Calculate the potential of a zinc-zinc ion electrode in which the zinc ion activity is 0.001M
`(E_(Zn^(2+)//Zn)^(@)=-0.76V,R=8.314KJ^(-1)mol^(-1),F=96,500" C "mol^(-1))`

Explanation

To calculate the potential of a zinc-zinc ion electrode where the zinc ion activity is 0.001 M, we will use the Nernst equation. The Nernst equation is given by:

\[ E = E^\circ - \frac{RT}{nF} \ln Q \]

Where:
- \( E \) is the electrode potential.
- \( E^\circ \) is the standard electrode potential.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T \) is the temperature in Kelvin (298 K for room temperature).
- \( n \) is the number of moles of electrons exchanged in the half-reaction.
- \( F \) is Faraday's constant (96500 C/mol).
- \( Q \) is the reaction quotient, which can be represented as the activity of the ions involved.

### Step-by-step Solution:

1. **Identify the given values**:
- Standard electrode potential \( E^\circ_{Zn^{2+}/Zn} = -0.76 \, V \)
- Activity of zinc ions \( [Zn^{2+}] = 0.001 \, M \)
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( F = 96500 \, C/mol \)
- Temperature \( T = 298 \, K \)
- Number of electrons \( n = 2 \) (since \( Zn^{2+} + 2e^- \rightarrow Zn \))

2. **Convert the activity to a logarithmic form**:
- The concentration \( [Zn^{2+}] = 0.001 \, M = 10^{-3} \, M \)
- Therefore, \( \log[Zn^{2+}] = \log(10^{-3}) = -3 \)

3. **Substitute values into the Nernst equation**:
\[ E = -0.76 \, V - \frac{(8.314 \, J/(mol \cdot K))(298 \, K)}{(2)(96500 \, C/mol)} \ln(0.001) \]

4. **Calculate the term \( \frac{RT}{nF} \)**:
\[ \frac{RT}{nF} = \frac{(8.314)(298)}{(2)(96500)} \approx 0.00414 \, V \]

5. **Calculate the potential**:
\[ E = -0.76 \, V - (0.00414 \, V)(-3) \]
\[ E = -0.76 \, V + 0.01242 \, V \]
\[ E \approx -0.74758 \, V \]

6. **Final rounding**:
\[ E \approx -0.748 \, V \]

### Final Answer:
The potential of the zinc-zinc ion electrode is approximately \( -0.748 \, V \).