A zinc electrode is placed in `0.1M` solution of `ZnSO_(4)` at `25^(@)C`. Assuming salt is dissociated to the extent of `20%` at this dilution. The potential. The potential of this electrode at this temperature is `:`
`(E^(c-)._(Zn^(2+)|Zn)=-0.76V)`
`a. 0.79V" ".b. -0.79V" "c. -0.81V." "d. 0.81V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the concentration of Zn²⁺ ions
Given that the zinc sulfate (ZnSO₄) solution is 0.1 M and it dissociates to the extent of 20%, we can calculate the concentration of Zn²⁺ ions.

- **Dissociation of ZnSO₄**:
\[ \text{ZnSO}_4 \rightarrow \text{Zn}^{2+} + \text{SO}_4^{2-} \]
- **Concentration of Zn²⁺**:
\[ \text{Concentration of Zn}^{2+} = 0.1 \, \text{M} \times 0.20 = 0.02 \, \text{M} \]

### Step 2: Use the Nernst equation
The Nernst equation relates the cell potential to the standard electrode potential and the concentrations of the reactants and products.

- **Nernst equation**:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where:
- \(E^\circ\) is the standard reduction potential,
- \(n\) is the number of moles of electrons transferred,
- \(Q\) is the reaction quotient.

### Step 3: Identify the parameters
From the problem:
- \(E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}\)
- \(n = 2\) (since \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \))
- \(Q = \frac{1}{[\text{Zn}^{2+}]} = \frac{1}{0.02} = 50\)

### Step 4: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:

\[ E = -0.76 - \frac{0.0591}{2} \log(50) \]

### Step 5: Calculate \(\log(50)\)
Using logarithmic properties:
\[ \log(50) = \log(5 \times 10) = \log(5) + \log(10) \approx 0.7 + 1 = 1.7 \]

### Step 6: Substitute \(\log(50)\) back into the equation
Now substituting \(\log(50)\) back into the equation:

\[ E = -0.76 - \frac{0.0591}{2} \times 1.7 \]
\[ E = -0.76 - 0.02955 \times 1.7 \]
\[ E = -0.76 - 0.050235 \]
\[ E \approx -0.81 \, \text{V} \]

### Final Answer
Thus, the potential of the zinc electrode in the given conditions is approximately:
\[ \boxed{-0.81 \, \text{V}} \]