Problem 67 - Chemistry Principles and Practice

Question

Calculate the value of the solubility product constant for \(\mathrm{Cd}(\mathrm{OH})_{2}\) from the half-cell potentials. $$ \begin{array}{lr} \mathrm{Cd}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(\mathrm{s}) & E^{\circ}=-0.403 \mathrm{~V} \\ \mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) & \\ E^{\circ} & =-0.83 \mathrm{~V} \end{array} $$

Step-by-Step Solution

Answer

The solubility product constant, \( K_{sp} \), for \( \text{Cd(OH)}_2 \) is \( 2.5 \times 10^{-15} \).

Step 1: Understand the Reactions

We are given two half-cell reactions involving cadmium and hydroxide ions. The first reaction is the reduction of \( \text{Cd}^{2+} \) to \( \text{Cd(s)} \), and the second involves the dissolution of \( \text{Cd(OH)}_2 \) solid to give \( \text{Cd(s)} \) and \( \text{OH}^- \).

Step 2: Write the Net Reaction

The net reaction we are interested in is the dissolution of \( \text{Cd(OH)}_2 \) into \( \text{Cd}^{2+} \) and \( \text{OH}^- \). To find this, perform the subtraction of the standard potentials: \[ \text{Cd(OH)}_2(s) \rightarrow \text{Cd}^{2+}(aq) + 2\text{OH}^-(aq) \]

Step 3: Use the Nernst Equation

The Nernst equation relates the cell potential at standard conditions to the Gibbs free energy: \( \Delta G = -nFE \), where \( n \) is the number of electrons (2 in this case), \( F \) is the Faraday's constant, and \( E \) is the cell potential. Use the difference in potentials to find \( \Delta G \).

Step 4: Calculate the Solubility Product Constant

The solubility product constant \( K_{sp} \) can be derived from the Gibbs free energy change: \( \Delta G = -RT \ln K_{sp} \). Rearrange to find \( K_{sp} \): \[ K_{sp} = e^{-\frac{\Delta G}{RT}} \]Use \( R = 8.314 J\,mol^{-1}K^{-1} \) and temperature in Kelvin (usually assumed to be 298 K unless specified).

Step 5: Perform the Calculations

Given \( E^\circ \text{ for the cell} = E^\circ(\text{Cd}^{2+}/\text{Cd}) - E^\circ(\text{Cd(OH)}_2/\text{Cd}) = -0.403 - (-0.83) = 0.427 \text{V} \). Calculate \( \Delta G = -2 \times 96485 \times 0.427 \approx -82322 \ J/mol \). Then, insert into the expression for \( K_{sp} \): \[ K_{sp} = e^{-\frac{-82322}{(8.314 \times 298)}} \approx 2.5 \times 10^{-15} \]

Step 6: Conclusion

The calculated value of the solubility product constant for \( \text{Cd(OH)}_2 \) is \( 2.5 \times 10^{-15} \).

Key Concepts

Solubility Product Constant

The solubility product constant, often denoted as \( K_{sp} \), represents the extent to which a compound will dissolve in water. It is a special type of equilibrium constant used for sparingly soluble salts. In the context of electrochemistry, it is particularly useful when dealing with slightly soluble ionic compounds like cadmium hydroxide, \( \text{Cd(OH)}_2 \).

Understanding \( K_{sp} \) involves looking at the overall dissolution reaction. For cadmium hydroxide, this reaction can be represented as:
\[ \text{Cd(OH)}_2(s) \rightleftharpoons \text{Cd}^{2+}(aq) + 2\text{OH}^-(aq) \]
From this reaction, the expression for the solubility product is:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
It's crucial to remember that solids are not included in the expression.

This expression shows how the concentrations of the ions are related in a saturated solution. It's a key factor when predicting or explaining the behavior of ionic compounds in solution.

Half-Cell Potentials

Half-cell potentials, also known as standard electrode potentials, indicate a half-reaction's tendency to occur as a reduction. These values are provided in volts and are determined relative to the standard hydrogen electrode. In the provided exercise, we have:

• The reduction potential for \( \text{Cd}^{2+} + 2\text{e}^- \rightarrow \text{Cd(s)} \) is \( -0.403 \text{ V} \).
• The reduction potential for \( \text{Cd(OH)}_2 + 2\text{e}^- \rightarrow \text{Cd(s)} + 2\text{OH}^- \) is \( -0.83 \text{ V} \).

By comparing these potentials, we can deduce which species is more likely to be reduced or oxidized under standard conditions.

Subtracting these potentials helps establish the cell potential for the net reaction under study. For the dissolution reaction of \( \text{Cd(OH)}_2 \), this difference provides insight into the feasibility and direction of the reaction.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that relates the concentration of ions to the electrode potential of a cell. It allows for understanding how changes in ion concentration affect the cell potential, thus offering insight into dynamic electrochemical reactions.

In a simplified form for a redox reaction, the Nernst equation is expressed as:
\[ E = E^ ext{°} - \frac{RT}{nF} \ln Q \]
where:

• \( E \) is the cell potential at non-standard conditions,
• \( E^ ext{°} \) is the standard cell potential,
• \( R \) is the ideal gas constant, \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \),
• \( T \) is the temperature in Kelvin,
• \( n \) is the number of moles of electrons exchanged,
• \( F \) is the Faraday constant, \( 96485 \text{ C mol}^{-1} \),
• and \( Q \) is the reaction quotient.

The Nernst equation is indispensable for calculating the cell potentials in non-standard conditions and is used to find the Gibbs free energy in the discussed exercise.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G \), indicates the spontaneity of a reaction. In electrochemistry, it's linked to cell potential through the relationship:
\[ \Delta G = -nFE \]
where \( n \) is the number of moles of electrons, \( F \) is Faraday's constant, and \( E \) is the cell potential. A negative \( \Delta G \) implies a spontaneous process under standard conditions.

Using the calculated cell potential from earlier steps, Gibbs free energy helps find the solubility product. The equation:
\[ \Delta G = -RT \ln K_{sp} \]
allows us to rearrange and solve for \( K_{sp} \):
\[ K_{sp} = e^{-\frac{\Delta G}{RT}} \]
This relationship highlights how \( \Delta G \) connects the thermodynamics of a reaction with its equilibrium expression, serving as a bridge between tabulated values and real-world conditions.