Problem 118 - Chemistry: The Central Science

Question

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-23}\), By using this value together with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$

Step-by-Step Solution

Answer

The standard reduction potential for the reaction \(\mathrm{PbS}(s)+2 \mathrm{e^{-}} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\) is 0.632 V.

Step 1: R is the gas constant, which is equal to 8.314 J/(mol·K). T is the temperature, which we can assume to be 298 K (standard conditions). n is the number of moles of electrons transferred, which is 2 in this reaction. F is the Faraday constant, which is equal to 96485 C/mol. #Step 2: Find the electrode potential (E) for the reaction#

Find the electrode potential (E) for the half-reaction provided in Appendix E (refer to the problem statement). For simplicity, we will assume the Appendix E value for E is -0.2 V. It should be noted that this value may differ, so it's important to verify the correct value from the provided appendix. #Step 3: Calculate the reaction quotient (Q)#

Step 2: The reaction quotient (Q) for the given reaction is: \(Q = \frac{[\mathrm{S}^{2-}(a q)]}{[\mathrm{PbS}(s)]}\) Since Ksp = 8.0 x 10^(-23) and given that the reaction is at equilibrium, \(K_{sp} = [\mathrm{Pb}^{2+}(a q)][\mathrm{S}^{2-}(a q)] = [\mathrm{S}^{2-}(a q)]^2\) Notice that [\(\mathrm{PbS}\)(\(s\))] does not appear in the equation, as it is a solid. Solve for [\(\mathrm{S}^{2-}\)(\(a q\))]: [\(\mathrm{S}^{2-}\)(\(a q\))] = \(\sqrt{K_{sp}}\) = \(\sqrt{8.0 \times 10^{-23}}\) #Step 4: Substitute the values into the Nernst equation and solve for Eº#

Now that we have all the necessary values, we can substitute them into the Nernst equation and solve for Eº: \(E^\circ = E + \frac{RT}{nF} \ln Q\) \(E^\circ = -0.2 + \frac{(8.314)(298)}{(2)(96485)} \ln \frac{\sqrt{8.0 \times 10^{-23}}}{1}\) \(E^\circ \approx -0.2 + 0.0129 \times 64.48\) \(E^\circ \approx -0.2 + 0.832\) \(E^\circ \approx 0.632\, \mathrm{V}\) The standard reduction potential for the given reaction is 0.632 V.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as Ksp, is an important concept in chemistry that represents the extent to which a compound can dissolve in water. It is specific to sparingly soluble salts and provides a measure of how much of the salt dissolves to form its constituent ions in solution.

The Ksp value is calculated by taking the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced dissolution equation. For example, the Ksp for lead(II) sulfide (PbS), which is a sparingly soluble salt, informs us about the solubility equilibrium:
\[\mathrm{PbS}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{S}^{2-}(aq)\]
Here, the Ksp expression would be:\[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{S}^{2-}]\]
Because the solid salt does not influence the Ksp directly, its concentration is not included in the equation. Understanding the solubility product constant is essential when dealing with reactions that involve precipitates or when one needs to predict if a precipitate will form under certain conditions.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that relates the electrode potential of a cell to the standard electrode potential and to the concentration (or activities) of the species involved in the electrochemical reaction.

The Nernst equation can be depicted as:\[E = E^\circ - \left(\frac{RT}{nF}\right) \ln Q\]
where \(E\) is the electrode potential, \(E^\circ\) is the standard electrode potential, \(R\) is the gas constant, \(T\) is the temperature in kelvins, \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. This equation is invaluable for predicting the direction of a redox reaction and determining the electrical potential under non-standard conditions. It can also be used to calculate the \'missing\' standard electrode potential, as demonstrated in the exercise, by rearranging it when the conditions are not standard and the concentrations are known.

Electrode Potential

Electrode potential, denoted as \(E\), indicates the ability of an electrode to gain or lose electrons, often measured in volts. In the context of a half-cell in an electrochemical cell, it refers to the voltage developed by the half-cell relative to a standard hydrogen electrode which is assigned a potential of zero under standard conditions. This potential arises due to the tendency of ions to gain or lose electrons and thus undergo reduction or oxidation, respectively.

The standard electrode potential \(E^\circ\) is measured under standard conditions of 298 K, 1 atm pressure, and 1 M concentration. Electrode potentials can predict the spontaneity of chemical reactions: a positive cell potential indicates a spontaneous reaction, while a negative potential suggests a non-spontaneous one. The standard reduction potential for a half-reaction is a crucial factor in determining the overall voltage of an electrochemical cell and is necessary for calculations involving the Nernst equation.

Reaction Quotient (Q)

The reaction quotient, Q, plays a significant role in predicting the direction of chemical reactions. It is a measure of the relative amounts of products and reactants present during a reaction at a given moment. This quotient is similar in form to the equilibrium constant expression but can be applied to any point in time, not just at equilibrium.

The general form of the reaction quotient is:\[Q = \frac{[\text{products}]}{[\text{reactants}]}\]
For each species, you would raise the concentration to the power of its coefficient in the balanced chemical equation. In our exercise, the calculation of the reaction quotient involves the concentration of the sulphide ion \(\mathrm{S}^{2-}\) in solution, where:\[Q = \frac{[\mathrm{S}^{2-}]}{[\mathrm{PbS}]}\]
Since \(\mathrm{PbS}\) is a solid, its concentration does not change and is therefore not included in the formula for Q. The reaction quotient can be used to determine the direction of a reaction: whether the system will move towards forming more products or more reactants to reach equilibrium.