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 \(\mathrm{B}\) 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 $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ 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 \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

Short answer

The solubility product constant, \(K_{sp}\), for MSO4 at 25°C can be calculated using the determined concentration of M2+ in solution B at equilibrium in the expression: \(K_{sp} = [M^{2+}][SO_4^{2-}]\).

Step-by-step solution

Determine the reduction potential of M2+/M for both compartments

Using the standard electrode potential, we can find the electrode potential for M2+ compartment and B compartment. For solution A, the standard reduction potential of M2+ is given as -0.31 V. The concentration of M2+ in solution A is 1.0 M. Therefore, using the Nernst equation, we can find the electrode potential of compartment A: \[E_A = E^0_A - \frac{0.0592}{n} \log{\frac{0.010}{1}}\] where \(E^0_A=-0.31 \mathrm{V}\), n=2, and the concentration of M2+ in solution A is 1M. For solution B, we are given the concentration of M(NO3)2 and Na2SO4 which are both 0.0100 moles in 1.0 L of water. At equilibrium, the reaction: \[M^{2+}(aq)+SO_4^{2-}(aq) \rightleftharpoons MSO_4(s)\] occurs. Let the concentration of M2+ remaining at equilibrium be x. Using an ICE table, we get: \[M^2+: 0.010 - x\] \[SO_4^{2-}: 0.010 - x\] Using the Nernst equation again, we can find the electrode potential of compartment B: \[E_B = E^0_B - \frac{0.0592}{n} \log{\frac{x^2}{(0.010 - x)^2}}\] where \(E^0_B=-0.31 \mathrm{V}\), n=2.

Calculate the cell potential and find the concentration of M2+ in solution B

As given, the cell potential is 0.44 V. The cell potential can be calculated by the difference in electrode potentials in compartments A and B: \[E_{cell} = E_B - E_A\] \[0.44 = -0.31 - \frac{0.0592}{2} \log{\frac{x^2}{(0.010 - x)^2}} + 0.31 - \frac{0.0592}{2} \log{\frac{0.010}{1}}\] Solve for x, which represents the concentration of M2+ in solution B at equilibrium.

Calculate the solubility product constant, Ksp

Now that we have determined the concentration of M2+ in solution B at equilibrium, we can calculate the solubility product constant, Ksp, for the reaction: \[M^{2+}(aq)+SO_4^{2-}(aq) \rightleftharpoons MSO_4(s)\] At equilibrium, the concentrations of M2+ and SO4^2- are both x (from Step 2). The Ksp expression for this reaction is: \[K_{sp} = [M^{2+}][SO_4^{2-}]\] Substitute the concentrations obtained in Step 2, and calculate Ksp for MSO4 at 25°C.

Key concepts

Nernst Equation

The Nernst Equation plays a fundamental role in electrochemistry by allowing us to calculate the electrode potential of an electrochemical cell under non-standard conditions. It relates the concentration of ions in solution to the cell potential and is key to understanding concentration cells.
The general form of the Nernst Equation is:

• \( E = E^0 - \frac{0.0592}{n} \log{Q} \)

Here, \( E \) represents the electrode potential, \( E^0 \) is the standard electrode potential, \( n \) is the number of electrons transferred in the redox reaction, and \( Q \) is the reaction quotient.
In the given exercise, the Nernst Equation is used to determine the potential difference between two compartments of a concentration cell, thereby assessing the driving force behind the redox reaction. The cell potential is influenced by the concentration of \( M^{2+} \) ions, showing that small changes in concentration can greatly affect the electrode potential.

Solubility Product Constant

The Solubility Product Constant, denoted as \( K_{sp} \), is vital for understanding the solubility of sparingly soluble salts. It measures the extent to which a compound can dissolve in solution and is crucial for predicting precipitate formation when different ions are mixed together.
In the formula:

• \( K_{sp} = [M^{2+}][SO_4^{2-}] \)

\( K_{sp} \) describes the equilibrium constants of the dissolution of salts like \( MSO_4 \), given as \( M^{2+} \) and \( SO_4^{2-} \). For the concentration cell being discussed, \( K_{sp} \) is determined using the concentrations of ions that remain in solution after equilibrium has been reached.
By calculating \( K_{sp} \), we can infer the stability of the solid form of \( MSO_4 \) in water, helping us understand the chemical environment within the cell and assess the likelihood of salts precipitating out of solution.

Electrode Potential

Electrode potential, also known as standard reduction potential, is the measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is represented by \( E^0 \), and knowing it is critical for understanding oxidative and reductive processes in electrochemistry.
In our exercise, the electrode potential for the reduction reaction \( M^{2+} + 2e^- \rightarrow M \) is provided as \(-0.31 V\). This value indicates the ease with which \( M^{2+} \) ions can gain electrons and be reduced to metal \( M \).
The electrode potential plays a key role in determining the overall voltage of an electrochemical cell. A positive cell potential means the redox reaction is favorable. For concentration cells, like the one considered, the difference in potential between solutions A and B is crucial, as it drives the flow of electrons and explains the cell's operating voltage.

Redox Reaction

A Redox Reaction involves the transfer of electrons between two species, where one undergoes oxidation by losing electrons, and the other undergoes reduction by gaining electrons. These reactions are fundamental to electrochemistry and energy conversion in cells.
In the exercise under consideration, the reaction \( M^{2+} + 2e^- \rightarrow M \) takes place. Here, \( M^{2+} \) ions in solution are reduced by gaining electrons and becoming the metal \( M \), thus exemplifying a redox process.
Understanding redox reactions involves recognizing how the redistribution of electrons contributes to energy changes within the cell. In a concentration cell, these changes in energy are harnessed to do electrical work, as electrons are transferred from areas of higher concentration to lower concentration, effectively generating an electric current.