Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{O}_{2}(g)+4 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q) $$ (a) Calculate \(E^{\circ}\) (b) Write the Nernst equation for the cell. (c) Calculate \(E\) under the following conditions: \(\left[\mathrm{NO}_{3}{ }^{-}\right]=0.750 M\), \(P_{\mathrm{NO}}=0.993 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.515 \mathrm{~atm}, \mathrm{pH}=2.85\)

Let's write down the half reactions to find \(E^{\circ}\): Reduction half-reaction (oxygen is reduced): $$ \frac{1}{2} O_{2}(g) + 2e^− + 2 H^+(aq) \longrightarrow H_2O(l) $$ Oxidation half-reaction (nitric oxide is oxidized): $$ NO(g) + H_2O(l) \longrightarrow NO_3^-(aq) + 2H^+(aq) + e^− $$ Now we need to look up the standard reduction potentials for each half-reaction. Source them from a reliable table to find: - Oxygen reduction potential: \(E^{\circ}_{O_2/H_2O} = 1.23 \,\text{V}\) - Nitric oxide reduction potential: \(E^{\circ}_{NO_3^-/NO} = 0.96 \,\text{V}\) Since the oxidation half-reaction involves the reverse of the reduction of nitric oxide, we need to flip the sign of the corresponding reduction potential: - Nitric oxide oxidation potential: \(E^{\circ}_{NO/NO_3^-} = -0.96 \,\text{V}\) Now, we can calculate the standard cell potential using the formula: $$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} $$ In our case, the cathode is the reduction of oxygen, and the anode is the oxidation of nitric oxide. So we have: $$ E^{\circ}_{cell} = 1.23 \,\text{V} - (-0.96 \,\text{V}) = 2.19 \,\text{V} $$

The Nernst equation calculates the cell potential under non-standard conditions and has the following form: $$ E = E^{\circ} - \frac{RT}{nF} \times \ln{Q} $$ where: - \(E\) is the cell potential under non-standard conditions - \(E^{\circ}\) is the standard cell potential (we already calculated it in Step 1) - \(R\) is the gas constant (\(8.314 \,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\)) - \(T\) is the temperature in Kelvin (\(25^{\circ}\mathrm{C} = 298\,\mathrm{K}\)) - \(n\) is the number of moles of electrons transferred - \(F\) is the Faraday constant (\(9.648 \times 10^{4}\,\text{C}\,\text{mol}^{-1}\)) - \(Q\) is the reaction quotient For our given reaction, 4 electrons are transferred. Now the Nernst equation for our cell looks like this: $$ E = E^{\circ} - \frac{RT}{4F} \times \ln{Q} $$

To calculate the cell potential (\(E\)) under the conditions given in the problem, we need to find the reaction quotient \(Q\). Based on the balanced chemical equation and ideal gas behavior: $$ Q = \frac{[NO_3^-]^4[H^+]^4}{[O_2]^{3/2}[NO]^4} $$ Using the conditions given in the exercise: 1. \([NO_3^-] = 0.750\,\text{M}\) 2. \(P_{NO} = 0.993\,\text{atm}\) 3. \(P_{O_2} = 0.515\,\text{atm}\) 4. \(\text{pH}=2.85 \Rightarrow [H^+] = 10^{-2.85} = 1.41 \times 10^{-3}\,\text{M}\) To convert the pressures to concentrations, we will use the ideal gas equation: $$ PV = nRT \Rightarrow \frac{n}{V} =\frac{P}{RT} $$ Assuming a volume of 1, we have: $$ [NO] = \frac{0.993\,\text{atm}}{0.0821 \, \text{L}\, \text{atm}\, \text{K}^{-1}\, \text{mol}^{-1} \cdot 298\,\text{K}} = 0.0407\,\text{M} $$ $$ [O_2] = \frac{0.515\,\text{atm}}{0.0821 \, \text{L}\,\text{atm}\, \text{K}^{-1}\,\text{mol}^{-1} \cdot 298\,\text{K}} = 0.0210\,\text{M} $$ Now, we can plug the concentrations into the Q formula: $$ Q = \frac{(0.75)^4(1.41 \times 10^{-3})^4}{(0.021)^{(3/2)}(0.0407)^4} = 1582.61 $$ Finally, we plug the \(Q\) value and the other known values into the Nernst equation and compute \(E\): $$ E = 2.19 \,\text{V} - \frac{8.314\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1} \cdot 298\,\text{K}}{4 \times 9.648 \times 10^4\, \text{C}\,\text{mol}^{-1}} \times \ln{(1582.61)} = 2.14\,\text{V} $$ The cell potential under the given conditions is \(2.14\,\text{V}\)