Question

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). 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 $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). 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 \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

Short answer

The solubility product \(K_{sp}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\) is \(3.31 \times 10^{-5}\).

Step-by-step solution

Write the Nernst equation for the given cell reaction

The Nernst equation is given by: \(E=E^{∘} - \frac{0.0591}{n} \log Q\) Since no other redox process occurs in the cell other than \(\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}\), we can use the standard reduction potential for the mentioned process (-0.31 V), and n=2 since, two electrons are exchanged. Furthermore, now the potential of the cell is given for the equilibrium (0.44 V). Fast Equilibrium is established in the compartment containing solution B, which gives: \(E = E^{∘} - \frac{0.0591}{2} \log \frac{\mathrm{M}^{2+}_B \mathrm{SO}_{4}^{2-}_B}{[\mathrm{M}^{2+}_{B_0}][\mathrm{SO}_4^{2-}_{B_0}] }\) Where 𝑀𝑏^(2^+) , 𝑆𝑂_𝑏^(4^–) are equilibrium concentrations and 𝑀^(2+)_\(𝐵_0\) , 𝑆𝑂_4^(2–)_\(𝐵_0\) are initial concentrations.

Identify the initial concentrations of ions in compartment B

Initially, 0.0100 mol of M(NO3)2 and 0.0100 mol of Na2SO4 are dissolved in 1 L of the solution. Therefore, the initial concentrations are: \([\mathrm{M}^{2+}_{B_0}] = \frac{0.0100\,\text{mol}}{1.00\,\text{L}}= 0.0100\,\text{M}\) \([\mathrm{SO}_4^{2-}_{B_0}] = \frac{0.0100\,\text{mol}}{1.00\,\text{L}}= 0.0100\,\text{M}\)

Insert the known values into the Nernst equation and solve for the equilibrium concentration of M^2+ in compartment B

Plug the values into the Nernst equation: \(0.44 = (-0.31) - \frac{0.0591}{2} \log \frac{\mathrm{M}^{2+}_B \mathrm{SO}_4^{2-}_B}{(0.0100)(0.0100)}\) Solving for \(\mathrm{M}^{2+}_B\) we get: \(\mathrm{M}^{2+}_B = 3.31 \times 10^{-3}\,\mathrm{M}\)

Calculate the value of K_sp for MSO4

Since the reaction is at equilibrium, the solubility product K_sp can be calculated as: \(K_{sp} = [\mathrm{M}^{2+}][\mathrm{SO}_{4}^{2-}]\) Since the concentration of sulfate ions remains constant, we can assume that \([\mathrm{SO}_{4}^{2-}_B] = [\mathrm{SO}_4^{2-}_{B_0}]\): \(K_{sp} = (3.31 \times 10^{-3})(0.0100)\) \(K_{sp} = 3.31 \times 10^{-5}\) Thus, the solubility product \(K_{sp}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\) is \(3.31 \times 10^{-5}\).

Key concepts

Nernst Equation

The Nernst equation is essential for understanding electrochemistry and is particularly relevant when studying concentration cells. It provides a way to calculate the electrochemical cell's potential at any given concentration of reactants and products. The formula is:

equation here: \( E = E^{∘} - (RT/nF) \times \text{ln} Q \)

where \(E\) is the cell potential at non-standard conditions, \(E^{∘}\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. For practical calculations at 25°C, the equation simplifies to:

equation here: \( E = E^{∘} - \frac{0.0591}{n} \times \text{log} Q \)

This equation is used to calculate the value of \(Q\) when the cell potential at equilibrium and standard reduction potential are known, as shown in our concentration cell example. It helps students understand how variations in ion concentration affect the voltage of a cell and is a fundamental principle in the study of electrochemistry.

Solubility Product Constant (Ksp)

The solubility product constant, or \(K_{sp}\), represents the extent to which a compound dissolves in water. It's a special type of equilibrium constant for the dissolution of a sparingly soluble ionic compound.

For a general salt, AB, which dissociates into A+ and B- ions:

equation here: \( AB(s) \rightleftharpoons A^{+}(aq) + B^{-}(aq) \)

The expression for the solubility product constant is given by:

equation here: \( K_{sp} = [A^{+}][B^{-}] \)

In the case of our concentration cell example, the salt is \(MSO_{4}\). It is essential to calculate \(K_{sp}\) to understand the solubility of \(MSO_{4}\) and predict whether it will precipitate under certain conditions. The concept of \(K_{sp}\) is a critical piece for students learning about the solubility of compounds and the factors that affect it, including ion concentration and temperature.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants or products in a system at equilibrium. In the context of a chemical reaction, it's the point at which the rate of the forward reaction is equal to the rate of the reverse reaction, so the concentrations of the reactants and products remain constant over time.

When a concentration cell reaches equilibrium, as with the reaction between \(M^{2+}\) and \(SO_4^{2-}\) to form \(MSO_4\), the Nernst equation relates these equilibrium concentrations to the cell potential. This allows us to express the solubility product constant \(K_{sp}\) in terms of these concentrations:

equation here: \( K_{sp} = [M^{2+}][SO_4^{2-}] \)

Using the values derived from the Nernst equation, we can calculate the equilibrium concentrations of ions involved in the reaction. Understanding equilibrium concentration is vital for predicting the behavior of ionic compounds in solution and is a foundational topic in chemistry education.

Standard Reduction Potential

The standard reduction potential (\(E^{∘}\)) is a measure of the tendency of a chemical species to acquire electrons and be reduced. It’s denoted in volts and is determined under standard conditions, which includes all solutes at a 1 M concentration and all gases at 1 atmosphere pressure.

Each half-cell in an electrochemical cell has its standard reduction potential, and the difference between the potentials of the two half-cells gives the standard cell potential \(E^{∘}_{cell}\):

equation here: \(E^{∘}_{cell} = E^{∘}_{cathode} - E^{∘}_{anode} \)

In our example, the given standard reduction potential for the cell reaction is -0.31 V. This value is a key factor in calculating cell potential using the Nernst equation and provides insight into the spontaneity of redox reactions. It is an essential concept in electrochemistry related to the prediction and understanding of cell voltage and chemical reactivity.