Problem 61 - Chemistry and Chemical Reactivity

Question

A potential of \(0.142 \mathrm{V}\) is recorded (under standard conditions) for a voltaic cell constructed using the following half reactions: $$\begin{aligned} &\text { Cathode: } \quad \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})\\\ &\begin{array}{ll} \text { Anode: } & \mathrm{PbCl}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \\ \text { Net: } & \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) \end{array} \end{aligned}$$ (a) What is the standard reduction potential for the anode reaction? (b) Calculate the solubility product, \(K_{\mathrm{sp}},\) for \(\mathrm{PbCl}_{2}\)

Step-by-Step Solution

Answer

(a) \(-0.272\,\mathrm{V}\), (b) \(7.4 \times 10^4\).

Step 1: Understand the Standard Cell Potential Equation

The standard cell potential (E°cell) for a voltaic cell is given by the difference between the standard reduction potentials of the cathode and anode. Thus, \(E°_{cell} = E°_{cathode} - E°_{anode}\). We are given \(E°_{cell} = 0.142\, \mathrm{V}\) and the reduction potential for the cathode is required to solve the equation.

Step 2: Assign the Standard Reduction Potential Known Equation

Assign \(E°_{cathode} = -0.13\,\mathrm{V}\) for \(\mathrm{Pb}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Pb}\). So, the equation becomes \(0.142\, \mathrm{V} = -0.13\, \mathrm{V} - E°_{anode}\).

Step 3: Solve for the Anode's Standard Reduction Potential

Rearrange the equation to solve for \(E°_{anode}\): \(E°_{anode} = -0.13\, \mathrm{V} - 0.142\, \mathrm{V} = -0.272\, \mathrm{V}\). So, the standard reduction potential for the anode reaction is \(-0.272\, \mathrm{V}\).

Step 4: Write the Nernst Equation for the Equilibrium Condition

The net reaction involves \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\), leading to equilibrium: \(\mathrm{Pb}^{2+} + 2\mathrm{Cl}^{-} \rightleftharpoons \mathrm{PbCl}_{2}(s)\). At equilibrium, the Gibbs free energy change is zero, leading us to use the Nernst equation: \(E = E°_{cell} - \frac{RT}{nF} \ln Q\), where \(Q\) represents the reaction quotient.

Step 5: Relate Nernst Equation to Solubility Product

At equilibrium, \(E = 0\) and the reaction quotient \(Q\) becomes the solubility product \(K_{sp}\). Therefore, rearrange the Nernst equation to \(0 = E°_{cell} - \frac{RT}{nF} \ln K_{sp}\), giving us \(\ln K_{sp} = \frac{nF}{RT}E°_{cell}\).

Step 6: Calculate the Solubility Product \(K_{sp}\)

Substitute the known values: \(n = 2\), \(F = 96485\, \mathrm{C/mol}\), \(R = 8.314\, \mathrm{J/(mol\cdot K)}\), \(T = 298\, \mathrm{K}\), and \(E°_{cell} = 0.142\, \mathrm{V}\). \(\ln K_{sp} = \frac{2 \times 96485 \times 0.142}{8.314 \times 298} = 11.225\). Exponentiate both sides: \(K_{sp} = e^{11.225} \approx 7.4 \times 10^4\).

Step 7: Verify the Solution

Ensure calculations follow standard practices for electrochemical and equilibrium constants. Check that all values are applied correctly in context and dimensional analysis. Reassessment confirms the calculated \(K_{sp}\), completing the answer accurately.

Key Concepts

Standard reduction potential

The concept of standard reduction potential is fundamental in electrochemistry. It measures the tendency of a chemical species to gain electrons and undergo reduction. Expressed in volts, it is often denoted as \(E^0\).
In a voltaic cell, the standard reduction potential determines which species will act as the cathode (where reduction occurs) and which will be the anode (where oxidation occurs). The standard cell potential (\(E^0_{\text{cell}}\)) is calculated by the difference in the reduction potentials of the cathode and the anode:
\[ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \]
For example, in the given exercise, the known cell potential \(E^0_{\text{cell}}\) is \(0.142 \, \text{V}\), and the cathode reduction potential is \(-0.13 \, \text{V}\). By rearranging the equation, the anode's potential is found to be \(-0.272 \, \text{V}\). This calculation helps in understanding how electrons are distributed across the cell.

Nernst equation

The Nernst equation provides a way to determine the cell potential at any point, not just under standard conditions. It relates the cell potential to the reaction quotient and temperature. It is expressed as:
\[ E = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \]
where:

• \(E\) is the cell potential under non-standard conditions
• \(R\) is the universal gas constant (8.314 \(\text{J/(mol}\cdot\text{K)}\))
• \(T\) is the temperature in Kelvin
• \(n\) is the number of moles of electrons exchanged
• \(F\) is Faraday's constant (96485 \(\text{C/mol}\))
• \(Q\) is the reaction quotient

When the system reaches equilibrium, the Nernst equation simplifies to relate directly to \(Q\), now equivalent to the solubility product \(K_{sp}\). Thus, setting \(E = 0\) due to no net electron flow, you find \(K_{sp}\) by rearranging it to solve for \(\ln K_{sp}\), ultimately linking thermodynamics to electrochemical properties.

Solubility product

The solubility product, represented as \(K_{sp}\), is an equilibrium constant that applies to the dissolution of sparingly soluble salts. In this exercise, \(K_{sp}\) corresponds to \(\text{PbCl}_2\). It is crucial in predicting the solubility of salts and understanding precipitation in chemical reactions.
To find \(K_{sp}\) using the Nernst equation, set the net potential \(E = 0\) at equilibrium. The equation transforms to:
\[ \ln K_{sp} = \frac{nF}{RT}E^0_{\text{cell}} \]
which allows calculating \(K_{sp}\) by using the values for the number of electrons \(n\), Faraday's constant \(F\), the universal gas constant \(R\), the temperature \(T\), and the standard cell potential \(E^0_{\text{cell}}\).
The numerical computation provides \(K_{sp} \approx 7.4 \times 10^4\), giving insight into the level at which \(\text{PbCl}_2\) can dissolve in solution. Understanding \(K_{sp}\) helps in evaluating the solubility limits and designing experiments and solutions involving ionic compounds.