To find the cell potential under non-standard conditions, we can use the Nernst equation, which is given as:

$$ E = E^o - \frac{RT}{nF} \ln(Q) $$

Where:

• $$E^o$$ is the standard cell potential (1.67 V).
• $$R$$ is the gas constant (8.314 J/mol·K).
• $$T$$ is the temperature in Kelvin (298 K for 25°C).
• $$n$$ is the number of moles of electrons transferred per mole of reaction (4 in this case).
• $$F$$ is the Faraday constant (96485 C/mol).
• $$Q$$ is the reaction quotient.

The reaction quotient, $$Q$$, can be calculated based on the given conditions and the reaction:

$$ Q = \frac{[\text{Fe}^{2+}]^2}{[\text{H}^+]^4 \cdot P(O_2)} $$

Substituting the given values:

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

$$ Q = \frac{(10^{-3})^2}{(10^{-3})^4 \cdot 0.1} $$

$$ Q = \frac{10^{-6}}{10^{-12} \cdot 0.1} $$

$$ Q = \frac{10^{-6}}{10^{-13}} $$

$$ Q = 10^{7} $$

Now, substituting the values into the Nernst equation:

$$ E = 1.67 \text{ V} - \frac{8.314 \times 298}{4 \times 96485} \ln(10^{7})$$

Calculating the term $$\frac{RT}{nF}$$;

$$ \frac{RT}{nF} = \frac{8.314 \times 298}{4 \times 96485} $$

$$ \approx \frac{2476}{385940} $$

$$ \approx 0.0064 \text{ V} $$

The natural logarithm of $$10^{7}$$ is about 16.1 (since $$\ln(10) \approx 2.303$$, so $$\ln(10^{7}) = 7 \times \ln(10) \approx 16.1$$).

Thus,

$$ E = 1.67 \text{ V} - 0.0064 \times 16.1$$

$$ E = 1.67 \text{ V} - 0.103 \text{ V} $$

$$ E = 1.567 \text{ V} $$

Therefore, the cell potential under the given conditions is approximately 1.57 V, so option D (1.57 V) is the correct answer.