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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 \(\mathrm{E}\), 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

Expert verified
The standard reduction potential for the half-cell reaction \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\) is approximately -0.414 V, as calculated using the Nernst equation and the given solubility product constant, \(K_{sp}\), value of \(8.0 \times 10^{-28}\).

Step by step solution

01

Write the balanced equation

First, we need to write the balanced equation for the dissolution of PbS(s) into its ions: \(\mathrm{PbS}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{S}^{2-}(a q)\)
02

Define the half-cell reaction

For the given half-cell reaction, we have: \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\)
03

Use the Nernst equation

The Nernst equation relates the standard reduction potential (Eº) of a half-cell reaction to the reduction potential (E) under non-standard conditions, the number of electrons involved in the reaction (n), the reaction quotient (Q) and the solubility product constant (Ksp). The Nernst equation is: \[E = Eº -\dfrac{RT}{nF} * \ln Q\] Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin (assumed to be 298 K), n is the number of electrons involved in the reaction (in this case, 2), and F is the Faraday constant (96485 C/mol).
04

Determine the reaction quotient (Q)

For our half-cell reaction, the reaction quotient (Q) can be written as: \[Q = \dfrac{[\mathrm{S^{2-}}]}{[\mathrm{Pb^{2+}}]}\] Since the reaction involves the dissolution of PbS, the concentrations of each ion are equal at equilibrium. Therefore, we can express the reaction quotient as: \[Q = \dfrac{[\mathrm{S^{2-}}]}{[\mathrm{S^{2-}}]} = 1\]
05

Substitute Q and Ksp into the Nernst equation

Now we can substitute the value of Q into the Nernst equation: \[E = Eº - \dfrac{RT}{2F} * \ln 1\] Since \(\ln 1 = 0\), it simplifies to: \[E = Eº\] Now, we can use the solubility product constant (Ksp) to find the value of Eº: \[K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{S^{2-}}] = [\mathrm{Pb^{2+}}]^{1}[\mathrm{S^{2-}}]^{1} \] Since Ksp is equal to \(8.0 \times 10^{-28}\), we can write: \([S^{2-}] = \sqrt{K_{sp}} = \sqrt{8.0 \times 10^{-28}}\)
06

Find the standard reduction potential (Eº)

Now we can find the standard reduction potential (Eº) using the value of E: \[E = Eº = \dfrac{RT}{2F} * \ln \sqrt{K_{sp}}\] Plug in the value of Ksp and the constants to find the standard reduction potential: \[Eº = \dfrac{(8.314)(298)}{2(96485)} * \ln \sqrt{8.0 \times 10^{-28}}\] Calculate the value of Eº: \[Eº \approx -0.414 V\] So, the value of the standard reduction potential for the half-cell reaction is -0.414 V.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant
The solubility product constant, denoted as \(K_{sp}\), is a fundamental concept in understanding the dissolution of sparingly soluble ionic compounds. These are substances that do not dissolve completely in water, but establish an equilibrium between the undissolved solid and the dissolved ions. The \(K_{sp}\) value helps predict how much of the ionic compound will dissolve in water, resulting in a saturated solution.
  • For a general ionic compound \(AB\) which dissolves in water, the equilibrium can be represented as \(AB(s) \rightleftharpoons A^+(aq) + B^-(aq)\).
  • The solubility product expression is given by \(K_{sp} = [A^+][B^-]\), where the concentrations are those of the ions formed at equilibrium.
  • A low \(K_{sp}\) value indicates very limited solubility.
In the case of lead(II) sulfide, \(\text{PbS} (s)\), the \(K_{sp}\) is \(8.0 \times 10^{-28}\). This shows that only an extremely small amount of lead(II) and sulfide ions will dissolve in water to form a saturated solution.
Standard Reduction Potential
The standard reduction potential, denoted as \(E^\circ\), is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is expressed in volts and is measured under standard conditions: 25°C (298 K), 1 atmosphere pressure, and 1 M concentration for each ionic species involved in the reaction.
  • A positive \(E^\circ\) value means the species is more likely to gain electrons and be reduced.
  • A negative \(E^\circ\) value indicates a lesser tendency to be reduced.
  • This value is crucial for determining the direction of redox reactions. It allows us to predict whether a particular half-reaction will occur as a reduction or as an oxidation in an electrochemical cell.
In the context of the exercise, we're calculating the standard reduction potential for the half-reaction involving lead(II) sulfide, which uses its \(K_{sp}\) to eventually find \(E^\circ\) that is \(-0.414 V\). This negative value suggests a reduced tendency for reduction under standard conditions.
Electrode Potential
Electrode potential is a broader term that includes both reduction and oxidation potentials. It reflects the potential difference between the electrode and its surroundings as electrons move to or from a solution in an electrochemical cell. The effective electrode potential is affected by several factors, including the concentration of ions, temperature, and the inherent properties of the elements and compounds involved.
  • Electrode potentials are measured in volts, similar to standard reduction potentials.
  • In practice, they help us understand how efficiently an electrode can gain or lose electrons, forming the foundation for electrochemical cell function.
  • The Nernst equation relates standard electrode potentials to actual conditions (not necessarily standard), incorporating concentrations, and thereby allowing for adjustments when calculating real-world potentials.
For the lead sulfide reaction present in the exercise, using the Nernst equation enables us to relate the electrode potential under the given conditions to the standard reduction potential derived from the solubility product constant \(K_{sp}\). Understanding this interplay is critical for predicting the behavior of compounds in varying electrochemical environments.

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Most popular questions from this chapter

A cell has a standard cell potential of \(+0.177 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction (a) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) halfcell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right) .\) The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{a}\) of benzoic acid?

A common shorthand way to represent a voltaic cell is anode|anode solution \(\|\) cathode solution \(\mid\) cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag} ;\) sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2}\); sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.

(a) Under what circumstances is the Nernst equation applicable? (b) What is the numerical value of the reaction quotient, Q, under standard conditions? (c) What happens to the emf of a cell if the concentrations of the reactants are increased?

A plumber's handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

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