The problem asks to calculate the cell potential (E) under non-standard conditions given the standard cell potential (E°) and the concentrations of reactants and products. We will use the Nernst equation for this purpose.

Step 1: Write down the Nernst Equation The Nernst equation relates the cell potential (E) to the standard cell potential (E°) and the reaction quotient (Q). At 25°C (298 K), the equation is: $E = E° - \frac{0.0591}{n} \log Q$ where:

• $E$ is the cell potential under non-standard conditions.
• $E°$ is the standard cell potential, given as 1.67 V.
• $n$ is the number of moles of electrons transferred in the balanced reaction.
• $Q$ is the reaction quotient.

Step 2: Determine the number of electrons transferred (n) The overall cell reaction is: $2Fe(s) + O_2(g) + 4H^+(aq) \to 2Fe^{2+} (aq) + 2H_2O (l)$ We can split this into oxidation and reduction half-reactions to find 'n'.

• Oxidation half-reaction: Iron is oxidized from Fe(s) to $Fe^{2+}(aq)$. $2Fe(s) \to 2Fe^{2+}(aq) + 4e^-$
• Reduction half-reaction: Oxygen is reduced in the presence of acid. $O_2(g) + 4H^+(aq) + 4e^- \to 2H_2O(l)$ In both half-reactions, 4 electrons are involved. Thus, the number of electrons transferred, $n = 4$.

Step 3: Calculate the reaction quotient (Q) The expression for the reaction quotient Q is the ratio of the activities of the products to the activities of the reactants, raised to the power of their stoichiometric coefficients. For pure solids and liquids, the activity is taken as 1. $Q = \frac{[Fe^{2+}]^2}{P_{O_2} \times [H^+]^4}$ We are given the following information:

• $[Fe^{2+}] = 10^{-3} M$
• $P(O_2) = 0.1$ atm
• pH = 3

First, we calculate the hydrogen ion concentration, $[H^+]$, from the pH: $pH = -\log[H^+]$ $3 = -\log[H^+]$ $[H^+] = 10^{-3} M$ Now, we substitute the given values into the expression for Q: $Q = \frac{(10^{-3})^2}{(0.1) \times (10^{-3})^4} = \frac{10^{-6}}{10^{-1} \times 10^{-12}} = \frac{10^{-6}}{10^{-13}} = 10^7$

Step 4: Calculate the cell potential (E) Now we substitute all the values into the Nernst equation: $E = E° - \frac{0.0591}{n} \log Q$ $E = 1.67 - \frac{0.0591}{4} \log(10^7)$ $E = 1.67 - \frac{0.0591}{4} \times 7$ $E = 1.67 - \frac{0.4137}{4}$ $E = 1.67 - 0.103425$ $E \approx 1.566575 V$ Rounding to two decimal places, we get $E = 1.57$ V.

Step 5: Compare with options The calculated value $1.57$ V matches option D.

Therefore, the cell potential at 25°C under the given conditions is 1.57 V.