Question

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 \(25^{\circ} \mathrm{C}\)

Short answer

The solubility product constant $K_{sp}$ for $\mathrm{MSO}_4$ at 25°C is approximately $2.17 \times 10^{-6}$.

Step-by-step solution

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.

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: $$ T(K) = 25 + 273.15 = 298.15 K $$

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: $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$

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: $$ E^{\circ} = 0 $$

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: $$ Q = \frac{(0.0100)^2}{(1)^2(0.0100)} = 1 $$

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.

Key concepts

Standard Cell Potential

The standard cell potential, symbolized as \(E^0\), is a measure of the voltage, or electrical potential, provided by a cell under standard conditions, which typically include a temperature of 298 K (25°C), a pressure of 1 atm, and 1 M concentration for all ions involved.

The concept is crucial in electrochemistry as it tells us the potential difference between the cathode and anode at standard state.

• **Cathode**: The electrode where reduction takes place.
• **Anode**: The electrode where oxidation occurs.

Since the standard cell potential of a concentration cell with identical electrodes, like in our exercise, is equal to zero, the system relies on concentration differences to generate a potential.

This means that for a concentration cell of this type, the observed voltage is purely due to differences in concentration, not the inherent potential of the electrodes.
This deviation from standard conditions is part of what we're measuring with the Nernst equation.

Reaction Quotient

The reaction quotient, denoted as **Q**, describes the ratio of the concentrations of the products to the reactants at any point in time and is not necessarily at equilibrium.

It follows the same form as the equilibrium constant but reflects real-time conditions, giving insight into how far a reaction has gone toward completion.

• For a general reaction, the formula is given by:\[Q = \frac{[products]^{coefficients}}{[reactants]^{coefficients}}\]
• If Q < K (equilibrium constant), the forward reaction is favored.
• If Q > K, the reverse reaction is favored.
• If Q = K, the system is at equilibrium.

In the given exercise, using the reaction quotient allows us to substitute into the Nernst equation to assess how concentrations influence the cell potential.

It's particularly useful for predicting whether a spontaneous reaction is likely to occur or not based on current conditions.

Solubility Product Constant

The solubility product constant, often noted as \(K_{sp}\), is a special type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds.

• It describes the maximum amount of solid that can dissolve in solution before it becomes saturated.
• The smaller the \(K_{sp}\), the less soluble the compound in water.

In context with our problem, upon the settlement of equilibrium during our experiment, the dissolution and precipitation are balanced, allowing us to calculate \(K_{sp}\).

This requires the measured cell potential, \(E = 0.44\text{ V}\), and utilizes the Nernst equation.
Substituting our gathered data into this equation, we establish the figure for \(K_{sp}\) of \(MSO_4\).
This constant is important for understanding the extent of a reaction in solutions and can be crucial when predicting whether precipitation will occur in given conditions.