Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with $\left[{\mathrm{Ni}}^{2+}\right]=1.0M$ separated by a porous disk from an aluminum metal electrode immersed in a solution with $\left[{\mathrm{Al}}^{3+}\right]=1.0M.$ 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 $\mathrm{Al}(\mathrm{OH}{)}_3$. $$\mathrm{Al}(\mathrm{OH}{)}_3(s)\rightleftharpoons {\mathrm{Al}}^{3+}(aq)+3{\mathrm{OH}}^{-}(aq)K_{\mathrm{sp}}=?$$

Short answer

The solubility product constant ($K_{sp}$) for aluminum hydroxide, $\mathrm{Al}(\mathrm{OH}{)}_3$, is approximately $2.4\times {10}^{-33}$.

Step-by-step solution

Identify the half-reactions

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

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 ($E_{cell}^{\circ }$) : Cell reaction: $$2{\mathrm{Ni}}^{2+}(aq)+3\mathrm{Al}(s)\to 3{\mathrm{Al}}^{3+}(aq)+2\mathrm{Ni}(s)$$ Standard cell potential: $$E_{cell}^{\circ }=E_{{\mathrm{Ni}}^{2+}/\mathrm{Ni}}^{\circ }-E_{{\mathrm{Al}}^{3+}/\mathrm{Al}}^{\circ }$$ Since we are given the measured cell potential (1.82V), we can use the Nernst equation to find $E_{cell}^{\circ }$.

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: $$1.82=E_{cell}^{\circ }-\frac{0.0592}{6}\log \frac{{\left[{\mathrm{Al}}^{3+}\right]}^3{\left[{\mathrm{Ni}}^{2+}\right]}^0}{{\left[{\mathrm{Ni}}^{2+}\right]}^2}$$ Since the nickel concentration is constant and equal to 1.0 M, the equation simplifies to: $$1.82=E_{cell}^{\circ }-\frac{0.0592}{6}\log {\left[{\mathrm{Al}}^{3+}\right]}^3$$ Now it's time to find $\left[{\mathrm{Al}}^{3+}\right]$.

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: $$1.82=E_{cell}^{\circ }-\frac{0.0592}{6}\log {(3.0\times {10}^{-4})}^3$$ Now, solve for $E_{cell}^{\circ }$: $$E_{cell}^{\circ }\approx 2.56V$$

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 $2.4\times {10}^{-33}$.

Key concepts

Nernst Equation

The Nernst Equation is crucial in electrochemical cell calculations. It helps determine the cell potential under non-standard conditions. By understanding how concentration influences cell performance, we can predict cell behavior in varying environments.
The Nernst Equation is expressed as: $$E=E^{i}ght{.}_{cell}-\frac{0.0592}{n}\log Q$$

• $E$ is the cell potential at non-standard conditions.
• $E^{ight{.}_{cell}}$ is the standard cell potential.
• $n$ is the number of moles of electrons exchanged.
• $Q$ is the reaction quotient, representing the ratio of product and reactant concentrations.

Using the Nernst Equation, you can correct for real-world conditions, which deviates from standard ones, by factoring in changes in ion concentration. This formula not only aids in calculating cell potential but also provides insight into how small shifts in ion concentration can have significant impacts on electrochemical reactions.

Solubility Product Constant

The Solubility Product Constant, often denoted as $K_{sp}$, is a way to measure the solubility of a compound in water. It's particularly vital for slightly soluble salts like ${\text{Al(OH)}}_3$. $K_{sp}$ is defined as the product of the concentrations of the ions raised to the power of their coefficients in the dissolution equation.
For $$\text{A}\rightleftharpoons {\text{B}}^{a+}+{\text{C}}^{b-}$$The equation is: $$K_{sp}=[\text{B}{]}^a[\text{C}{]}^b$$Understanding $K_{sp}$ is important as it forecasts whether a compound will precipitate under a given set of conditions. In the context of ${\text{Al(OH)}}_3$, a low $K_{sp}$ value indicates that only a small amount of ${\text{Al}}^{3+}$ and ${\text{OH}}^{-}$ ions are present in solution before precipitation occurs. Thus, $K_{sp}$ becomes essential when examining the relationship between solubility and equilibrium within an ionic mixture.

Half-reactions

Half-reactions break down the complex electrochemical processes into smaller steps, each detailing the oxidation or reduction occurring at an electrode. They are vital in understanding how electrons are transferred in electrochemical cells.
A half-reaction consists of two main parts:- The oxidation half-reaction represents the loss of electrons; for example, the reaction of ${\text{Ni}}^{2+}(aq)+2e^{-}\to \text{Ni}(s)$.- The reduction half-reaction involves the gain of electrons; for instance, ${\text{Al}}^{3+}(aq)+3e^{-}\to \text{Al}(s)$.By adding the half-reactions together, you form the full cell reaction which showcases the overall electron flow. This process is key to understanding how different elements interact within an electrochemical circuit and is especially important when determining cell potentials and tracking emergency electric charge.

Standard Cell Potential

Standard Cell Potential, denoted $E_{cell}^{\circ }$, is the potential difference between two electrodes at standard conditions. It's a measure of the driving force behind an electrochemical reaction and is always measured in volts.
$E_{cell}^{\circ }$ is calculated by subtracting the standard reduction potentials of the electrodes: $$E_{cell}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$$This value tells us under standard conditions how spontaneous a reaction is.

• If $E_{cell}^{\circ }$ is positive, the reaction is spontaneous.
• If $E_{cell}^{\circ }$ is negative, the reaction is non-spontaneous.

Pure components and 1M solutions at 25°C are often standard conditions used in these calculations.
The standard cell potential serves as a benchmark to gauge how variations in concentration, temperature, and pressure impact the operational voltage of a cell. It's key for manipulating cell configurations to optimize efficiency and understand potential energy changes during reactions.