Question

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is \(1.00 M\) and the cell generates an emf of \(+0.22 \mathrm{~V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? \((\mathbf{b})\) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

Short answer

In short, the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell is \(6.0\times10^{7}\,\mathrm{M}\), and the solubility product constant (\(K_{sp}\)) for \(\mathrm{PbSO}_{4}\) is \(6.0\times10^{7}\).

Step-by-step solution

Write the Nernst equation

The Nernst equation relates the concentration of the species involved in the reaction to the cell potential (emf): \(E_{cell} = E_{cell}^{0} - \dfrac{0.0592}{n} \log_{10}{\dfrac{[\mathrm{Pb}^{2+}]}{[\mathrm{Sn}^{2+}]}}\) We need to substitute the given values and calculate \([\mathrm{Pb}^{2+}]\), i.e., the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell.

Substitute the given values into the Nernst equation

We have: \(E_{cell}= +0.22\,\mathrm{V}\), \([\mathrm{Sn}^{2+}] = 1.00 \,\mathrm{M}\). The standard cell potential, \(E_{cell}^{0}\), can be found from the standard reduction potentials of \(\mathrm{Sn}^{2+}\) and \(\mathrm{Pb}^{2+}\). Since both ions have a +2 charge, \(n=2\). For the reaction \(\mathrm{Sn}^{2+} + 2e^{-} \longrightarrow \mathrm{Sn}(s)\), the standard reduction potential is \(E^{0}(\mathrm{Sn}) = -0.14\,\mathrm{V}\). For the reaction \(\mathrm{Pb}^{2+} + 2e^{-} \longrightarrow \mathrm{Pb}(s)\), the standard reduction potential is \(E^{0}(\mathrm{Pb}) = -0.13\,\mathrm{V}\). The standard cell potential can be determined as: \(E_{cell}^{0}= E^{0}(\mathrm{Sn}) - E^{0}(\mathrm{Pb}) = -0.14-(-0.13)= -0.01\,\mathrm{V}\). Now, substitute these values into the Nernst equation: \(0.22 = -0.01 - \dfrac{0.0592}{2} \log_{10}{\dfrac{[\mathrm{Pb}^{2+}]}{1}}\)

Solve for the concentration of \(\mathrm{Pb}^{2+}\)

Now, we can solve for \([\mathrm{Pb}^{2+}]\): \(0.23= \dfrac{0.0592}{2} \log_{10}{[\mathrm{Pb}^{2+}]}\) \(7.776=\log_{10}{[\mathrm{Pb}^{2+}]}\) \(10^{7.776}=[\mathrm{Pb}^{2+}]\) Therefore, \([\mathrm{Pb}^{2+}]=6.0\)x\(10^{7}\,\mathrm{M}\).

Write the expression for the equilibrium constant, \(K_{sp}\)

The reaction for \(\mathrm{PbSO}_{4}(s)\) dissolution is: \(\mathrm{PbSO}_{4}(s) \longleftrightarrow \mathrm{Pb}^{2+}(a q) + \mathrm{SO}_{4}^{2-}(a q)\) Now we can write the expression for the solubility product constant, \(K_{sp}\): \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{SO}_{4}^{2-}]\)

Substitute the values and find \(K_{sp}\)

We have: \([\mathrm{Pb}^{2+}] = 6.0\)x\(10^{7}\,\mathrm{M}\) and \([\mathrm{SO}_{4}^{2-}] = 1.00\,\mathrm{M}\). Now, plug in the values in the expression for \(K_{sp}\): \(K_{sp} = (6.0\times10^{7})(1.00)\) Hence, \(K_{sp} = 6.0\)x\(10^{7}\).

Key concepts

Nernst Equation

The Nernst equation is a crucial tool in electrochemistry, especially when dealing with voltaic or galvanic cells. It connects the potential of an electrochemical cell to the concentrations of the reacting ions. Understanding this equation allows us to predict how changes in concentration affect the electromotive force (emf) of the cell.

The Nernst equation is written as:

• \( E_{cell} = E_{cell}^{0} - \frac{0.0592}{n} \log_{10}{\frac{[\mathrm{Pb}^{2+}]}{[\mathrm{Sn}^{2+}]}} \)

Here, \( E_{cell} \) is the actual cell potential and \( E_{cell}^{0} \) is the standard cell potential. The number \( n \) is the number of electrons transferred in the reaction. The concentrations are in molar units.

In our case, knowing the emf is \(+0.22 \, \mathrm{V}\) and the concentrations allow us to rearrange this equation and solve for unknown values like the \([\mathrm{Pb}^{2+}]\). This equation is essential for predicting cell behavior under non-standard conditions.

Standard Reduction Potential

Standard reduction potentials measure the tendency of a chemical species to gain electrons and be reduced. These potentials are determined under standard conditions (1 M concentration, 1 atm pressure, and 25°C), and they play a pivotal role in understanding which way a redox reaction will proceed.

In a voltaic cell, reactions occur in two half-cells, each with its own standard reduction potential. The equation for the cell potential is:

• \( E_{cell}^{0} = E_{reduction}^{0} - E_{oxidation}^{0} \)

In the provided exercise, the standard reduction potential for the reaction of \(\mathrm{Sn}^{2+}\) to \(\mathrm{Sn}(s)\) is \(-0.14\, \mathrm{V}\), while that for \(\mathrm{Pb}^{2+}\) to \(\mathrm{Pb}(s)\) is \(-0.13\, \mathrm{V}\). The negative reduction potential indicates that these are less likely to occur spontaneously, meaning they favor oxidation.

The difference between these potentials lets us calculate the standard cell potential, \(E_{cell}^{0}\), which is often used in the Nernst equation.

Solubility Product Constant

The solubility product constant, known as \(K_{sp}\), is an equilibrium constant for the dissolution of a sparingly soluble compound. It provides insight into the solubility of ionic compounds and helps predict precipitation reactions.

The dissolution reaction for \(\mathrm{PbSO}_{4}(s)\) is:

• \( \mathrm{PbSO}_{4}(s) \leftrightarrow \mathrm{Pb}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \)

Therefore, the \(K_{sp}\) expression is:

• \( K_{sp} = [\mathrm{Pb}^{2+}]\times[\mathrm{SO}_{4}^{2-}] \)

Understanding and calculating \(K_{sp}\) is crucial for predicting whether a salt will dissolve or precipitate in solution. In the exercise, we use calculated concentrations from the Nernst equation to determine \(K_{sp}\) for \(\mathrm{PbSO}_{4}\), demonstrating the interconnectedness of electrochemical and solubility concepts.