Question

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-28} .\) By using this value together with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$

Short answer

The standard reduction potential for the reaction \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\) is -0.6891 V.

Step-by-step solution

Analyze the given reaction and the solubility product

The solubility product \(K_{sp}\) represents the equilibrium constant for the dissolution of an ionic compound in water, which in this case, is lead sulfide, PbS. The reaction for the solubility can be written as: $$ \mathrm{PbS}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{S}^{2-}(a q) $$ For this reaction, the solubility product constant is given as \(K_{sp} = 8.0 \times 10^{-28}\). From this relation, we can find the standard cell potential for the given electrochemical reaction.

Write the half-reactions for the electrochemical process

To find the standard cell potential for the given reaction: $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$ We need to break it down into two half-reactions: 1. Reduction half-reaction: $$ \mathrm{S}^{2-}(a q) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{S}(s) $$ 2. Oxidation half-reaction: $$ \mathrm{PbS}(s) \longrightarrow \mathrm{Pb}^{2+}(a q) + \mathrm{S}^{2-}(a q) + 2 \mathrm{e}^- $$

Use the Nernst equation to relate Ksp and standard reduction potential

According to the Nernst equation, $$ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln{Q} $$ Where: - \(E_{cell}\) is the cell potential under non-standard conditions - \(E^0_{cell}\) is the standard cell potential (the value we want to find) - R is the ideal gas constant (8.314 J/mol K) - T is the temperature in Kelvin (assuming 298.15 K) - n is the number of electrons transferred (2 for this reaction) - F is the Faraday's constant (\(96485\, C/mol\)) - Q is the reaction quotient. For the solubility equilibrium reaction, $$ Q = [\mathrm{Pb}^{2+}][\mathrm{S}^{2-}] = x^2, $$ where x is the solubility of the compound. At equilibrium, \(Q = K_{sp}\), and the \(E_{cell} = 0\), leading to: $$ 0 = E^0_{cell} - \frac{RT}{nF} \ln{K_{sp}} $$

Solve for standard cell potential

Rearrange the equation from step 3 to solve for \(E^0_{cell}\): $$ E^0_{cell} = \frac{RT}{nF} \ln{K_{sp}} $$ Plug in the values R, T, n, F, and \(K_{sp}\): $$ E_{cell}^0 = \frac{8.314 \times 298.15}{2 \times 96485} \ln{(8.0 \times 10^{-28})} $$ Calculate the standard cell potential: $$ E^0_{cell} =-0.6891 \, V $$

Find the standard reduction potential for the given reaction

The standard reduction potential for the PbS(s) + 2 e- → Pb(s) + S2- (aq) reaction is the same as the calculated standard cell potential, which is -0.6891 V. In conclusion, the standard reduction potential for the given reaction is -0.6891 V.

Key concepts

Understanding Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a critical concept in chemistry that helps us predict whether a compound will dissolve in water and to what extent. It pertains to the equilibrium that is established when an ionic solid dissolves in water, breaking down into its constituent ions.

For a salt like lead sulfide, or PbS, the dissolution can be represented by the equation: \[\begin{equation}\mathrm{PbS}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{S}^{2-}(aq)\end{equation}\]The Ksp is determined experimentally and is specific for each substance at a given temperature. It's expressed as the product of the concentrations of the resulting ions, each raised to the power of its stoichiometric coefficient. In simpler terms, for the compound PbS, the Ksp is the product of the concentration of lead ions and sulfide ions in the solution when the solid phase is in equilibrium with the dissolved phase.
When calculating Ksp, we usually assume that the solid phase, PbS in this case, has a constant activity of 1, which simplifies the equilibrium expression to only include the dissolved ions.

The Nernst Equation and its Application

The Nernst equation is a fundamental part of electrochemistry and enables us to calculate the electrical potential of an electrochemical cell under non-standard conditions. It determines how the potential changes as the concentration of reactants and products varies.

\[\begin{equation}E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln{Q}\end{equation}\]In this equation, E_cell is the cell potential, E⁰_cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday’s constant, and Q is the reaction quotient, which is a measure of the ratio of the products to reactants. The Nernst equation shows us that as a cell reaches equilibrium (when Q equals the equilibrium constant, K), the cell potential moves towards zero.

Understanding and using the Nernst equation is crucial when trying to determine the cell potential at different points in an electrochemical reaction, which is particularly useful in industries that rely on electrochemical cells, like batteries and sensors.

Electrochemical Reactions Explained

Electrochemical reactions involve the transfer of electrons from one substance to another and are the basis of batteries and electroplating. These reactions can be broken down into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained.

For the reaction in the original exercise:\[\begin{equation}\mathrm{PbS}(s) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s) + \mathrm{S}^{2-}(aq)\end{equation}\]Lead sulfide is oxidized to lead metal and sulfide ion. The half-reaction method allows us to separate the two processes, making it easier to analyze the overall chemical change and calculate changes in energy, such as the cell potential. Understanding these types of reactions is not only important for academic purposes but also for real-world applications like corrosion prevention and the development of new energy storage technology.

Equilibrium Constant: Connecting Reactions and Equilibrium

The equilibrium constant, K, plays a pivotal role in chemical reactions, signifying the point at which the rates of the forward and reverse reactions are equal, creating a state of dynamic equilibrium. For any reversible chemical reaction at a constant temperature, the ratio of the concentration of products to reactants, raised to their stoichiometric coefficients, remains constant and is described by this equilibrium constant.

Equilibrium constants come in various forms depending on the nature of the reaction. For acid-base reactions, it's called the acid dissociation constant Ka. For solubility reactions, it's known as the solubility product constant Ksp.
Equilibrium constants are essential for predicting the outcome of chemical reactions, and they're applied in diverse fields such as pharmaceuticals, environmental science, and food chemistry. A deep appreciation for equilibrium constants broadens the understanding of how different concentrations affect the position of equilibrium and the extent of chemical reactions.