Question

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 \(\mathrm{Al}(\mathrm{OH})_{3}\). $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

Short answer

The solubility product for Al(OH)₃ is \(K_{\mathrm{sp}} = 1.0 \times 10^{-41}\).

Step-by-step solution

Write the half-reactions and find the standard reduction potentials

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: \(E^0_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66 V \)

Write the balanced redox equation for the cell reaction

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: \(E^0_{cell} = E^0_{\mathrm{Ni}^{2+}/\mathrm{Ni}} - E^0_{\mathrm{Al}^{3+}/\mathrm{Al}} = -0.25 - (-1.66) = 1.41 V\)

Calculate the reaction quotient (Q)

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

Use the Nernst equation to calculate the cell potential

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: \( \Delta E = E_\text{measured} - E_\text{calculated} = 1.82 - 1.41 = 0.41 V \)

Solve for Ksp using the concentrations and cell potential

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 \(1.0 \times 10^{-41}\).

Key concepts

Nickel Electrode

The nickel electrode plays a crucial role in an electrochemical cell. It acts as the conductor within which nickel ions are reduced and deposited as solid nickel. This process is known as a reduction reaction and is characterized by the gain of electrons by nickel ions - The reaction for nickel can be written as: \( \mathrm{Ni}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Ni}(s) \) - The standard reduction potential for this reaction is \( E^{0}_{\mathrm{Ni}^{2+}/\mathrm{Ni}} = -0.25 \text{ V} \). Nickel electrodes are commonly used due to their moderate reactivity and ability to conduct electricity efficiently. In this electrochemical cell, the nickel electrode contributes to the overall cell potential, serving as one half of the redox equation that governs the cell's functionality.

Aluminum Electrode

In the electrochemical cell, the aluminum electrode serves as the site where aluminum ions are reduced to solid aluminum. This reduction occurs as part of the cell's overall redox process. - The reaction at the aluminum electrode is: \( \mathrm{Al}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Al}(s) \)- The standard reduction potential for aluminum is \( E^{0}_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66 \text{ V} \). Aluminum electrodes are notable for their highly negative standard reduction potential, indicating a strong tendency to lose electrons. This makes aluminum effective as an anode, allowing it to participate in oxidation processes when paired with other metals in an electrochemical cell. Its role in creating a cell potential is critical to driving the electrochemical reactions.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry used to determine the cell potential under non-standard conditions. It takes into account the concentration of ions involved in the half-reactions, allowing us to predict how the cell potential changes when these concentrations vary.
The basic form of the Nernst equation is:\[ E = E^0_{\text{cell}} - \frac{0.0592}{n} \log Q \]- Here, \(E^0_{\text{cell}}\) is the standard cell potential.- \(n\) is the number of moles of electrons transferred in the reaction.- \(Q\) represents the reaction quotient, a measure of the current state of the chemical process relative to its equilibrium state. For the nickel and aluminum electrochemical cell, the Nernst equation helped relate the observed cell potential to the underlying ionic concentrations. Since the reaction quotient \(Q\) was 1, the change in potential was solely dependent on the external conditions, which included the effects of the precipitation reaction.

Precipitation Reaction

A precipitation reaction occurs when a solid forms and separates from the solution as a precipitate. In electrochemical cells, such reactions can influence the cell potential significantly.
In the given exercise, the aluminum hydroxide \(\mathrm{Al(OH)}_{3}\) precipitation reaction occurs due to the addition of sodium hydroxide to the aluminum compartment:\[ \mathrm{Al}^{3+}(aq) + 3\mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{Al(OH)}_{3}(s) \]This reaction reduces the concentration of \(\mathrm{Al}^{3+}\) ions available in the solution, affecting the Nernst equation and altering the cell's potential. The extent of this precipitation is often quantified by the solubility product constant \(K_{\text{sp}}\), which is calculated based on the equilibrium state of the reaction:
- For \(\mathrm{Al(OH)}_{3}\), its solubility product expression is: \[K_{\text{sp}} = [\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^3\]The presence of the precipitate directly impacts the electrochemical reactions by decreasing the concentration of ions that can participate in the reaction, leading to an observed change in the cell potential that is greater than predicted by standard calculations.