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 \(1.0 \times 10^{-4} M\) 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 value of Ksp for Al(OH)₃ is approximately \(7.91\times10^{-27}\).

Step-by-step solution

Write down the half-reactions and the overall cell reaction.

First, let's write both half reactions and the overall cell reaction: 1. Nickel Half-reaction: Ni²⁺(aq) + 2e⁻ → Ni(s) (Reduction) 2. Aluminum Half-reaction: Al³⁺(aq) + 3e⁻ → Al(s) (Oxidation) Overall Cell Reaction: 2Al³⁺(aq) + 3Ni²⁺(aq) → 2Al(s) + 3Ni(s)

Write the Nernst Equation for the cell reaction.

Now, we can write the Nernst equation for the cell reaction: \( E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q \) Where: - \(E_{cell}\) is the measured cell potential (given), - \(E^0_{cell}\) is the standard cell potential, - n is the number of electrons transferred in the overall reaction (6 in this case), - Q is the reaction quotient.

Calculate the standard cell potential and reaction quotient Q.

First, we need to find the standard cell potential. Given that the oxidation potential of Aluminum is -1.66 V and the reduction potential of Nickel is -0.25 V: \(E^0_{cell} = E^0_{cathode} - E^0_{anode} = (-0.25) - (-1.66) = 1.41 V \) Secondly, we need to find the reaction quotient (Q) to substitute into the Nernst equation. Q can be calculated using the following expression: \( Q = \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Ni}^{2+}]^3} \) Since the concentration of Al³⁺ ions is not given directly, let's assume it to be x: \( Q = \frac{x^2}{(1.0)^3} = x^2 \)

Substitute the known values in the Nernst equation and solve for x (Al³⁺).

Substituting the known values in the Nernst equation and solving for x (Al³⁺): \( 1.82 = 1.41 - \frac{0.0592}{6} \log x^2 \) \(0.41 = -\frac{0.0592}{6} \log x^2 \) \(\log x^2 = -41.409 \) \(x = \sqrt{1\times10^{-41.409}} = 7.91\times10^{-15} M \)

Use Al³⁺ concentration and given OH⁻ concentration to find Ksp for Al(OH)₃.

The reaction for the dissolution of Al(OH)₃ is: \[ Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq) \] Now, we will use the concentrations of Al³⁺ and OH⁻ to find Ksp: \( K_{sp} = [\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^3 \) \( K_{sp} = (7.91\times10^{-15})(1.0\times10^{-4})^3 \) \( K_{sp} = 7.91\times10^{-27} \) So, the value of Ksp for Al(OH)₃ is approximately \(7.91\times10^{-27}\).

Key concepts

Understanding Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy or vice versa through redox reactions. These cells consist of two half-cells, each containing an electrode and an electrolyte. The electrodes are often metals, and the electrolyte is a solution of ions that participate in the redox reaction.

In the exercise provided, we find a nickel electrode in a solution of nickel ions and an aluminum electrode in a solution of aluminum ions, connected by a porous disk allowing ion flow without mixing the solutions directly. Each metal electrode goes through specific half-reactions of reduction or oxidation, contributing to the overall cell potential – the measure of the cell's ability to produce an electric current.

Educational Tip: To understand electrochemical cells, it's beneficial to visualize them as batteries with a flow of electrons from the anode (oxidation) to the cathode (reduction).

The Nernst Equation: Predicting Cell Potentials

The Nernst equation is critical for predicting the potential of an electrochemical cell under non-standard conditions. This equation adjusts the standard cell potential to account for the actual concentrations of reactants and products at any given moment.

The equation is given by: \[ E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q \]
Where n is the number of moles of electrons transferred in the redox reactions, Q is the reaction quotient, and E^0_{cell} is the standard cell potential. By knowing these values and the measured cell potential, E_{cell}, we can compute unknown concentrations, as seen in the exercise.

Educational Tip: Remember, the Nernst equation allows for precise calculations even in dynamic conditions - a crucial aspect when dealing with real-life chemical scenarios.

Solubility Product Constant (Ksp): Predicting Precipitation

The solubility product constant (Ksp) is a value that helps us predict whether a solid will precipitate or dissolve in solution. For a general salt, AB, which dissociates into A⁺ and B⁻ ions, the Ksp is represented as:

\[ K_{sp} = [A⁺][B⁻] \]
It relates to the maximum product of the ionic concentrations of a dissolved species that can exist before precipitation occurs. Specific to our exercise, the Ksp for \(Al(OH)_3\) gives us insight into how much aluminum hydroxide can dissolve before it starts to form a solid.

Educational Tip: Ksp values are temperature-dependent and unique to each compound, which is why they are incredibly useful for chemists to understand reaction equilibria in different solutions.

Chemical Equilibrium: The Balance of Reactions

The concept of chemical equilibrium plays a central role in understanding reactions such as the dissolution of \(Al(OH)_3\) in the given exercise. Equilibrium occurs when the forward and reverse reactions happen at the same rate, leading to constant concentrations of reactants and products.

In an equilibrium expression, the concentrations of the products are divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. This ratio remains constant at a given temperature unless the system is disturbed.

Educational Tip: Knowing that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their rates of formation are equal, can be a beacon in understanding complex chemical behavior.