Problem 72 - Chemistry The Central Science

Question

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (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? (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 \(\mathrm{PbSO}_{4} ?\)

Step-by-Step Solution

Answer

The concentration of Pb²⁺ in the anode half-cell is approximately 0.015 M, and the Ksp of PbSO₄ is 0.015.

Step 1: Identify the two half-reactions

First, let's identify the two half-reactions occurring in the voltaic cell: 1. Cathode (reduction): Sn2+(aq) + 2e- → Sn(s) 2. Anode (oxidation): Pb(s) → Pb2+(aq) + 2e-

Step 2: Look up the standard reduction potentials

Now, we need to look up the standard reduction potentials for Sn2+(aq) and Pb(s). From standard reduction potential tables, we find: Eº(Sn2+/Sn) = -0.14 V Eº(Pb2+/Pb) = -0.13 V

Step 3: Calculate the overall cell potential

Using the Nernst equation, we can calculate the cell potential: Ecell = Eº(cell) + \(\frac{RT}{nF}\) * ln(Q) We know that Ecell is given as +0.22 V. The number of electrons transferred in the reaction, n, is 2. The reaction quotient Q is \(\frac{[Pb^{2+}]}{[Sn^{2+}]}\). We are also given that the concentration of Sn2+ is 1.00 M. Eº(cell) = Eº(cathode) - Eº(anode) = (-0.14 V) - (-0.13 V) = -0.01 V Now, we can plug in the values and solve for [Pb2+].

Step 4: Solve for the concentration of Pb2+

Plugging in the values in the Nernst equation: 0.22 V = -0.01 V + \(\frac{(RT)(2.303)}{2F}\) * log\(\frac{[Pb^{2+}]}{1}\) 0.23 V = \(\frac{(RT)(2.303)}{2F}\) * log([Pb2+]) Now, we can solve for [Pb2+]: log([Pb2+]) = \(\frac{0.23(2F)}{(2.303)(RT)}\) [Pb2+] = 10^(\(\frac{0.23(2F)}{(2.303)(RT)}\)) ≈ 0.015 M

Step 5: Find the Ksp value of PbSO4

Using the information provided in part (b), we know that the concentration of SO42- is 1.00 M in equilibrium with PbSO4(s). The Ksp expression for PbSO4 is: Ksp = [Pb2+][SO42-] Using the concentration of [Pb2+] we found in step 4 and the given [SO42-]: Ksp = (0.015)(1.00) = 0.015 Therefore, the Ksp of PbSO4 is 0.015.

Key Concepts

Nernst Equation

The Nernst equation is a powerful tool in electrochemistry used for determining the cell potential of an electrochemical cell that operates under non-standard conditions. The standard conditions mean reacting concentrations of 1 M, a pressure of 1 atm, and a specified temperature, often 25°C.
The Nernst equation can be written as:
\[ E_{cell} = E^{\circ}_{cell} + \frac{RT}{nF} \ln(Q) \]In this equation:

• \(E_{cell}\) is the cell potential at non-standard conditions.
• \(E^{\circ}_{cell}\) is the standard cell potential, indicating the potential difference at standard conditions.
• \(R\) is the universal gas constant (8.314 J/mol K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the electrochemical reaction.
• \(F\) is the Faraday constant (approximately 96,485 C/mol).
• \(Q\) is the reaction quotient, representing the concentrations or activities of the products and reactants.

Understanding this equation is essential as it shows how the potential in an electrochemical cell can shift based on concentration changes.

Standard Reduction Potentials

Standard reduction potentials are a way to express the tendency of a chemical species to acquire electrons and be reduced. These values are measured in volts and are tabulated for various half-reactions, providing a straightforward way to predict the voltage of galvanic cells.
Every half-cell reaction has a standard reduction potential, \(E^{\circ}\), which is measured under standard conditions. For example:

• For \( \text{Sn}^{2+} + 2e^- \longrightarrow \text{Sn} \), the standard reduction potential \(E^{\circ} = -0.14 \text{ V}\).
• For \( \text{Pb}^{2+} + 2e^- \longrightarrow \text{Pb} \), \(E^{\circ} = -0.13 \text{ V}\).

These values help determine the direction of electron flow in a cell. A more positive \(E^{\circ}\) means a greater tendency for the species to be reduced.
To find the cell potential, you subtract the potential for the anode (oxidation) from the cathode (reduction):
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]This calculation assists in understanding whether the reaction will proceed spontaneously, which is indicated by a positive cell potential.

Equilibrium Constant

In complete chemical equilibrium, certain reactions maintain a balance between reactants and products. For solids like salts, this balance is represented by the equilibrium constant, \(K_{sp}\), specifically for sparingly soluble salts.
The expression for \(K_{sp}\) involves the molar concentrations of the constituent ions in a saturated solution. For example, the dissolution of \(\text{PbSO}_4\) can be represented as:
\[ \text{PbSO}_4(s) \leftrightarrow \text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]The corresponding solubility product expression would be:
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]In this expression, each term represents the concentration of that ion at equilibrium.
For the reaction given in the exercise, we calculated \([\text{Pb}^{2+}]\) to be 0.015 M and with \([\text{SO}_4^{2-}] = 1.00 \text{ M}\), the \(K_{sp}\) can be determined as:
\[ K_{sp} = (0.015)(1.00) = 0.015 \]This value provides crucial insights into the solubility and stability of salts in aqueous solutions.