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We need to calculate the \(\left[\mathrm{H}^{+}\right]\) using the buffer system. The buffer system is composed of sodium acetate and acetic acid. The dissociation reaction for acetic acid is given by: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}^{+}$$ Using the given concentrations and the \(K_{\mathrm{a}}\) value, we can apply the Henderson-Hasselbalch equation to find the proton concentration: $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log{\frac{\left[\mathrm{CH}_{3}\mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3}\mathrm{COOH}\right]}}$$ Plug in the given values: $$\mathrm{pH} = -\log{(1.8 \times 10^{-5})} + \log{\frac{0.100}{0.100}} = -\log{(1.8 \times 10^{-5})}$$ Calculate the value of \(\mathrm{pH}\): $$\mathrm{pH} \approx 4.74$$ Now, we can calculate \(\left[\mathrm{H}^{+}\right]\): $$\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}} = 10^{-4.74} \approx 1.84 \times 10^{-5}\,\mathrm{M}$$

Using the standard reduction potentials for the half-reactions, we can calculate the standard cell potential and equilibrium constant for the overall reaction at \(25^{\circ}\mathrm{C}\). The half-reactions are: $$\mathrm{O}_{2}(g) + 4 \ \mathrm{H}^{+}(aq) + 4 \ \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) \quad \quad E^{\circ} = 1.23\,\mathrm{V}$$ $$2 \mathrm{Br}^{-}(aq) \longrightarrow \mathrm{Br}_{2}(l) + 2 \ \mathrm{e}^{-} \quad \quad E^{\circ} = 1.09\,\mathrm{V}$$ The standard cell potential is: $$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ} = (1.23 - 1.09)\,\mathrm{V} = 0.14\,\mathrm{V}$$ Now, we can calculate the equilibrium constant \(K\) using the Nernst equation at \(25^{\circ}\mathrm{C}\): $$\Delta G_{\mathrm{cell}}^{\circ} = -nFE^{\circ} = -RT \ln{K}$$ Rearrange to solve for \(K\): $$K = \mathrm{e}^{\frac{-\Delta G_{\mathrm{cell}}^{\circ}}{RT}}$$ At \(25^{\circ}\mathrm{C}\), we have \(R = 8.314\,\mathrm{J\cdot mol^{-1} \cdot K^{-1}}\) and \(T = 298\,\mathrm{K}\). Keep in mind that the balanced overall reaction has \(n=4\) moles of electrons transferred. Plugging in the values, we can find \(K\): $$K = \mathrm{e}^{\frac{-(4)(96485\,\mathrm{C \cdot mol^{-1}})(0.14\,\mathrm{V})}{(8.314\,\mathrm{J \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}} \approx 36.1$$

Finally, we can use the Nernst equation to calculate the cell voltage: $$E_{\mathrm{cell}} = E_{\mathrm{cell}}^{\circ} - \frac{RT}{nF} \ln{Q}$$ The reaction quotient \(Q\) is given by: $$Q = \frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2\left[\mathrm{Br}_{2}\right]^2}{\left[\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^4\left[\mathrm{Br}^{-}\right]^4}$$ Since the concentrations of \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{Br}_{2}\) are quite high as products, the reaction quotient simplifies to: $$Q \approx \frac{1}{\left[\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^4\left[\mathrm{Br}^{-}\right]^4} = \frac{1}{(1.00\,\mathrm{atm})(1.84 \times 10^{-5}\,\mathrm{M})^4(0.100\,\mathrm{M})^4}$$ Replace the pressure of oxygen gas with its concentration using the ideal gas law: $$\left[\mathrm{O}_{2}\right] = \frac{n}{V} = \frac{P}{RT} = \frac{(1.00\,\mathrm{atm})(0.0821\,\mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}{1\,\mathrm{L}} \approx 0.0245\,\mathrm{M}$$ Insert the new value of \(\left[\mathrm{O}_{2}\right]\) into the reaction quotient: $$Q \approx \frac{1}{(0.0245\,\mathrm{M})(1.84 \times 10^{-5}\,\mathrm{M})^4(0.100\,\mathrm{M})^4} \approx 5.56 \times 10^{-12}$$ Now, we can evaluate the Nernst equation: $$E_{\mathrm{cell}} = 0.14\,\mathrm{V} - \frac{(8.314\,\mathrm{J \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}{(4)(96485\,\mathrm{C \cdot mol^{-1}})} \ln{(5.56 \times 10^{-12})} \approx 0.36\,\mathrm{V}$$ Thus, the calculated cell voltage is approximately \(0.36\,\mathrm{V}\).