Chapter 18: Problem 67
Calculate the value of the solubility product constant for \(\mathrm{Cd}(\mathrm{OH})_{2}\) from the half-cell potentials. $$ \begin{array}{lr} \mathrm{Cd}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(\mathrm{s}) & E^{\circ}=-0.403 \mathrm{~V} \\ \mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) & \\ E^{\circ} & =-0.83 \mathrm{~V} \end{array} $$
Short Answer
Step by step solution
Understand the Reactions
Write the Net Reaction
Use the Nernst Equation
Calculate the Solubility Product Constant
Perform the Calculations
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
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
- 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 \( -0.83 \text{ V} \).
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
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.
Gibbs Free Energy
\[ \Delta G = -nFE \]
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.