A voltaic cell utilizes the following reaction: $4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when $\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\( \)0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=50.7 \mathrm{kPa},\( and the \)\mathrm{pH}$ of the solution in the cathode half-cell is \(3.50 ?\)

First, we need to identify the half-reactions for the oxidation and reduction processes: Oxidation half-reaction: \[\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\] Reduction half-reaction: \[\frac{1}{2}\text{O}_2(g) + 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2\text{O}(l)\] Now, let's multiply the oxidation half-reaction by 4 to have the same number of \(e^-\) as in the reduction half-reaction, which gives us the overall balanced redox reaction: \[4 \text{Fe}^{2+}(aq) + \text{O}_{2}(g) + 4 \text{H}^{+}(aq) \rightarrow 4 \text{Fe}^{3+}(aq) + 2 \text{H}_{2}\text{O}(l)\]

E° of the cell can be calculated by subtracting E° of oxidation (anode) from E° of reduction (cathode): \[E°_{cell} = E°_{red} - E°_{ox}\] For both half reactions, E° of Fe\(^{3+}\)/Fe\(^{2+}\): \(0.77 V\) E° of O\(_2\)/H\(_2\)O: \(1.23 V\) Now, we can calculate E° of the cell: \[E°_{cell} = 1.23 V - 0.77 V = 0.46 V\]

The Nernst equation allows us to find the emf under non-standard conditions: \[E = E° - \frac{RT}{nF} \ln(Q)\] Here, R is the universal gas constant, T is the temperature, n is the number of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient. We will use the following values: T = 298 K (approximately 25°C, room temperature) R = 8.314 J/molK (gas constant) F = 96,485 C/mol (Faraday constant) n = 4 electrons (as determined in Step 1) Now, let's write the expression for the reaction quotient Q: \[Q=\frac{\left[\text{Fe}^{3+}\right]^4}{(\left[\text{Fe}^{2+}\right]^4)([\text{O}_2])}\] We have the following given values: \(\left[\text{Fe}^{2+}\right] = 1.3 \text{M}\) \(\left[\text{Fe}^{3+}\right] = 0.010 \text{M}\) \(P_{\text{O}_2} = 50.7 \text{kPa}\) We can convert the partial pressure of O\(_2\) to concentration using Henry's Law: \([O_2] = \frac{P_{O_2}}{k_H}\) Given that for O\(_2\) at 25°C, \(k_H = 769 \text{kPa}/\text{M}\), \([\text{O}_{2}] = \frac{50.7 \text{kPa}}{769 \text{kPa}/\text{M}} = 0.066 \text{M}\) Now, we can calculate Q: \[Q=\frac{(0.010 \text{M})^4}{(1.3 \text{M})^4(0.066 \text{M})} = 1.16 \times 10^{-11}\] Using the values above and the Nernst equation, we can calculate the emf under non-standard conditions: \[E = 0.46 V - \frac{(8.314 \text{J/molK})(298 \text{K})}{(4)(96,485 \text{C/mol})} \ln(1.16 \times 10^{-11})\] \[E \approx 0.99 V\]

(a) The emf of this cell under standard conditions is: \(E°_{cell} = 0.46 V\) (b) The emf of this cell under non-standard conditions is: \(E \approx 0.99 V\)