Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni²⁺] = 1.0 M separated by a porous disk from an aluminum metal electrode immersed in a solution with [Al³⁺] = 1.0 M. Sodium hydroxide is added to the aluminum compartment, causing Al(OH)₃(s) to precipitate. After precipitation of Al(OH)₃has ceased, the concentration of OH is 1.0 x 10^-4 M, and the measured cell potential is 1.82 V. Calculate the K_sp value for Al(OH)₃.

$\mathbf{Al}\mathbf{(}\mathbf{OH}{\mathbf{)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{Al}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{?}$

Short answer

TheK_sp value for Al (OH)₃ is $4.32\times {10}^{-31}.$

Step-by-step solution

Half-cell reaction at anode and cathode

The half-cell reaction at anode and cathode can be denoted as,

$\begin{array}{rcl}2\mathrm{Al}& \to & 2{\mathrm{Al}}^{3+}+6{\mathrm{e}}^{-}\\ 3{\mathrm{Ni}}^{2+}+6{\mathrm{e}}^{-}& \to & 3\mathrm{Ni}\end{array}$

The overall cell reaction can be denoted as,

$2\mathrm{Al}+3{\mathrm{Ni}}^{2+}\to 2{\mathrm{Al}}^{3+}+3\mathrm{Ni}$

Finding standard electrode potential for a cell

The expression is as follows,

$$\begin{gathered} {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={\mathrm{E}}^{\mathrm{o}}\left(\mathrm{cathode}\right)-{\mathrm{E}}^{\mathrm{o}}\left(\mathrm{anode}\right) \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=-0.23 \mathrm{V}-(-1.66 \mathrm{V}) \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=1.43 \mathrm{V} \end{gathered}$$

Nernst equation expression for net reaction

The expression of Nernst equation can be denoted as,

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \left(\frac{{\left[{\mathrm{Al}}^{3+}\right]}^2}{{\left[{\mathrm{Ni}}^{2+}\right]}^3}\right)\\ \log {\left[{\mathrm{Al}}^{3+}\right]}^2& =& \frac{\frac{\mathrm{n}\left({{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-{\mathrm{E}}_{\mathrm{cell}}\right)}{0.0591}+3\log \left[{\mathrm{Ni}}^{2+}\right]}{2}\end{array}$

$\begin{array}{rcl}\log \left[{\mathrm{Al}}^{3+}\right]& =& \frac{\frac{6(1.43-1.82)}{0.0591}+3 \log }{2}\\ \log \left[{\mathrm{Al}}^{3+}\right]& =& -19.80\\ \left[{\mathrm{Al}}^{3+}\right]& =& 1.6\times {10}^{-20}\end{array}$

Then,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& \left[{\mathrm{Al}}^{3+}\right]\left({\left[{\mathrm{OH}}^{-}\right]}^3\right)\\ {\mathrm{K}}_{\mathrm{sp}}& =& (1.6\times {10}^{-20})\left((3\times {10}^{-4}{)}^3\right)\\ {\mathrm{K}}_{\mathrm{sp}}& =& 4.32\times {10}^{-31}\end{array}$