Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

Step by step solution

01

Write down the Nernst equation for cell potential

The Nernst equation relates the cell potential to the reduction potentials of the half-reactions and the concentrations of the involved species. The Nernst equation is given by: $$ E=E^{0}-\frac{RT}{nF} \ln Q $$ where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the absolute temperature in Kelvin, n is the number of moles of electrons transferred per mole of product, F is the Faraday's constant, and Q is the reaction quotient.

02

Calculate the temperature in Kelvin

We need to convert the temperature given (25°C) in Celsius to Kelvin. To do so, we use the following conversion: $$ T(K) = T(°C) + 273.15 $$ Substitute the given temperature: $$ T(K) = 25 + 273.15 = 298.15 K $$

03

Determine the half-reaction that occurs in the cell

We know that no other redox process occurs in the cell other than the reduction of M²⁺ to metal M. We have the half-reaction: $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$

04

Calculate the standard cell potential (E⁰)

Using the standard reduction potential of M²⁺/M, we can calculate the standard cell potential: $$ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} $$ Since both electrodes are made of metal M, E⁰_cathode and E⁰_anode are equal. As a result: $$ E^{\circ} = 0 $$

05

Calculate the reaction quotient (Q)

The reaction quotient Q for the half-reaction is represented by the ratio of the concentrations of the products and reactants raised to the power of their stoichiometric coefficients: $$ Q = \frac{[\mathrm{M}^{2+}]^2}{[A]^{2}[\mathrm{SO}_4^{2-}]} $$ We know the initial concentrations of M²⁺ and SO₄²⁻ ions in the cell. Since the volume of solution B is 1.0 L, the initial concentration of M²⁺ ions is 0.0100 M and the initial concentration of SO₄²⁻ ions is 0.0100 M as well. Substituting these values into the equation for Q gives: $$ Q = \frac{(0.0100)^2}{(1)^2(0.0100)} = 1 $$

06

Use the Nernst equation to calculate the value of Ksp

Now that we have all the required values, we can substitute them into the Nernst equation and solve for Ksp. The Nernst equation for this exercise is: $$ E=E^{0}-\frac{RT}{nF} \ln Q $$ Substitute the values into the equation: E = 0.44 V (given) E⁰ = 0 (calculated in Step 4) R = 8.314 J/(mol·K) T = 298.15 K (calculated in Step 2) n = 2 (from the reduction half-reaction) F = 96485 C/mol Q = 1 (calculated in Step 5) Solve for Ksp: $$ 0.44 = 0 - \frac{(8.314)(298.15)}{(2)(96485)} \ln K_{sp} $$ Solving for Ksp gives: $$ K_{sp} = \exp \left( \frac{(2)(96485)(0.44)}{(8.314)(298.15)} \right) = \exp(12.292) \approx 2.17 \times 10^{-6} $$ Thus, the solubility product constant Ksp for MSO₄ at 25°C is approximately 2.17 × 10⁻⁶ at 25°C.

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