Question

The \(K_{s}\) 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 $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$

Short answer

The standard reduction potential for the reaction \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\) is approximately \(-0.101 \, \text{V}\).

Step-by-step solution

Write the solubility product constant equation for PbS

The solubility product constant, Ks, is given by the product of the equilibrium concentrations of the dissolved ions raised to the power of their coefficients. For the given reaction, the Ks equation for PbS will be: \(K_s = [\mathrm{Pb}^{2+}][\mathrm{S}^{2-}]\)

Write the equations for equilibrium concentrations of Pb2+ and S2-

Let the equilibrium concentration of Pb2+ and S2- be represented by x: \([\mathrm{Pb}^{2+}] = x\) \([\mathrm{S}^{2-}] = x\)

Substitute the equilibrium concentrations into the Ks equation

Now we will substitute the equilibrium concentrations into the Ks equation: \(K_s = (x)(x)\)

Calculate the equilibrium concentrations

Now we'll solve for x using the given Ks value: \(8.0 \times 10^{-28} = x^2\) From here, we can determine the value of x by taking the square root of both sides of the equation: \(x = \sqrt{8.0 \times 10^{-28}}\) \(x \approx 2.83 \times 10^{-14}\) Now that we have the equilibrium concentration of both Pb2+ and S2-, we can move on to determine the standard reduction potential using the Nernst equation.

Write the Nernst equation

The Nernst equation relates the electrode potential, E, to the standard electrode potential, \(E^\circ\), using the reaction's stoichiometric coefficients, the concentrations of the ions involved and the temperature: \(E = E^\circ - \frac{RT}{nF} \ln Q\) For this reaction at standard conditions (T=298 K) and considering that R (gas constant) = \(8.314 \,\text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\) and F (Faraday's constant) = \(9.6485 × 10^4 \,\text{C} \cdot \text{mol}^{-1}\), the equation becomes: \(E = E^\circ - \frac{0.0592}{2} \log \frac{[\mathrm{Pb}^{2+}][\mathrm{S}^{2-}]}{[\mathrm{PbS}]}\) Since PbS is a solid, its concentration is not included in the equation. Notice that we have used the logarithm with base 10 (log) instead of the natural logarithm (ln) by replacing \(\frac{2.303 RT}{nF}\) with \(\frac{0.0592}{n}\). The coefficient n is equal to 2 since there are 2 moles of electrons involved in the reaction. Now we will substitute the calculated equilibrium concentrations into the Nernst equation: \(E = E^\circ - \frac{0.0592}{2} \log [(\mathrm{2.83} \times 10^{-14})^{2}]\)

Use the electrode potential from Appendix E

Using Appendix E, we can find the standard electrode potential for the reaction: \(\mathrm{Pb}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Pb}\): \(E^\circ = -0.13 \, \text{V}\) Now we substitute this value back into the Nernst equation: \(E = (-0.13) - \frac{0.0592}{2} \log [(\mathrm{2.83} \times 10^{-14})^{2}]\)

Calculate the standard reduction potential

Finally, we can now calculate the standard reduction potential for the given reaction: \(E \approx (-0.13) - \frac{0.0592}{2} \log [(\mathrm{2.83} \times 10^{-14})^{2}]\) \(E \approx -0.13 - (-0.02896) = -0.10104 \, \text{V}\) Therefore, the standard reduction potential for the reaction is approximately \(-0.101 \, \text{V}\).

Key concepts

Solubility Product Constant

The solubility product constant, represented as \(K_s\), is an essential concept in electrochemistry. It helps us understand the equilibrium between a solid and its ions in a solution. In simple terms, it's a measure of how much of a solid dissolves in a solution to form ions. Let's take a closer look using our example of lead(II) sulfide, \(\text{PbS}\). Lead(II) sulfide is a sparingly soluble salt. When it dissolves partially in water, it breaks down into lead ions, \(\text{Pb}^{2+}\), and sulfide ions, \(\text{S}^{2-}\). The balanced chemical equilibrium equation is given by:

• \( \text{PbS}(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{S}^{2-}(aq) \)

The solubility product constant \(K_s\) for lead(II) sulfide is calculated with the expression:

• \( K_s = [\text{Pb}^{2+}][\text{S}^{2-}] \)

In this case, the product of the concentrations of these ions remains constant at equilibrium for a given temperature. With \(K_s = 8.0 \times 10^{-28}\), you see that \(\text{PbS}\) is not very soluble, since the equilibrium concentrations of the ions are quite low, calculated to be about \(2.83 \times 10^{-14}\, \text{mol/L}\) for each ion. Understanding \(K_s\) can help predict whether a precipitate will form in a reaction under specific conditions.

Nernst Equation

The Nernst equation is a powerful tool in electrochemistry for calculating the cell potential under non-standard conditions. It connects the standard electrode potential to the actual conditions (concentrations and temperatures) of a reaction. The equation is formulated as:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]Here, \(E\) is the electrode potential, \(E^\circ\) is the standard electrode potential, \(R\) is the universal gas constant \(8.314\,\text{J/mol·K}\), \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged in the reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.
For reactions at room temperature (298 K), a common simplification of the Nernst equation is:

• \( E = E^\circ - \frac{0.0592}{n} \log Q \)

In the context of our \(\text{PbS}(s)\) reaction, we observe that \(n = 2\) due to the two electrons involved. The reaction quotient \(Q\) is the ratio of the concentrations of the products and reactants, raised to the power of their stoichiometric coefficients, excluding solids or pure liquids:

• \( Q = \frac{[\text{Pb}^{2+}][\text{S}^{2-}]}{[\text{PbS}]} \), \([\text{PbS}]\) is constant since it's a solid.

By substituting the calculated concentrations \(2.83 \times 10^{-14}\) into the Nernst equation, and using the \(E^\circ = -0.13\,\text{V}\) for lead, we derive the new cell potential under specific conditions.

Standard Reduction Potential

The standard reduction potential, denoted \(E^\circ\), is a measure of the tendency of a chemical species to gain electrons, thus undergoing reduction. Standard conditions entail a concentration of 1 M, a pressure of 1 atm, and a temperature of 298 K. Each half-reaction in electrochemistry has a standard reduction potential, which can be found in electrochemical series tables.
The standard reduction potential for a reaction offers insight into the reaction's spontaneity and directionality. An essential reference is the comparison of \(E^\circ\) values:

• Species with higher \(E^\circ\) values are more likely to get reduced under standard conditions.
• Conversely, species with lower \(E^\circ\) values are more likely to get oxidized, acting as reducing agents.

In our exercise, the standard reduction potential for the lead half-reaction, \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}(s)\), was given as \(-0.13\,\text{V}\). To find the \(E^\circ\) for the overall reaction, involving \(\text{PbS}\), we used the Nernst equation to adjust it based on non-standard conditions. This gave us the calculated value of approximately \(-0.101\,\text{V}\). Thus, knowing \(E^\circ\) helps predict whether the reaction will occur and its propensity to do so under standard and non-standard conditions.