Problem 68 - Chemistry Principles and Practice

Question

Calculate the value of the solubility product constant for \(\mathrm{PbSO}_{4}\) from the half-cell potentials. $$ \begin{aligned} \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) & E^{\circ}=-0.356 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) & E^{\circ}=-0.126 \mathrm{~V} \end{aligned} $$

Step-by-Step Solution

Answer

The solubility product constant \( K_{sp} \) for \( \mathrm{PbSO}_4 \) is approximately \( 6.1 \times 10^{-8} \).

Step 1: Write the Net Electrochemical Reaction

For the dissolution of \( \mathrm{PbSO}_4 \) in water, the net reaction is:\[ \mathrm{PbSO}_4(s) \rightarrow \mathrm{Pb^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq) \] This occurs via the reduction of \( \mathrm{PbSO}_4 \) to \( \mathrm{Pb} \) and \( \mathrm{SO_4^{2-}} \).

Step 2: Use Given Half-Reactions

To write the net reaction and represent the process electrochemically, consider the given half-reactions:1. \( \mathrm{PbSO}_4(s) + 2 e^- \rightarrow \mathrm{Pb}(s) + \mathrm{SO_4^{2-}}(aq) \), with \( E^\circ = -0.356 \, \text{V} \)2. \( \mathrm{Pb^{2+}}(aq) + 2 e^- \rightarrow \mathrm{Pb}(s) \), with \( E^\circ = -0.126 \, \text{V} \)These will be combined to find the standard cell potential.

Step 3: Calculate the Standard Cell Potential (E°cell)

The standard cell potential is calculated by subtracting the anode potential from the cathode potential:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]Substituting the given values:\[ E^\circ_{\text{cell}} = (-0.126 \, \text{V}) - (-0.356 \, \text{V}) = 0.230 \, \text{V} \]

Step 4: Use Relationship between E°cell and Ksp

The relationship between the standard cell potential \( E^\circ_{\text{cell}} \) and the solubility product constant \( K_{sp} \) is given by:\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} = RT \ln(K_{sp}) \]Where \( n \) is the number of moles of electrons transferred \( (n = 2) \), \( F \) is Faraday's constant \( (96485 \, \text{C/mol}) \), \( R \) is the ideal gas constant \( (8.314 \, \text{J/mol\cdot K}) \), and \( T \) is the temperature in Kelvin \( (298 \, \text{K} \)).

Step 5: Calculate Ksp using the Equation

First, calculate \( \Delta G^\circ \):\[ \Delta G^\circ = -2 \times 96485 \, \text{C/mol} \times 0.230 \, \text{V} \]\[ \Delta G^\circ = -44383.1 \, \text{J/mol} \]Substitute \( \Delta G^\circ \) into the equation for \( K_{sp} \):\[ 44383.1 \, \text{J/mol} = 8.314 \, \text{J/mol\cdot K} \times 298 \, \text{K} \times \ln(K_{sp}) \]\[ \ln(K_{sp}) = \frac{44383.1}{2475.772} \]\[ \ln(K_{sp}) = 17.922 \]Therefore, calculate \( K_{sp} \):\[ K_{sp} = e^{17.922} \approx 6.1 \times 10^{-8} \]

Step 6: Short Answer

The solubility product constant \( K_{sp} \) for \( \mathrm{PbSO}_4 \) is approximately \( 6.1 \times 10^{-8} \).

Key Concepts

Electrochemical Reactions

Electrochemical reactions involve the transfer of electrons between chemical species. These reactions are fundamental in devices like batteries, where they convert chemical energy into electrical energy. In the reaction process, oxidation and reduction occur simultaneously, which is why they are often referred to as redox reactions.

In a typical electrochemical cell, there are two electrodes submerged in an electrolyte solution. One electrode undergoes oxidation (loses electrons) and is called the anode. The other electrode undergoes reduction (gains electrons) and is called the cathode. This flow of electrons through a conductor creates an electric current.

• Oxidation: Loss of electrons.
• Reduction: Gain of electrons.
• Redox: Combination of reduction and oxidation processes.

Understanding these processes is key to determining how cells create electrical energy and perform calculations related to their efficiency.

Half-Cell Potentials

Half-cell potentials measure a single electrode's tendency to be reduced, given as a voltage. It indicates how strongly a species wants to gain or lose electrons.

When half-cells are combined to form a full electrochemical cell, the potential difference reflects the ability of the cell to do work. For any electrochemical system, half-cell potentials are critical because:

• They offer insight into how electrons will flow between species.
• They help determine which electrode will serve as the anode or cathode.
• They are used in calculating the standard cell potential (E°).

Achieving accurate values for these potentials can reveal much about the electrochemical properties of a system and assist in predicting the direction and feasibility of redox reactions.

Standard Cell Potential

The standard cell potential, denoted as \( E^\circ_{\text{cell}} \), is the voltage or electric potential generated by a cell under standard conditions (1M concentrations, 1 atm pressure, and 25°C).

It's calculated from the half-cell potentials using the following formula:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

This value tells us how forceful the electrochemical reaction is.

• A positive \( E^\circ_{\text{cell}} \) indicates a spontaneous reaction under standard conditions.
• A negative \( E^\circ_{\text{cell}} \) suggests a non-spontaneous process that requires external energy to occur.

The standard cell potential plays a crucial role in assessing a cell's efficiency and determining its practicality for applications.

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), measures the maximum amount of work done by a thermodynamic system at constant temperature and pressure. It's a valuable concept in chemistry because it predicts the spontaneity of a reaction.

In electrochemistry, the relationship between Gibbs Free Energy and the standard cell potential is given by:\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

Where:

• \( n \) is the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant, approximately \( 96485 \) C/mol.
• \( \Delta G^\circ \) indicates the change in free energy under standard conditions.

If \( \Delta G^\circ \) is negative, the reaction is spontaneous. If positive, it's non-spontaneous under the given conditions. Understanding this helps explain why certain reactions prefer to change states and offers insights into reaction feasibilities.

Faraday's Law

Faraday's Law is foundational in electrochemistry. It describes the relationship between the quantity of electric charge and the amount of substance that undergoes oxidation or reduction.

Faraday's Law states:

• The amount of a substance altered at an electrode during electrolysis is proportional to the quantity of electricity that passes through the circuit.
• For every mole of electrons transferred, a predictable amount of substance is deposited or dissolved at an electrode.

These concepts are expressed in the equation:\[ Q = nF \]Where:

• \( Q \) is the total charge in coulombs.
• \( n \) is the number of moles of particles oxidized or reduced.
• \( F \) is Faraday’s constant.

This law is used for calculating the amount of electricity needed to drive an electrochemical reaction and is essential for understanding the quantitative aspects of electrolysis and electroplating.