An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M\). Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

We can look up the reduction half-reactions and their standard reduction potentials in a table: For nickel: \( \mathrm{Ni}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Ni}(s) \) Reduction potential: \(E^0_{\mathrm{Ni}^{2+}/\mathrm{Ni}} = -0.25 V \) For aluminum: \( \mathrm{Al}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Al}(s) \) Reduction potential: \(E^0_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66 V \)

Multiplying the Ni half-reaction by 3 and the Al half-reaction by 2 (to equalize the number of electrons exchanged) and adding them together, we get: \( 3\mathrm{Ni}^{2+}(aq) + 6e^{-} + 2\mathrm{Al}^{3+}(aq) + 6e^{-} \rightarrow 3\mathrm{Ni}(s) + 2\mathrm{Al}(s) \) This simplifies to: \( 3\mathrm{Ni}^{2+}(aq) + 2\mathrm{Al}^{3+}(aq) \rightarrow 3\mathrm{Ni}(s) + 2\mathrm{Al}(s) \) The standard cell potential is the difference between the reduction potentials of the two half-reactions: \(E^0_{cell} = E^0_{\mathrm{Ni}^{2+}/\mathrm{Ni}} - E^0_{\mathrm{Al}^{3+}/\mathrm{Al}} = -0.25 - (-1.66) = 1.41 V\)

Now we find the reaction quotient, Q, using the given concentrations: \(Q = \frac{\left[\mathrm{Ni}^{2+}\right]^3}{\left[\mathrm{Al}^{3+}\right]^2} = \frac{(1.0)^3}{(1.0)^2} = 1\)

Using the Nernst equation, we can relate the cell potential, E, at the given conditions to the standard cell potential, E⁰: \( E = E^0_{cell} - \frac{0.0592}{n} \log{Q} \) The number of electrons transferred in the balanced cell equation is 6, so n = 6. Since Q = 1, the logarithm term vanishes: \( E = 1.41 - 0 = 1.41 V \) However, the measured cell potential is given as 1.82 V. The difference between the calculated and measured potentials must be due to the effect of the precipitation of Al(OH)₃: \( \Delta E = E_\text{measured} - E_\text{calculated} = 1.82 - 1.41 = 0.41 V \)

Now we need to find the solubility product, Ksp, for Al(OH)₃ using the equation: \( K_{\mathrm{sp}} = \left[\mathrm{Al}^{3+}(a q)\right] \left[\mathrm{OH}^{-}(a q)\right]^3 \) We are given the concentration of OH⁻ ions in the solution as 1.0 × 10⁻⁴ M. We can relate the concentration of Al³⁺ ions to the concentration of OH⁻ ions using the balanced equation for Al(OH)₃ dissolution: \[ \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{OH}^-]^6} = K_{\mathrm{sp}} \] Now we can use the Nernst equation to include the effect of the precipitation of Al(OH)₃ on the cell potential: \[ \Delta E = -\frac{0.0592}{n}\log\frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{OH}^-]^6} = -\frac{0.010}( \log K_{\mathrm{sp}}) \] Plug in the calculated value of ΔE = 0.41 V: \[ 0.41 = -\frac{0.010}( \log K_{\mathrm{sp}}) \] Now we can solve for Ksp: \[ \log K_{\mathrm{sp}} = \frac{0.41}{-0.010} = -41 \] \[K_{\mathrm{sp}} = 10^{-41} \] So the solubility product for Al(OH)₃ is \(1.0 \times 10^{-41}\).