Calculate the emf of the following cell at `25^(@)C`.
`Zn|Zn^(2+)`(0.001 M) || `H^(+)(0.01 M) | H_(2)`(1 bar) | Pt(s). Given that `E_((Zn^(2+)//Zn))^(@)=-0.76V, E_((H^(+)//H_(2)))^(@)=0.00" V "`

Explanation

To calculate the emf of the given electrochemical cell at 25°C, we will follow these steps:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
1. **Anode (oxidation)**: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
2. **Cathode (reduction)**: \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \)

### Step 2: Write the standard reduction potentials
Given:
- \( E^\circ (\text{Zn}^{2+}/\text{Zn}) = -0.76 \, \text{V} \)
- \( E^\circ (\text{H}^+/H_2) = 0.00 \, \text{V} \)

### Step 3: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.00 \, \text{V} - (-0.76 \, \text{V}) = 0.76 \, \text{V} \]

### Step 4: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( Q \) is the reaction quotient.

### Step 5: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) for the reaction:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]
is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \]
Substituting the concentrations and pressure:
- \( [\text{Zn}^{2+}] = 0.001 \, \text{M} \)
- \( [\text{H}^+] = 0.01 \, \text{M} \)
- \( P_{\text{H}_2} = 1 \, \text{bar} \)

Thus,
\[ Q = \frac{0.001}{(0.01)^2 \cdot 1} = \frac{0.001}{0.0001} = 10 \]

### Step 6: Substitute values into the Nernst equation
Now substituting into the Nernst equation:
\[ E_{\text{cell}} = 0.76 \, \text{V} - \frac{0.0591}{2} \log(10) \]
Calculating \( \log(10) = 1 \):
\[ E_{\text{cell}} = 0.76 \, \text{V} - \frac{0.0591}{2} \cdot 1 \]
\[ E_{\text{cell}} = 0.76 \, \text{V} - 0.02955 \, \text{V} \]
\[ E_{\text{cell}} = 0.7305 \, \text{V} \]

### Final Answer
The emf of the cell at 25°C is:
\[ E_{\text{cell}} = 0.7305 \, \text{V} \]

---