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 \(1.00 \mathrm{M}\) and the cell generates an emf of \(+0.22 \mathrm{~V}\), what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}{ }^{2-}\right]=1.00 \mathrm{M}\) in equilibrium with \(\mathrm{PbSO}_{4}(s)\), what is the \(K_{4 p}\) of \(\mathrm{PbSO}_{4}\) ? Batteries and Fuel Cells (Section 20.7)

Short answer

(a) The concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell is \(1.32 \times 10^{-2} M\). (b) The \(K_{sp}\) of \(\mathrm{PbSO}_4\) is approximately \(1.32 \times 10^{-2}\).

Step-by-step solution

Write down the balanced half-reactions and the Nernst equation.

First, we separate the given cell reaction into half-reactions. For the cathode (reduction) half-cell: \[ \mathrm{Sn}^{2+}(aq) + 2e^- \rightarrow \mathrm{Sn}(s) \] For the anode (oxidation) half-cell: \[ \mathrm{Pb}(s) \rightarrow \mathrm{Pb}^{2+}(aq) + 2e^- \] Next, we write down the Nernst equation: \[ E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10}\frac{Q}{1} \] where \(E^0_{cell}\) is the standard cell potential, \(E_{cell}\) is the cell potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient.

Calculate the standard cell potential.

Using the standard reduction potentials from the textbook, we can find the standard cell potential for the given reaction: Standard Reduction potential for cathode (\(E^0_{Sn^{2+}/Sn}\)) = -0.14 V Standard Reduction potential for anode (\(E^0_{Pb^{2+}/Pb}\)) = -0.13 V Calculate the cell potential difference: \[ E^0_{cell} = E^0_{Sn^{2+}/Sn} - E^0_{Pb^{2+}/Pb} = (-0.14) - (-0.13) = -0.01 V \]

Solve for the concentration of Pb²⁺.

Plug in the known values into the Nernst equation: \( E_{cell} = E^0_{cell} - \frac{0.0592}{n}\log_{10} \frac{[\mathrm{Sn}^{2+}]}{[\mathrm{Pb}^{2+}]} \) \[ 0.22 V = -0.01 V - \frac{0.0592}{2} \log_{10} \frac{1.00 M}{[\mathrm{Pb}^{2+}]}\] Solve for \([\mathrm{Pb}^{2+}]\): \( [\mathrm{Pb}^{2+}] = 1.32 \times 10^{-2} M\)

Write the balanced equation for the solubility reaction and the \(K_{sp}\) expression.

For the solubility reaction of \(\mathrm{PbSO}_4\): \[ \mathrm{PbSO}_4 (s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{SO}_4^{2-}(aq) \] The solubility product constant (\(K_{sp}\)) expression is: \[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{SO}_4^{2-}] \]

Calculate the \(K_{sp}\) of \(\mathrm{PbSO}_4\).

Using the given concentration of \(\mathrm{SO}_4^{2-}\) and the concentration of \(\mathrm{Pb}^{2+}\) calculated in step 3: \[ K_{sp} = (1.32 \times 10^{-2} M)(1.00 M) = 1.32 \times 10^{-2} \] The \(K_{sp}\) of \(\mathrm{PbSO}_4\) is approximately \(1.32 \times 10^{-2}\).

Key concepts

Half-Reactions

In a voltaic cell, the overall reaction is broken into two **half-reactions**: one representing oxidation and the other reduction. These processes occur at different electrodes and are crucial in understanding cell operations. - **Reduction Half-Reaction**: Occurs at the cathode. Electrons are gained. In the problem, the reduction half-reaction for tin is: \[ \mathrm{Sn}^{2+}(aq) + 2e^- \rightarrow \mathrm{Sn}(s) \]- **Oxidation Half-Reaction**: Occurs at the anode. Electrons are lost. In this exercise, the oxidation half-reaction for lead is: \[ \mathrm{Pb}(s) \rightarrow \mathrm{Pb}^{2+}(aq) + 2e^- \]Understanding these reactions individually helps in analyzing electron flow, reaction stoichiometry, and the overall cell operation.

Nernst Equation

The **Nernst equation** is fundamental in calculating cell potentials under non-standard conditions. This equation accounts for the concentrations of the ions involved in the electrochemical reaction. It is expressed as:\[ E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10}Q \]Where:- \( E^0_{cell} \) is the standard cell potential, a measure of the potential difference under standard conditions.- \( n \) is the number of electrons exchanged per mole of reaction.- \( Q \) is the reaction quotient, calculated from the concentrations of the reactants and products.Applying the Nernst equation allows us to determine actual cell potential, helping to predict how concentration changes affect the electromotive force (emf) of the cell.

Cell Potential

The **cell potential** or electromotive force (emf) indicates the voltage produced by a voltaic cell. This potential is derived from the energy difference between the two half-cells in the reaction.1. **Standard Cell Potential (\( E^0_{cell} \))**: This is the potential measured when concentrations are at standard conditions (usually 1 M, 1 atm, and 25°C). - For our reaction, the cell potential is: \( E^0_{cell} = -0.01 \) V, derived from the difference in standard reduction potentials of the two half-reactions.2. **Non-Standard Cell Potential (\( E_{cell} \))**: This potential is when the conditions deviate from standard. - The actual cell potential of 0.22 V indicates a slightly different condition than standard.Understanding the cell potential helps us analyze the feasibility and efficiency of the cell reactions.

Solubility Product Constant

The **solubility product constant (\( K_{sp} \))** applies to sparingly soluble salts, representing the product of the ion concentrations at equilibrium. It's an essential concept in predicting the extent to which a compound will dissolve.For **lead sulfate (PbSO₄)** in our cell:- The solubility equilibrium is: \[ \mathrm{PbSO}_4(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{SO}_4^{2-}(aq) \]- The \( K_{sp} \) is given by: \[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{SO}_4^{2-}] \]Using calculated values, the \( K_{sp} \) for PbSO₄ is approximately \( 1.32 \times 10^{-2} \). Knowing the \( K_{sp} \) helps ascertain the solubility of lead sulfate under given conditions and whether the salt will precipitate in the solution.