Electrochemistry and Solubility Product (Ksp) Calculations

Electrochemical Cells and Solubility Product Calculations

Electrochemical Cells and Standard Cell Potential

Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. The standard cell potential ($E^\circ_{cell}$) is a measure of the driving force behind the electrochemical reaction under standard conditions (1 M, 1 atm, 25°C).

• Half-Cell Reactions: Each half-cell contains a redox couple, and the overall cell reaction is the sum of the two half-reactions.

• Standard Cell Potential ($E^\circ_{cell}$): The difference in potential between the two half-cells when all species are in their standard states.

• Example: In the problem, the following half-reactions are given:

  • $2\text{Cu}(s) \rightarrow 2\text{Cu}^+(aq) + 2e^-$

  • $\text{Cu}_2\text{S}(s) + 2e^- \rightarrow 2\text{Cu}(s) + \text{S}^{2-}(aq)$

  The net reaction is: $\text{Cu}_2\text{S}(s) \rightleftharpoons 2\text{Cu}^+(aq) + \text{S}^{2-}(aq)$

Solubility Product Constant ($K_{sp}$)

The solubility product constant ($K_{sp}$) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It represents the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation.

• Definition: For $\text{Cu}_2\text{S}(s) \rightleftharpoons 2\text{Cu}^+(aq) + \text{S}^{2-}(aq)$, the solubility product is: $K_{sp} = [\text{Cu}^+]^2[\text{S}^{2-}]$

• Application: $K_{sp}$ can be determined using electrochemical data by relating the cell potential to the equilibrium constant.

Relationship Between Cell Potential and Equilibrium Constant

The standard cell potential is related to the equilibrium constant ($K$) of the cell reaction by the Nernst equation at standard conditions:

• Equation: $E^\circ_{cell} = \frac{0.0592}{n} \log K$ or $\Delta G^\circ = -nFE^\circ_{cell}$ and $\Delta G^\circ = -RT \ln K$ Therefore, $nFE^\circ_{cell} = RT \ln K$ $\log K = \frac{nE^\circ_{cell}}{0.0592}$ (at 25°C)

• Where:

  • $n$ = number of moles of electrons transferred

  • $F$ = Faraday's constant ($96485$ C/mol)

  • $R$ = gas constant ($8.314$ J/mol·K)

  • $T$ = temperature in Kelvin (usually $298$ K)

• Example Calculation:

  • Given $E^\circ_{cell} = -1.41$ V, $n = 2$

  • $\log K_{sp} = \frac{2 \times -1.41}{0.0592} = -47.6$

  • $K_{sp} = 10^{-47.6} \approx 2.5 \times 10^{-48}$

  • However, the answer choices suggest a positive exponent, indicating a possible sign error in the cell potential or reaction direction. Always ensure the cell is written as spontaneous (positive $E^\circ_{cell}$) for $K > 1$.

Worked Example: Calculating $K_{sp}$ from Cell Potential

1. Write the net ionic equation: $\text{Cu}_2\text{S}(s) \rightleftharpoons 2\text{Cu}^+(aq) + \text{S}^{2-}(aq)$

2. Identify $n$ (number of electrons transferred): $n = 2$

3. Use the Nernst equation at standard conditions: $E^\circ_{cell} = \frac{0.0592}{n} \log K_{sp}$ Rearrange: $\log K_{sp} = \frac{nE^\circ_{cell}}{0.0592}$

4. Plug in values: $\log K_{sp} = \frac{2 \times -1.41}{0.0592} = -47.6$ $K_{sp} = 10^{-47.6} \approx 2.5 \times 10^{-48}$

5. Compare to answer choices: The closest is $2.3 \times 10^{-48}$ (option e).

Summary Table: Key Relationships

Quantity	Symbol	Equation	Description
Standard Cell Potential	$E^\circ_{cell}$	$E^\circ_{cell} = \frac{0.0592}{n} \log K$	Relates cell potential to equilibrium constant
Solubility Product	$K_{sp}$	$K_{sp} = [\text{Cu}^+]^2[\text{S}^{2-}]$	For $\text{Cu}_2\text{S}$ dissolution
Gibbs Free Energy	$\Delta G^\circ$	$\Delta G^\circ = -nFE^\circ_{cell}$	Relates free energy to cell potential

Additional info: The sign of $E^\circ_{cell}$ determines whether the reaction is spontaneous ($K > 1$) or nonspontaneous ($K < 1$). In this problem, the negative cell potential indicates a nonspontaneous reaction as written, so $K_{sp}$ is very small, consistent with the low solubility of $\text{Cu}_2\text{S}$.