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 \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) \(\mathrm{mol} \mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \operatorname{MSO}_{4}(s)$$ 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 \(K_{\mathrm{sp}}\) for. \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

Short answer

The value of \(K_{sp}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\) is \(2.09 \times 10^{-4}\).

Step-by-step solution

Write the cell's half-reactions

The cell's half reactions are given as: \[ \begin{aligned} \text{Left half-cell:} \quad \mathrm{M}^{2+} + 2 \mathrm{e}^{-} &\longrightarrow \mathrm{M} \\ \text{Right half-cell:} \quad \mathrm{M} &\longrightarrow \mathrm{M}^{2+} + 2 \mathrm{e}^{-} \end{aligned} \] This represents a concentration cell, as both the reduction and oxidation processes occur at the metal M.

Express cell potential in terms of equilibrium constants

According to the Nernst equation, the cell potential is given by: \[ E_{cell} = E^0 - \frac{RT}{2F} \ln \frac{[\mathrm{M}^{2+}_{right}]}{[\mathrm{M}^{2+}_{left}]} \] For a concentration cell, the standard potential, \(E^0=0\). So, the equation becomes: \[ E_{cell} = - \frac{RT}{2F} \ln \frac{[\mathrm{M}^{2+}_{right}]}{[\mathrm{M}^{2+}_{left}]} \]

Find equilibrium concentrations

We are given that the equilibrium cell potential \(E_{cell} = +0.44 \mathrm{~V}\) and the left compartment's concentration of \(\mathrm{M}^{2+}\) is \(1.0 \mathrm{~M}\). Using the Nernst equation, we can find the concentration of \(\mathrm{M}^{2+}(aq)\) in the right cell compartment: \[ 0.44 = - \frac{(8.314 \mathrm{~J\,mol^{-1}K^{-1}})(298 \mathrm{~K})}{2(96485 \mathrm{~C\,mol^{-1}})} \ln \frac{ [\mathrm{M}^{2+}_{right}]}{[\mathrm{M}^{2+}_{left}]} \] Now, solve for \([\mathrm{M}^{2+}_{right}]\): \[ [\mathrm{M}^{2+}_{right}] = 1.0 \mathrm{~M} \cdot e^{\frac{2(96485 \mathrm{~C\,mol^{-1}})(0.44 \mathrm{~V})}{(8.314 \mathrm{~J\,mol^{-1}K^{-1}})(298 \mathrm{~K})}} = 0.0209 \mathrm{~M} \]

Find the \(K_{sp}\) expression and input the equilibrium concentrations

Given the precipitation reaction and the standard reduction potential: \[ \mathrm{M}^{2+}(aq) +\mathrm{SO}_{4}^{2-}(aq) \rightleftharpoons \mathrm{MSO}_{4}(s) \] The solubility product constant, \(K_{sp}\), is: \[ K_{sp} = [\mathrm{M}^{2+}][\mathrm{SO}_{4}^{2-}] \] Now, find the equilibrium concentration of \(\mathrm{SO}_{4}^{2-}(aq)\) by considering that initially there are \(0.0100 \,\mathrm{mol}\) of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\). Since \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) dissociates completely, it will give \(0.0100 \,\mathrm{mol} \, \mathrm{SO}_{4}^{2-}\). Thus, \([\mathrm{SO}_{4}^{2-}]_{eq} = 0.0100 \,\mathrm{M}\).

Calculate the value of \(K_{sp}\)

Now that we have the equilibrium concentrations of \(\mathrm{M}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\), we can find the value of \(K_{sp}\): \[ K_{sp} = (0.0209 \mathrm{~M})(0.0100 \mathrm{~M}) = 2.09 \times 10^{-4} \] The value of \(K_{sp}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\) is \(2.09 \times 10^{-4}\).

Key concepts

Concentration Cell

A concentration cell is a type of electrochemical cell where both electrodes are made of the same material, but they are immersed in solutions of different concentrations. This difference in concentration creates a potential difference, or cell potential, between the two electrodes. The driving force behind the cell potential in a concentration cell is the tendency for ions to move from an area of higher concentration to one of lower concentration in order to reach equilibrium across the two compartments.

In our textbook example, both electrodes are made of metal M, but they are in different concentrations of M²⁺ ions in their respective solutions, leading to the establishment of a cell potential. As no other redox process occurs, analyzing the concentration cell provides insights into fundamental electrochemistry and allows for the calculation of various important thermodynamic quantities, such as the solubility product constant (K_sp).

Nernst Equation

The Nernst equation plays a crucial role in electrochemistry as it relates the cell potential to the concentrations of the reactants and products at any point in time. It is especially useful in non-standard conditions where concentrations might not be at standard state values. The equation is given as:

\[ E = E^0 - \frac{RT}{nF} \ln\left(\frac{[\text{products}]}{[\text{reactants}]^n}\right) \]

In context with the concentration cell, it allows us to calculate the cell potential given the known concentrations of the metal ions involved. The value of \(E^0\) for a concentration cell becomes 0 because both the electrodes are of the same material. Thus, any potential measured is solely due to the concentration gradient between the two solutions.

Cell Potential

Cell potential, often symbolized as \(E_{cell}\), is the measure of the electrical potential difference between two half-cells in an electrochemical cell. It is a critical concept that reflects the cell's ability to produce an electric current when a circuit is completed. The cell potential is influenced by factors such as the nature of the electrodes, the concentration of ions in solution, and temperature.

In the given exercise, a positive cell potential of +0.44 V indicates that the cell is capable of doing work, as the electrons are spontaneously moving from the higher concentration side to the lower concentration side. The cell potential, when at standard conditions (1 M concentration and 25°C), is referred to as the standard cell potential, which in the example is used to calculate the standard reduction potential for the reduction of M²⁺.

Equilibrium Constant

The equilibrium constant, denoted K, is a number that expresses the ratio of the concentrations of the products to the reactants for a reversible chemical reaction at equilibrium. In the context of a solubility equilibrium, which involves a solid dissolving in a liquid to form a saturated solution, the equilibrium constant is known as the solubility product constant, K_sp. This particular constant represents the level at which a compound dissolves in solution and is specific to each solute-solvent combination at a given temperature.

By measuring cell potential and relating it to the Nernst equation as seen in our textbook example, we can determine the equilibrium constant without directly measuring solute concentrations. This calculation shows us how electrochemical methods can reveal information about chemical equilibria.

Standard Reduction Potential

The standard reduction potential, or standard electrode potential, is a measure of the tendency of a chemical species to be reduced, and is represented by \(E^0\) in electrochemistry. It's determined under standard conditions, which are 25°C, 1 atmosphere pressure, and 1 M concentration of each ion. These values are tabulated for half-reactions in which electrons are accepted by a chemical species (reduction).

In our exercise, the standard reduction potential for the process \(\mathrm{M}^{2+} + 2 e^{-} \longrightarrow \mathrm{M}\) is given as -0.31 V. It signifies the ease with which the metal ion M²⁺ can be reduced to metal M. This value is used to calculate other electrochemical values, such as the cell potential, when conditions vary from the standard state. Understanding standard reduction potentials allows us to predict the direction of electron flow and the feasibility of redox reactions.