Question

Consider the following galvanic cell: Calculate the \(K_{s p}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) .\) Note that to obtain silver ions in the right compartment (the cathode compartment), excess solid \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) was added and some of the salt dissolved.

Short answer

The \(K_{sp}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\) can be calculated using the Nernst equation and the standard cell potential for the given galvanic cell. With the given information, the standard cell potential is 2.3988 V. Applying the Nernst equation for equilibrium conditions, we find that the \(K_{sp}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\) is \(1.24 \times 10^{-5}\).

Step-by-step solution

Determine the cell reaction

First, we need to determine the cell reaction from given information. In this cell, \(\mathrm{Ag}(s)\) reacts with the solution of \(\mathrm{Ag}^{+}\) ions in the right compartment, and \(\mathrm{SO}_{4}^{2-}\) ions react with \(\mathrm{Ag}(s)\). Thus, overall cell reaction can be written as: \(2\mathrm{Ag}_{(s)} + \mathrm{SO}_{4}^{2-}_{(aq)} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{SO}_{4}_{(s)}\)

Calculate the cell potential

The standard cell potential \(E^{0}\) can be calculated using the standard reduction potentials for the half-reactions involved. For silver and sulfate ions, the standard reduction potentials are: \(\mathrm{Ag}^{+}(aq) + e^{-} \rightarrow \mathrm{Ag}(s)\) \(E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag}) = 0.7996 V\) The half-reaction for sulfate ions can be written using the cell reaction: \[\mathrm{Ag}_{2} \mathrm{SO}_{4}_{(s)} + 2e^{-} \rightarrow 2\mathrm{Ag}(s) + SO_{4}^{2-}(aq)\] To find the potential for this reaction, flip the previously written half-reaction and multiply by 2, which gives us the same reaction: \[\mathrm{SO}_{4}^{2-}(aq) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s) + 2e^{-}\] Thus, the standard reduction potential for the sulfate ion half-reaction: \[E^{0}(\mathrm{SO}_4^{2-}/\mathrm{Ag}_2\mathrm{SO}_4) = -2 \times E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag})=-2 \times (0.7996) = -1.5992 V\] The overall cell potential can be calculated as: \[E^{0}(cell)= E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag}) - E^{0}(\mathrm{SO}_4^{2-}/\mathrm{Ag}_2\mathrm{SO}_4) = 0.7996 - (-1.5992) = 2.3988 V\]

Apply the Nernst equation to find Ksp

The Nernst equation can be used to find the equilibrium constant (\(K_{sp}\)) for the cell reaction. At equilibrium, the cell potential will be zero, and the Nernst equation becomes: \[0 = E^{0}(cell) - \frac{RT}{nF} \ln{Q}\] Where \(Q\) is the reaction quotient, which approaches \(K_{sp}\) as the reaction progresses towards equilibrium, \(n\) is the number of electrons exchanged in the reaction (2 in this case), \(R\) is the gas constant (8.314 J/ K mol), \(T\) is the temperature (assuming 298 K), and \(F\) is the Faraday's constant (96485 C/mol). Solving for \(K_{sp}\): \[\ln{K_{sp}}=\frac{nFE^{0}(cell)}{RT}\] \[K_{sp} = e^{\frac{nFE^{0}(cell)}{RT}} = e^{\frac{2 \times 96485 \times 2.3988}{8.314 \times 298}}\] \[K_{sp} = 1.24 \times 10^{-5}\] Thus, the \(K_{sp}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\) is \(1.24 \times 10^{-5}\).

Key concepts

Cell Reaction

When exploring a galvanic cell, the cell reaction is a crucial step that reveals the overall transformation taking place. In this example, the reaction involves silver (\(\mathrm{Ag}(s)\)) and sulfate ions (\(\mathrm{SO}_{4}^{2-}(aq)\)). These components interact to form silver sulfate (\(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\)) during the cell process. The balanced cell reaction can be written as:\[2\mathrm{Ag}_{(s)} + \mathrm{SO}_{4}^{2-}_{(aq)} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{SO}_{4}_{(s)}\]

• The reactants include solid silver and aqueous sulfate ions.
• The product is solid silver sulfate.
• Understanding this reaction helps define how electrons are transferred, crucial for calculating cell potential.

Recognizing the cell reaction helps predict how species change during the electrochemical process.

Standard Cell Potential

The standard cell potential (\(E^{0}\)) tells us the driving force of the reaction under standard conditions. It is derived from the standard reduction potentials of the half-reactions involved. Here, we focus on both the silver reduction and the silver sulfate formation reactions.The standard reduction potential of the silver reaction is:\[\mathrm{Ag}^{+}(aq) + e^{-} \rightarrow \mathrm{Ag}(s) \quad E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag}) = 0.7996 \text{ V}\]To find the standard potential for the overall cell:\[E^{0}(cell) = E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag}) - (E^{0}(\mathrm{SO}_4^{2-}/\mathrm{Ag}_2\mathrm{SO}_4))\]Substitute values:\[E^{0}(cell) = 0.7996 - (-1.5992) = 2.3988 \text{ V}\]

• The positive value indicates spontaneous reaction direction.
• A large cell potential suggests a strong tendency for the reaction to occur.

Nernst Equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions and is essential for understanding reaction dynamics as the cell moves towards equilibrium. For this reaction, it's applied as:\[E = E^{0}(cell) - \frac{RT}{nF} \ln{Q}\]When equilibrium is reached (\(E = 0\)), this equation helps calculate the equilibrium constant (\(K_{sp}\)). Rearranging gives:\[\ln{K_{sp}} = \frac{nFE^{0}(cell)}{RT}\]

• \(n\) is the number of electrons transferred (2 here).
• Substituting known values allows us to find \(K_{sp}\).
• This reveals the reaction's position and the solubility of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\).

Equilibrium Constant

The equilibrium constant (\(K_{sp}\)) is fundamental for quantifying how far a reaction goes to form products at equilibrium. A key insight into solubility, especially in contexts like this galvanic cell example, it quantifies the equilibrium concentrations of ions.For \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\), we found:\[K_{sp} = e^{\frac{2 \times 96485 \times 2.3988}{8.314 \times 298}}\]Calculation results in:\[K_{sp} = 1.24 \times 10^{-5}\]

• This small value indicates low solubility of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)\).
• It delineates how much solid dissolves to reach equilibrium.
• Equilibrium constants are crucial for predicting the extent of reactions in different conditions.