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

Short Answer

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

Step by step solution

01

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 \( \mathrm{SO_4^{2-}} \).
02

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.
03

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:\[ E^\circ_{\text{cell}} = (-0.126 \, \text{V}) - (-0.356 \, \text{V}) = 0.230 \, \text{V} \]
04

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 \( (298 \, \text{K} \)).
05

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 \( K_{sp} \):\[ K_{sp} = e^{17.922} \approx 6.1 \times 10^{-8} \]
06

Short Answer

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

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Key Concepts

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

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.

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Most popular questions from this chapter

A Consider the standard reduction potentials of cesium and lithium. $$ \begin{array}{ll} \mathrm{Cs}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Cs}(\mathrm{s}) & E^{\circ}=-3.026 \mathrm{~V} \\\ \mathrm{Li}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Li}(\mathrm{s}) & E^{\circ}=-3.095 \mathrm{~V} \end{array} $$ The periodic trends in the properties of the element indicate that fluorine is the most chemically reactive nonmetal, so perhaps it is not surprising that the standard reduction potential of fluorine has the highest positive value for a nonmetallic element. However, periodic properties of the elements also indicate that cesium should be the most reactive metal. Comparison of the voltage of the cesium half-reaction with that of lithium shows that the standard reduction potential of lithium is less negative than that of cesium, indicating that lithium is a better oxidizer than is cesium. (a) Calculate the standard cell voltages of the voltaic cells based on the reaction $$ 2 \mathrm{M}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{M}^{+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq}) $$ where \(\mathrm{M}\) is \(\mathrm{Cs}\) and \(\mathrm{Li}\). (b) Assuming that the pressure of \(\mathrm{F}_{2}(\mathrm{~g})\) stays at \(1.00 \mathrm{~atm}\), what concentration does \(\mathrm{Li}^{+}(\mathrm{aq})\) have to be for the voltage of the \(\mathrm{Li} / \mathrm{F}_{2}\) voltaic cell to equal the standard voltage of the \(\mathrm{Cs} / \mathrm{F}_{2}\) voltaic cell? (c) Can you suggest a reason why the standard reduction potential of lithium is lower than that of cesium, even though periodic trends indicate that cesium is the more reactive metal? (d) Calculate \(\Delta G^{o}\) for both the \(\mathrm{Li} / \mathrm{F}_{2}\) and the \(\mathrm{Cs} / \mathrm{F}_{2}\) voltaic cells from their \(E^{\circ} \mathrm{s}\). Compare this with the Gibbs' free energies of formation of \(2 \mathrm{~mol} \mathrm{LiF}\) and CsF. Can you explain the difference? (e) Given the fact that alkali metals react rather violently with water, it would be unlikely that any voltaic cell can be constructed using Li(s) or \(\mathrm{Cs}(\mathrm{s})\) in the presence of water. A more likely scenario is that the voltaic cell would have no solvent, so that the voltaic cell reaction would be $$ 2 \mathrm{M}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MF}(\mathrm{xtal}) $$ where \(\mathrm{M}\) is \(\mathrm{Cs}\) or \(\mathrm{Li}\). What would be the \(E^{\circ}\) s of the two different voltaic cells if this were the reaction? (Hint: See your answer to part d.)

List the halogens in order of increasing oxidizing power.

Balance the following reactions, and specify which species is oxidized and which is reduced. (a) \(\mathrm{Na}+\mathrm{FeCl}_{3} \rightarrow \mathrm{Fe}+\mathrm{NaCl}\) (b) \(\mathrm{SnCl}_{2}+\mathrm{FeCl}_{3} \rightarrow \mathrm{SnCl}_{4}+\mathrm{FeCl}_{2}\) (c) \(\mathrm{CO}+\mathrm{Cr}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Cr}+\mathrm{CO}_{2}\)

Another type of battery is the al 71 in which the cell reaction is $$ \mathrm{Zn}(\mathrm{s})+\mathrm{HgO}(\mathrm{s}) \rightarrow \mathrm{Hg}(\ell)+\mathrm{ZnO}(\mathrm{s}) $$ \(E^{o}=+1.35 \mathrm{~V}\) (a) What is the standard free energy change for this reaction? (b) The standard free energy change in a voltaic cell is the maximum electrical energy that the cell can produce. If the reaction in a zinc-mercury cell consumes \(1.00 \mathrm{~g}\) mercury oxide, what is the standard free energy change? (c) For how many hours could a mercury cell produce a \(10-\mathrm{mA}\) current if the limiting reactant is \(3.50 \mathrm{~g}\) mercury oxide?

A half-cell that consists of a silver wire in a \(1.00 M\) \(\mathrm{AgNO}_{3}\) solution is connected by a salt bridge to a \(1.00 \mathrm{M}\) thallium(I) acetate solution that contains a metallic T1 electrode. The voltage of the cell is \(1.136 \mathrm{~V}\), with the silver as the positive electrode. (a) Write the half-reactions and the overall chemical equation for the spontaneous reaction. (b) Use the standard potential of the silver half-reaction, with the voltage of the cell, to calculate the standard reduction potential for the thallium half-reaction.

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