Calculate the half-cell potential at 298 K for the reaction,
`Zn^(2+)+2e^(-) rarr Zn`
If `[Zn^(2+)]=0.1 M` and `E^(@)=-0.76` volt.

Answer

To calculate the half-cell potential for the reaction \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) at 298 K, we will use the Nernst equation. The Nernst equation is given by:

\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{1}{[\text{Zn}^{2+}]} \right) \]

Where:
- \( E \) is the half-cell potential we want to calculate.
- \( E^0 \) is the standard electrode potential, which is given as -0.76 V.
- \( n \) is the number of electrons transferred in the reaction (which is 2 for this reaction).
- \( [\text{Zn}^{2+}] \) is the concentration of zinc ions, which is given as 0.1 M.

### Step-by-Step Solution:

1. **Identify the values**:
- \( E^0 = -0.76 \, \text{V} \)
- \( n = 2 \)
- \( [\text{Zn}^{2+}] = 0.1 \, \text{M} \)

2. **Substitute the values into the Nernst equation**:
\[ E = -0.76 - \frac{0.059}{2} \log \left( \frac{1}{0.1} \right) \]

3. **Calculate the logarithm**:
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

4. **Substitute the logarithm value back into the equation**:
\[ E = -0.76 - \frac{0.059}{2} \times 1 \]

5. **Calculate \( \frac{0.059}{2} \)**:
\[ \frac{0.059}{2} = 0.0295 \]

6. **Final substitution**:
\[ E = -0.76 - 0.0295 \]

7. **Perform the final calculation**:
\[ E = -0.7895 \, \text{V} \]

8. **Round to appropriate significant figures**:
\[ E \approx -0.79 \, \text{V} \]

### Final Answer:
The half-cell potential at 298 K for the reaction \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) is approximately **-0.79 V**.