Consider the following cell reaction.
`2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq)+2H_(2)O(l),`
`E^(@)=1.67V`
At `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` is

Explanation

To find the cell potential for the given reaction, we can use the Nernst equation:

\[ E = E^0 - \frac{0.059}{n} \log Q \]

where:
- \(E\) is the cell potential,
- \(E^0\) is the standard cell potential,
- \(n\) is the number of moles of electrons transferred in the reaction,
- \(Q\) is the reaction quotient.

### Step 1: Identify the standard cell potential \(E^0\)
From the question, we have:
\[ E^0 = 1.67 \, \text{V} \]

### Step 2: Determine the number of electrons transferred \(n\)
In the given reaction:
\[ 2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l) \]
Each iron atom goes from an oxidation state of 0 to +2, meaning each iron atom loses 2 electrons. Since there are 2 iron atoms:
\[ n = 2 \times 2 = 4 \, \text{electrons} \]

### Step 3: Calculate the reaction quotient \(Q\)
The reaction quotient \(Q\) is given by the expression:
\[ Q = \frac{[Fe^{2+}]^2}{P(O_2) \cdot [H^+]^4} \]
Given:
- \([Fe^{2+}] = 10^{-3} \, M\)
- \(P(O_2) = 0.1 \, atm\)
- pH = 3, which gives \([H^+] = 10^{-3} \, M\)

Now substituting these values into the equation for \(Q\):
\[ Q = \frac{(10^{-3})^2}{(0.1) \cdot (10^{-3})^4} \]
Calculating \(Q\):
\[ Q = \frac{10^{-6}}{0.1 \cdot 10^{-12}} = \frac{10^{-6}}{10^{-13}} = 10^{7} \]

### Step 4: Substitute values into the Nernst equation
Now substituting \(E^0\), \(n\), and \(Q\) into the Nernst equation:
\[ E = 1.67 - \frac{0.059}{4} \log(10^7) \]

Calculating \(\log(10^7)\):
\[ \log(10^7) = 7 \]

Now substituting this value:
\[ E = 1.67 - \frac{0.059}{4} \cdot 7 \]
Calculating \(\frac{0.059}{4}\):
\[ \frac{0.059}{4} = 0.01475 \]
Now calculating the full expression:
\[ E = 1.67 - 0.01475 \cdot 7 = 1.67 - 0.10325 \]
\[ E = 1.56675 \approx 1.57 \, \text{V} \]

### Final Answer
The cell potential at \(25^{\circ}C\) is approximately:
\[ \boxed{1.57 \, V} \]