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 Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} M\) and the measured cell potential is 1.82 \(\mathrm{V}\) . Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=?$$

Step by step solution

01

Identify the half-reactions

We first need to identify the half-reactions that make up the electrochemical cell: Nickel: $$\mathrm{Ni}^{2+}(aq)+2 e^{-} \rightarrow Ni(s)$$ Aluminum: $$\mathrm{Al}^{3+}(aq)+3 e^{-} \rightarrow Al(s)$$

02

Determine the cell reaction and standard cell potential

Now, we will combine the half-reactions to form the cell reaction and find the standard cell potential (\(E^{\circ}_{cell}\)) : Cell reaction: $$2 \mathrm{Ni}^{2+}(aq)+3 \mathrm{Al}(s) \rightarrow 3 \mathrm{Al}^{3+}(aq)+2 \mathrm{Ni}(s)$$ Standard cell potential: $$E^{\circ}_{cell} = E^{\circ}_{\mathrm{Ni^{2+}/Ni}} - E^{\circ}_{\mathrm{Al^{3+}/Al}}$$ Since we are given the measured cell potential (1.82V), we can use the Nernst equation to find \(E^{\circ}_{cell}\).

03

Apply the Nernst equation

Applying the Nernst equation, we get: $$E=E^\circ - \frac{0.0592}{n} \log \frac{Q}{K}$$ For this electrochemical cell, where \(n=6\) and the cell potential is given as 1.82V: $$1.82 = E^{\circ}_{cell} - \frac{0.0592}{6} \log \frac{\left[\mathrm{Al}^{3+}\right]^{3} \left[\mathrm{Ni}^{2+}\right]^{0}}{\left[\mathrm{Ni}^{2+}\right]^{2}}$$ Since the nickel concentration is constant and equal to 1.0 M, the equation simplifies to: $$1.82 = E^{\circ}_{cell} - \frac{0.0592}{6} \log \left[\mathrm{Al}^{3+}\right]^{3}$$ Now it's time to find \(\left[\mathrm{Al}^{3+}\right]\).

04

Find the concentration of Al³⁺

We know that the concentration of \(\mathrm{OH}^{-}\) at equilibrium is \(1.0 \times 10^{-4} M\). We can use the relation between \(\mathrm{OH}^{-}\) and \(\mathrm{Al}^{3+}\) to find the concentration of \(\mathrm{Al}^{3+}\): At equilibrium: $$3 \left[\mathrm{OH}^{-}\right] = \left[\mathrm{Al}^{3+}\right]$$ $$\left[\mathrm{Al}^{3+}\right] = 3 \times 1.0 \times 10^{-4} M = 3.0 \times 10^{-4} M$$ Plug this value into the modified Nernst equation: $$1.82 = E^{\circ}_{cell} - \frac{0.0592}{6} \log \left(3.0 \times 10^{-4}\right)^{3}$$ Now, solve for \(E^{\circ}_{cell}\): $$E^{\circ}_{cell} \approx 2.56 V$$

05

Calculate the \(K_{sp}\) value

Finally, use the relationship between cell potential and solubility product constant (\(K_{sp}\)) to find the value of \(K_{sp}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\): $$1.82 = 2.56 - \frac{0.0592}{6} \log Q$$ Where \(Q = \left[\mathrm{Al}^{3+}\right] \left[\mathrm{OH}^{-}\right]^{3}\) and at equilibrium \(Q=K_{sp}\) Calculating for \(K_{sp}\), we get: $$K_{sp} \approx 2.4 \times 10^{-33}$$ Therefore, the solubility product constant for aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3}\), is approximately \(2.4 \times 10^{-33}\).

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