Calculate emf of the following cell at `25^(@)C`:
`Fe|Fe^(2+)(0.001m)||H^(+)(0.01M)|H_(2)(g)(1" bar")|Pt(s)`
`E^(@)(Fe^(2+)//Fe)=-0.44V,E^(@)(H^(+)//H_(2))=0.00V`

Explanation

To calculate the EMF (Electromotive Force) of the given electrochemical cell at 25°C, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- The oxidation half-reaction involves iron:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]
- The reduction half-reaction involves hydrogen:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

2. **Write the Overall Cell Reaction**:
- Combining the two half-reactions gives:
\[ \text{Fe} + 2\text{H}^+ \rightarrow \text{Fe}^{2+} + \text{H}_2 \]

3. **Calculate the Standard EMF (E°cell)**:
- The standard reduction potentials are given:
- \( E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44 \, \text{V} \)
- \( E^\circ(\text{H}^+/H_2) = 0.00 \, \text{V} \)
- The standard EMF of the cell can be calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ(\text{H}^+/H_2) - E^\circ(\text{Fe}^{2+}/\text{Fe}) \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = 0.00 - (-0.44) = 0.44 \, \text{V} \]

4. **Use the Nernst Equation**:
- The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
- Where \( n \) is the number of moles of electrons transferred (which is 2 in this case), and \( Q \) is the reaction quotient.

5. **Calculate the Reaction Quotient (Q)**:
- The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Fe}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \]
- Substituting the given concentrations:
- \([\text{Fe}^{2+}] = 0.001 \, \text{M}\)
- \([\text{H}^+] = 0.01 \, \text{M}\)
- \(P_{\text{H}_2} = 1 \, \text{bar}\)
- Therefore:
\[ Q = \frac{0.001}{(0.01)^2 \cdot 1} = \frac{0.001}{0.0001} = 10 \]

6. **Substitute Values into the Nernst Equation**:
- Now substituting into the Nernst equation:
\[ E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \log(10) \]
- Since \(\log(10) = 1\):
\[ E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \cdot 1 = 0.44 - 0.02955 = 0.41045 \, \text{V} \]

### Final Answer:
The EMF of the cell at 25°C is approximately **0.41045 V**.