The `E_(Call)^(@)=1.18V` for `Zn(s)||Zn^(+2)(1M)||Cu^(+2)(1M)|Cu(s)`. The value of `x` if when excess granulated zinc is added to `1M Cu^(+2)` solution the `[Cu^(+2)]_(eq)` becomes `10^(-x)M` is `(T=298 K ,(2.303RT)/(F)=0.059)`

Explanation

To solve the problem, we need to determine the value of \( x \) when excess granulated zinc is added to a 1M \( Cu^{2+} \) solution, resulting in an equilibrium concentration of \( Cu^{2+} \) given by \( [Cu^{2+}]_{eq} = 10^{-x} \) M.

### Step-by-Step Solution:

1. **Identify the Cell Reaction:**
The cell consists of zinc and copper half-reactions. The overall cell reaction can be written as:
\[ Cu^{2+} + Zn \rightarrow Zn^{2+} + Cu \]
Here, zinc is oxidized to \( Zn^{2+} \) and \( Cu^{2+} \) is reduced to copper.

2. **Use the Nernst Equation:**
The Nernst equation relates the cell potential (\( E_{cell} \)) to the standard cell potential (\( E^{\circ} \)) and the concentrations of the reactants and products:
\[ E_{cell} = E^{\circ} - \frac{2.303RT}{nF} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]}\right) \]
Given that \( E^{\circ} = 1.18 \, V \), \( R = 8.314 \, J/(mol \cdot K) \), \( T = 298 \, K \), and \( F = 96485 \, C/mol \), we can simplify \( \frac{2.303RT}{F} \) to \( 0.059 \, V \) (as given).

3. **Set Up the Nernst Equation:**
Since we have excess zinc, we can assume its concentration is constant and does not affect the equilibrium. Thus, we can simplify our equation:
\[ E_{cell} = 1.18 - 0.059 \log \left( \frac{1}{10^{-x}} \right) \]
This simplifies to:
\[ E_{cell} = 1.18 - 0.059 \log(10^{x}) = 1.18 - 0.059x \]

4. **At Equilibrium, \( E_{cell} = 0 \):**
At equilibrium, the cell potential becomes zero:
\[ 0 = 1.18 - 0.059x \]

5. **Solve for \( x \):**
Rearranging the equation gives:
\[ 0.059x = 1.18 \]
\[ x = \frac{1.18}{0.059} \approx 20 \]

### Final Answer:
The value of \( x \) is approximately **20**.