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.

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: \(2\mathrm{Ag}_{(s)} + \mathrm{SO}_{4}^{2-}_{(aq)} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{SO}_{4}_{(s)}\)

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: \[E^{0}(cell)= E^{0}(\mathrm{Ag}^{+}/\mathrm{Ag}) - E^{0}(\mathrm{SO}_4^{2-}/\mathrm{Ag}_2\mathrm{SO}_4) = 0.7996 - (-1.5992) = 2.3988 V\]

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 \(1.24 \times 10^{-5}\).