Calculate the electrode potential of the electrode `Zn//Zn^(2+)` (conc. `=0.1M`) at `25^(@)C`
Given that `E_(Zn//Zn^(2+))^(@)=0.7618` volt.

Explanation

To calculate the electrode potential of the electrode `Zn//Zn^(2+)` with a concentration of `0.1 M` at `25°C`, we will use the Nernst equation. The standard reduction potential for the half-cell reaction is given as `E°(Zn/Zn^(2+)) = 0.7618 V`.

### Step-by-Step Solution:

1. **Identify the Nernst Equation**:
The Nernst equation is given by:
\[ E = E° - \frac{0.0591}{n} \log Q \]
where:
- \(E\) = electrode potential
- \(E°\) = standard electrode potential
- \(n\) = number of moles of electrons transferred in the half-reaction
- \(Q\) = reaction quotient

2. **Determine the Number of Electrons Transferred (\(n\))**:
For the zinc half-reaction:
\[ Zn^{2+} + 2e^- \rightarrow Zn \]
Here, \(n = 2\) because 2 electrons are involved in the reduction of \(Zn^{2+}\) to \(Zn\).

3. **Calculate the Reaction Quotient (\(Q\))**:
The reaction quotient \(Q\) for the half-reaction can be expressed as:
\[ Q = \frac{[Zn^{2+}]}{[Zn]} \]
Since the concentration of solid zinc (\([Zn]\)) is constant, it is not included in the expression. Therefore:
\[ Q = [Zn^{2+}] = 0.1 \, M \]

4. **Substitute Values into the Nernst Equation**:
Now we can substitute the values into the Nernst equation:
\[ E = E° - \frac{0.0591}{n} \log Q \]
Substituting \(E° = 0.7618 \, V\), \(n = 2\), and \(Q = 0.1\):
\[ E = 0.7618 - \frac{0.0591}{2} \log(0.1) \]

5. **Calculate \(\log(0.1)\)**:
\[ \log(0.1) = \log(10^{-1}) = -1 \]

6. **Substitute \(\log(0.1)\) into the Equation**:
\[ E = 0.7618 - \frac{0.0591}{2} \times (-1) \]
\[ E = 0.7618 + \frac{0.0591}{2} \]
\[ E = 0.7618 + 0.02955 \]

7. **Final Calculation**:
\[ E = 0.79135 \, V \]
Rounding to three significant figures, we get:
\[ E \approx 0.791 \, V \]

### Final Answer:
The electrode potential of the electrode `Zn//Zn^(2+)` at `25°C` with a concentration of `0.1 M` is approximately **0.791 V**.