Problem 69 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

(a) The standard anode potential is -0.272 V. (b) The solubility product, \(K_{sp}\), is approximately 61,508.25.

Step 1: Write the cell potential formula

The cell potential, \(E^\circ_\text{cell}\), is given as 0.142 V according to the problem's statement. The standard cell potential is related to the cathode and anode reactions by the equation: \[E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}\]where \(E^\circ_\text{cathode}\) and \(E^\circ_\text{anode}\) are the standard reduction potentials for the cathode and anode, respectively.

Step 2: Identify known reduction potentials

The reduction potential for the cathode (\(Pb^{2+} + 2e^- \rightarrow Pb\)) is a known value from standard tables, typically \(E^\circ_\text{cathode} = -0.13 \text{ V}\). We'll use this information in our calculations.

Step 3: Calculate the anode potential

Using the formula derived in Step 1, solve for \(E^\circ_\text{anode}\):\[E^\circ_\text{anode} = E^\circ_\text{cathode} - E^\circ_\text{cell} = -0.13 \text{ V} - 0.142 \text{ V} = -0.272 \text{ V}\].Thus, the standard reduction potential for the anode is -0.272 V.

Step 4: Setup the expression for Ksp

The net reaction is \(Pb^{2+}(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s)\), indicating the dissociation of PbCl2 in water with the equilibrium constant represented as:\[K_{sp} = [Pb^{2+}][Cl^-]^2\].We need to compute \(K_{sp}\) based on the given cell potential.

Step 5: Relate standard cell potential to Ksp

The relationship between the cell potential and equilibrium constant is given by the Nernst equation at equilibrium:\[E^\circ_\text{cell} = \frac{RT}{nF}\ln(K_{sp})\]where \(R\) is the gas constant (8.314 J/mol∙K), \(T\) is temperature in Kelvin (298 K for standard conditions), \(n\) is the number of moles of electrons transferred in the net reaction (\(n=2\)), and \(F\) is Faraday's constant (96485 C/mol).

Step 6: Solve for Ksp

Rearrange the Nernst equation to solve for \(K_{sp}\):\[\ln(K_{sp}) = \frac{nFE^\circ_\text{cell}}{RT}\]Substitute the known values:\[\ln(K_{sp}) = \frac{2 \times 96485 \times 0.142}{8.314 \times 298} = \frac{27344.9}{2478.972} \approx 11.03\]Exponentiate to solve for \(K_{sp}\):\[K_{sp} = e^{11.03} \approx 61,508.25\].

Key Concepts

Voltaic cell

A voltaic cell is a device that generates electrical energy from a spontaneous redox reaction. It consists of two half-cells connected by a salt bridge. Each half-cell contains an electrode and an electrolyte.

In the exercise, the cell has two half reactions: the reduction of lead ions at the cathode and the discharge at the anode.

Here's how it works:

• Electrons move from the anode to the cathode, generating electric current.
• The salt bridge maintains electrical neutrality by allowing ions to flow between the half-cells.
• This flow of electrons and ions keeps the reaction running until one reactant is exhausted.

This setup is typical of many portable batteries we use daily, converting chemical energy to electrical power.

Reduction potential

Reduction potential measures a substance's tendency to gain electrons in a redox reaction. This is a key factor in determining the direction and spontaneity of chemical reactions.

Standard reduction potentials are measured under standard conditions (298 K, 1M concentration, 1 atm pressure) and are expressed in volts.

To understand the voltaic cell in the exercise:

• Cathode reaction: lead ions gain electrons, with a potential of -0.13 V.
• Anode reaction: calculated to have a potential of -0.272 V.

The difference between these potentials determines the overall cell potential, which tells us how much voltage the cell can produce.

Solubility product

Solubility product (K_{sp}) is a measure of the solubility of a compound. It represents the extent to which a compound can dissolve in water.

In the exercise, we calculate K_{sp} for lead chloride (PbCl_2), which dissociates into lead ions and chloride ions in water.

Here's how it works:

• K_{sp} is represented as \([\text{Pb}^{2+}][\text{Cl}^-]^2\).
• The higher the K_{sp}, the more soluble the compound.
• The concentration of ions at equilibrium gives us the solubility.

By understanding K_{sp}, we can predict whether a precipitate will form under certain conditions.

Nernst equation

The Nernst equation relates the cell potential to the concentrations of reactants and products, providing a way to calculate cell potential under non-standard conditions.

It is expressed as:

\[E = E^\circ - \frac{RT}{nF} \ln(Q)\]
where:

• E is the cell potential at non-standard conditions.
• E^\circ is the standard cell potential.
• R is the gas constant, 8.314 J/mol∙K.
• T is the temperature in Kelvin.
• n is the number of moles of electrons exchanged.
• F is Faraday's constant, 96485 C/mol.
• Q is the reaction quotient.

In the exercise, we use the Nernst equation at equilibrium to find K_{sp}, linking the reaction's electrical and chemical aspects. This connection is vital in predicting how cells function under different conditions.