Question

What is \(E^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction? $$ \begin{array}{c} \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s) \\ K_{\mathrm{sp}} \text { for } \mathrm{BaSO}_{4} \text { is } 1.1 \times 10^{-10} \end{array} $$

Short answer

Answer: The standard reduction potential (E°) cannot be determined for the given precipitation reaction, as it requires information from a redox reaction, which is not available in this case.

Step-by-step solution

Write down the Nernst equation

The Nernst equation relates the reduction potential of a redox reaction to the standard reduction potential, temperature, and concentrations of the species involved: $$ E = E^\circ - \frac{RT}{nF} \ln Q $$ where, E = Reduction potential of the reaction E° = Standard reduction potential R = Gas constant (8.314 J/mol·K) T = Temperature in Kelvin (25°C = 298.15 K) n = Number of electrons transferred in the reaction F = Faraday's constant (96,485 C/mol) Q = Reaction quotient

Determine the number of electrons transferred (n)

Since there is no redox reaction taking place in the equation, we can assume that the value of n is 1. This simplifies the Nernst equation.

Calculate the Reaction Quotient (Q) using Ksp

The reaction quotient (Q) for the given reaction can be expressed as: $$ Q = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}] $$ As the solubility product constant for BaSO₄ is given, we can write: $$ K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}] $$ So, the reaction quotient Q is equal to the solubility product constant Ksp: $$ Q = 1.1 × 10^{-10} $$

Substitute the values and solve for E°

Now we can substitute the values for T, n, F, and Q in the Nernst equation and solve for the standard reduction potential E°: $$ E = E^\circ - \frac{8.314\,\mathrm{J/mol\cdot K} \times 298.15\,\mathrm{K}}{1 \times 96,485\,\mathrm{C/mol}} \ln (1.1 \times 10^{-10}) $$ Rearrange the equation to solve for E°: $$ E^\circ = E + \frac{8.314\,\mathrm{J/mol\cdot K} \times 298.15\,\mathrm{K}}{1 \times 96,485\,\mathrm{C/mol}} \ln (1.1 \times 10^{-10}) $$ E° cannot be determined from the information provided, since we cannot determine the E value from a precipitation reaction. In redox reactions, E can be determined by the balance between reduction potential and oxidation potential, but in precipitation reactions these values are not available.