Problem 68 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) The concentration of \(\mathrm{Pb}^{2+}\) in the anode compartment is \(2.72 \times 10^{-3} \mathrm{M}\). (b) The solubility product constant \(K_{sp}\) for \(\mathrm{PbSO_{4}}\) is \(2.72 \times 10^{-3}\).

Step 1: Write down the redox reactions and the Nernst equation

First, let's write half-reactions for the given redox reaction: \[Sn^{2+} + 2e^{-} \rightarrow Sn\] \[Pb \rightarrow Pb^{2+} + 2e^{-}\] Now, let's write down the Nernst equation for this cell: \[E_{cell} = E_{0} - \frac{0.0592}{n} \log{Q}\] Where: - \(E_{cell}\): the cell potential - \(E_{0}\): the standard cell potential - n: number of electrons transferred (in this case, n = 2) - Q: the reaction quotient

Step 2: Calculate the standard cell potential

We need to find the standard reduction potentials for both half-reactions to find \(E_{0}\). From the standard electrode potential table, we obtain the following potentials: Sn\(^{2+}\)/Sn: -0.14 V Pb\(^{2+}\)/Pb: -0.13 V Now, given the reaction mentioned above: \[Sn^{2+} + Pb \rightarrow Sn + Pb^{2+}\] We have: \(E_{0} = E_{cathode}^0 - E_{anode}^0 = (-0.14) - (-0.13) = -0.01 \mathrm{V}\)

Step 3: Write the reaction quotient and obtain the concentration of Pb\(^{2+}\)

The reaction quotient (Q) for the reaction is: \[Q = \frac{[Pb^{2+}]}{[Sn^{2+}]}\] Now, we can use the Nernst equation and the given cell potential (0.22 V): \[0.22 \mathrm{V} = -0.01 \mathrm{V} - \frac{0.0592}{2} \log{\frac{[Pb^{2+}]}{1.00}}\] Solving for [Pb\(^{2+}\)]: \[[Pb^{2+}] = 2.72 \times 10^{-3} \mathrm{M}\]

Step 4: Calculate the solubility product constant

We have the equilibrium concentrations of Pb\(^{2+}\) and SO\(_4^{2-}\) ions in equilibrium with PbSO\(_4\). The solubility product constant \(K_{sp}\) can be calculated as follows: \[K_{sp} = [Pb^{2+}] \times [SO_4^{2-}]\] Using the obtained concentration of Pb\(^{2+}\) and given concentration of SO\(_4^{2-}\): \[K_{sp} = (2.72 \times 10^{-3})(1.00) = 2.72 \times 10^{-3}\] So, the solubility product constant \(K_{sp}\) for \(\mathrm{PbSO_{4}}\) is \(2.72 \times 10^{-3}\).

Key Concepts

Redox Reactions

Redox reactions are also known as oxidation-reduction reactions. They involve the transfer of electrons between two substances. One substance gets oxidized while the other gets reduced. In the given voltaic cell, we have two half-reactions happening:

• Reduction: \[\mathrm{Sn^{2+} + 2e^- \rightarrow Sn}\]This means the tin ion (\(\mathrm{Sn}^{2+}\)) gains electrons to become tin metal (\(\mathrm{Sn}\)).
• Oxidation: \[\mathrm{Pb \rightarrow Pb^{2+} + 2e^-}\]Here, lead metal gives up electrons to form lead ion (\(\mathrm{Pb}^{2+}\)).

The substance that gets reduced gains electrons and is called the oxidizing agent. Meanwhile, the substance that gets oxidized loses electrons and is known as the reducing agent. In our cell, \(\mathrm{Sn^{2+}}\) is the oxidizing agent and \(\mathrm{Pb}\) is the reducing agent. The whole purpose of these reactions is to generate an electromotive force (EMF) that can be used to do work, like in a battery.

Nernst Equation

The Nernst equation is essential when working with voltaic cells that do not operate under standard conditions. It allows us to calculate the cell potential using the concentrations of the reactants and products involved in the electrochemical reaction. The Nernst equation is given by:\[E_{cell} = E_{0} - \frac{0.0592}{n} \log{Q}\]Here's what those terms mean:

• \(E_{cell}\) is the EMF or cell potential we measure in volts.
• \(E_{0}\) is the standard cell potential, which can be obtained from electrode potential tables.
• \(n\) represents the number of moles of electrons transferred in the reaction. Here, \(n=2\) as two electrons are involved.
• \(Q\) is the reaction quotient and is calculated based on the concentrations of the ions in the cell.

In this exercise, we use the Nernst equation to find the concentration of \(\mathrm{Pb^{2+}}\), given the EMF and the concentration of \(\mathrm{Sn^{2+}}\). The equation shows us how the potential changes with non-standard concentrations, providing a more realistic picture of cell behavior in practical scenarios.

Electrode Potentials

Electrode potentials, or standard reduction potentials, tell us how likely a chemical species is to gain electrons in a redox reaction. These potentials can be found in standard electrode potential tables. For voltaic cells, the electrode potential of each half-reaction helps to determine the cell potential and the direction of electron flow.

The half-reaction with higher reduction potential will occur at the cathode, where reduction occurs.

The half-reaction with lower reduction potential will occur at the anode, where oxidation happens.

In our voltaic cell:

• Cathode reaction: \(\mathrm{Sn^{2+}/Sn}\) with a reduction potential of \(-0.14\, \mathrm{V}\)
• Anode reaction: \(\mathrm{Pb^{2+}/Pb}\) with a reduction potential of \(-0.13\, \mathrm{V}\)

Since the cathode reaction has a lower potential than the anode, we subtract the anode potential from the cathode potential to find the standard cell potential:\[E_{0} = -0.14 \, \text{V} - (-0.13 \, \text{V}) = -0.01 \, \text{V}\]This tells us the inherent energy difference driving the electron flow, which is crucial for calculating the Nernst equation later on.