Question

Consider a cell in which the reaction is $$ \mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g) $$ (a) Calculate \(E^{\circ}\) for this cell. (b) Chloride ions are added to the \(\mathrm{Pb} \mid \mathrm{Pb}^{2+}\) half- cell to precipitate \(\mathrm{PbCl}_{2}\). The voltage is measured to be $$ \begin{array}{l} +0.210 \mathrm{~V} \text { . Taking }\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M} \text { and } P_{\mathrm{H}_{2}}=1.0 \mathrm{~atm}, \text { cal- } \\\ \text { culate }\left[\mathrm{Pb}^{2+}\right] . \end{array} $$ (c) Taking [Cl ^ ] in (b) to be 0.10 M, calculate \(K_{\text {sp }}\) of \(\mathrm{PbCl}_{2}\).

Short answer

Question: Calculate the solubility product Ksp of PbCl₂ given the cell reaction: $$\mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g)$$, when the cell voltage is +0.210 V, [H⁺] = 1.0 M, PH₂ = 1.0 atm, and [Cl⁻] = 0.10 M. Answer: The solubility product Ksp of PbCl₂ is 8.23 * 10⁻⁷.

Step-by-step solution

Determine the half-reactions and their standard reduction potentials

We have the given cell reaction: $$ \mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g) $$ The half-reactions involved are: 1. Pb²⁺(aq) + 2e⁻ → Pb(s), with standard reduction potential E°(Pb²⁺/Pb) = −0.13 V (reduction half-reaction) 2. 2H⁺(aq) + 2e⁻ → H₂(g), with standard reduction potential E°(H⁺/H₂) = 0 V (Since it's hydrogen electrode)

Calculate the standard cell potential E°

The given cell reaction can be considered as the sum of the half-reactions, we can find the standard cell potential E° using the relation: $$ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} $$ From step 1 we have: E°(Pb²⁺/Pb) = -0.13 V (as reduction half-reaction) E°(H⁺/H₂) = 0 V (as oxidation half-reaction) Plugging these values into the equation: $$ E^\circ_{cell} = -0.13 - 0 $$ $$ E^\circ_{cell} = -0.13\, V $$ So, the standard cell potential for this reaction is -0.13 V. (b) Calculate the concentration of Pb²⁺ when the cell voltage is +0.210 V

Use the Nernst equation

We can now apply the Nernst equation to find the concentration of Pb²⁺ at cell voltage +0.210 V. The Nernst equation is given by: $$ E = E^\circ - \frac{RT}{nF} \ln Q $$ Here E = measured cell voltage = +0.210 V, E° = standard cell potential = -0.13 V, R = gas constant = 8.314 J/(mol K), T = temperature (assuming room temperature, 298 K) n = number of moles of electrons exchanged (in this case, 2), F = Faraday's constant = 96485 C/mol, Q = reaction quotient. Since we have the information of concentrations for other species involved: [H⁺] = 1.0 M and PH₂ = 1.0 atm, we can rewrite the Nernst equation as: $$ 0.210 = -0.13 - \frac{8.314 * 298}{2 * 96485} \ln\left(\frac{[\mathrm{Pb}^{2+}]}{(1)^2 * 1}\right) $$

Solve for [Pb²⁺]

Rearrange the Nernst equation and solve for [Pb²⁺]: $$ 0.340 = \frac{8.314 * 298}{2 * 96485} \ln\left([\mathrm{Pb}^{2+}]\right) $$ $$ \ln\left([\mathrm{Pb}^{2+}]\right) = \frac{2 * 96485}{8.314 * 298} * 0.340 $$ $$ [\mathrm{Pb}^{2+}] = e^{\frac{2 * 96485}{8.314 * 298} * 0.340} \approx 8.23 * 10^{-5}\,\text{M} $$ So, the concentration of Pb²⁺ ions is 8.23 * 10⁻⁵ M. (c) Calculate the solubility product Ksp of PbCl₂, given [Cl⁻] = 0.10 M

Write the solubility product expression

For the dissolution of PbCl₂, the reaction is $$\mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^-(a q) $$ The expression for the solubility product Ksp is given by: $$ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 $$

Calculate Ksp

Using the concentrations found in previous parts, we can find Ksp: $$ K_{sp} = (8.23 * 10^{-5})(0.10)^2 $$ $$ K_{sp} = 8.23 * 10^{-7} $$ Thus, the solubility product Ksp for PbCl₂ is 8.23 * 10⁻⁷.