Consider the following cell reation `:`
`2Fe(s)+O_(2)(g)+4H^(o+)(aq) rarr 2Fe^(2+)(aq)+2H_(2)O(l)" "E^(c-)=1.67V`
`At[Fe^(2+)]=10^(-3)M,p(O_(2))=0.1atm` and `pH=3` .
The cell potential at `25^(@)C` is
(a)`1.47V`
(b)`1.77V`
(c)`1.87V`
(d)`1.57V`

Options

A. `1.47V`
B. `1.77V`
C. `1.87V`
D. `1.57V`

Explanation

To calculate the cell potential at 25°C for the given reaction, we will use the Nernst equation. Here are the steps to solve the problem:

### Step 1: Determine the concentration of H⁺ ions
Given that pH = 3, we can find the concentration of H⁺ ions using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Identify the standard cell potential (E°)
The standard cell potential (E°) is given as:
\[ E° = 1.67 \, \text{V} \]

### Step 3: Identify the number of electrons transferred (n)
From the balanced reaction, we see that 4 electrons are involved in the reaction:
\[ 2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l) \]
Thus, \( n = 4 \).

### Step 4: Write the Nernst equation
The Nernst equation is given by:
\[ E = E° - \frac{0.059}{n} \log Q \]
where \( Q \) is the reaction quotient.

### Step 5: Calculate the reaction quotient (Q)
The reaction quotient \( Q \) can be calculated using the concentrations and pressures of the reactants and products:
\[ Q = \frac{[Fe^{2+}]^2}{[H^+]^4 \cdot p(O_2)} \]
Substituting the values:
- \([Fe^{2+}] = 10^{-3} \, \text{M}\)
- \([H^+] = 10^{-3} \, \text{M}\)
- \(p(O_2) = 0.1 \, \text{atm}\)

Thus,
\[ Q = \frac{(10^{-3})^2}{(10^{-3})^4 \cdot 0.1} = \frac{10^{-6}}{10^{-12} \cdot 0.1} = \frac{10^{-6}}{10^{-13}} = 10^{7} \]

### Step 6: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:
\[ E = 1.67 - \frac{0.059}{4} \log(10^7) \]
Calculating the logarithm:
\[ \log(10^7) = 7 \]
Now substituting this back:
\[ E = 1.67 - \frac{0.059}{4} \cdot 7 \]
Calculating the term:
\[ \frac{0.059 \cdot 7}{4} = \frac{0.413}{4} = 0.10325 \]
Finally, substituting this value:
\[ E = 1.67 - 0.10325 = 1.56675 \approx 1.57 \, \text{V} \]

### Final Answer
Thus, the cell potential at 25°C is approximately:
\[ \boxed{1.57 \, \text{V}} \]