A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

First, we need to calculate the cell potential under the conditions mentioned in part (a). To do this, we need to know the half-cell reaction and the standard reduction potential of the cell. The half-cell reaction is given in the problem statement as: \[ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} \] Since we are given the standard reduction potential (\(\mathscr{E}^{\circ}\)) for this half-reaction, we can use the following formula to calculate the cell potential: \[ \mathscr{E}_{\text {cell }}=\mathscr{E}^{\circ}_{\text {cell }}-\frac{RT}{nF} \ln \frac{\left[\mathrm{Ag}^+\right]^2}{\left[\mathrm{CrO}_{4}{ }^{2-}\right]} \] For part (a), we have \(\mathscr{E}^{\circ}_{\text {cell }}=0.446\text{ V}\) and \(\left[\mathrm{CrO}_{4} {}^{2-}\right]=1.00\text{ mol/L}\). Also, we know that the gas constant R = 8.314 J/(mol K), T = 25 °C = 298 K, F = 96,485 C/mol, and n = 2 (since there are two electrons in the half-reaction). Plugging the values into the formula, we get: \[ \mathscr{E}_{\text {cell }}=0.446\text{ V}-\frac{8.314\times 298}{2\times 96,485} \ln \frac{\left[1\right]^2}{1} \] Since the ln term becomes zero, \(\mathscr{E}_{\text {cell }}\) remains as 0.446 V. Now, we can use the cell potential to calculate the standard Gibbs free energy change: \[ \Delta G = -nFE_{\text {cell }} \] Plugging the values in the formula: \[ \Delta G = -(2)(96,485)(0.446)\text{ J/mol} \] This gives us \(\Delta G = -86,573\text{ J/mol}\).

The Nernst equation for this cell would be derived from the cell potential formula mentioned in step 1. Here is the Nernst equation for the given cell: \[ \mathscr{E}_{\text {cell }}=0.446\text{ V}-\frac{RT}{nF} \ln \frac{\left[\mathrm{Ag}^+\right]^2}{\left[\mathrm{CrO}_{4} {}^{2-}\right]} \]

For part (c), we are given that \(\left[\mathrm{CrO}_{4} {}^{2-}\right]=1.00 \times 10^{-5} \text{ M}\). We need to find the cell potential for this concentration. Using the Nernst equation from step 2: \[ \mathscr{E}_{\text {cell }}=0.446\text{ V}-\frac{8.314\times 298}{2\times 96,485} \ln \frac{\left[1\right]^2}{1.00 \times 10^{-5}} \] This gives us \(\mathscr{E}_{\text {cell }} = 0.619\text{ V}\).

For part (d), we are given that the measured cell potential is 0.504 V. We need to find the corresponding \(\left[\mathrm{CrO}_{4}{}^{2-}\right]\) concentration. Rearrange the Nernst equation in step 2 and plug in the given value of \(\mathscr{E}_{\text {cell }}\): \[ \left[\mathrm{CrO}_{4} {}^{2-}\right] = \frac{\left[1\right]^2}{\exp\left(\frac{2\times 96,485}{8.314\times 298}\left(0.446-0.504\right)\right)} \] This gives us \(\left[\mathrm{CrO}_{4} {}^{2-}\right] = 6.22 \times 10^{-5} \text{ M}\).

Now we need to calculate the solubility product (Ksp) for silver chromate, Ag_2CrO_4. We can use the following formula for Ksp: \[ K_{sp}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_{4}{ }^{2-}\right] \] From step 3, we know the cell potential when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right] = 1.00 \times 10^{-5}\text{ M}\) is 0.619 V. We can use this cell potential to find the corresponding \(\left[\mathrm{Ag}^{+}\right]\) concentration using the Nernst equation in step 2: \[ \left[\mathrm{Ag}^{+}\right] = \exp\left(\frac{2\times 96,485}{8.314\times 298}\left(0.446-0.619\right)\right) \] This gives us \(\left[\mathrm{Ag}^{+}\right] = 1.02 \times 10^{-3}\text{ M}\). Now we can calculate Ksp: \[ K_{sp}=\left[1.02 \times 10^{-3}\right]^2\left[1.00 \times 10^{-5}\right] \] This gives us \(K_{sp} = 1.04 \times 10^{-11}\).