibrium constant

The equation can be written as:

$$\begin{gathered} \log K=\frac{n{E^o}_{cell}}{0.0591}---\left(1\right) \\ {K}_{reverse}=\frac{1}{K} \\ \log K=-20.81 \\ K=1.54\times {10}^{-21} \\ {K}_{reverse}=\frac{1}{1.54\times {10}^{-21}} \\ {K}_{reverse}=6.5\times {10}^{20} \end{gathered}$$