For a cell reaction :
`A (s) + 2 B^(+) (aq) to A^(2+) (aq) + 2 B(s) ` the equilibrium constant `1 xx 10^(4)` . Calculate `E_(cell)^(Theta)` .

Explanation

To calculate the standard cell potential \( E_{\text{cell}}^\Theta \) for the given cell reaction:

\[ A (s) + 2 B^{+} (aq) \rightarrow A^{2+} (aq) + 2 B(s) \]

with an equilibrium constant \( K = 1 \times 10^{4} \), we can use the relationship between the standard cell potential and the equilibrium constant, which is derived from the Nernst equation.

### Step-by-Step Solution:

1. **Identify the Reaction and Its Components**:
- The reaction involves the conversion of solid \( A \) to \( A^{2+} \) ions and the reduction of \( B^{+} \) ions to solid \( B \).
- The oxidation state of \( A \) changes from 0 (in \( A \)) to +2 (in \( A^{2+} \)), indicating that \( A \) loses 2 electrons.

2. **Determine the Number of Electrons Transferred (n)**:
- From the reaction, we see that \( A \) is oxidized to \( A^{2+} \), which involves the loss of 2 electrons.
- Thus, \( n = 2 \).

3. **Use the Relationship Between \( E_{\text{cell}}^\Theta \) and \( K \)**:
- The relationship is given by:
\[ E_{\text{cell}}^\Theta = \frac{0.0591}{n} \log K \]
- Here, \( K = 1 \times 10^{4} \) and \( n = 2 \).

4. **Calculate \( \log K \)**:
- Calculate \( \log(1 \times 10^{4}) \):
\[ \log(1 \times 10^{4}) = 4 \]

5. **Substitute Values into the Equation**:
- Now substitute \( n \) and \( \log K \) into the equation:
\[ E_{\text{cell}}^\Theta = \frac{0.0591}{2} \times 4 \]

6. **Perform the Calculation**:
- Calculate:
\[ E_{\text{cell}}^\Theta = \frac{0.0591 \times 4}{2} = \frac{0.2364}{2} = 0.1182 \text{ V} \]

7. **Final Answer**:
- Rounding to three significant figures, we get:
\[ E_{\text{cell}}^\Theta \approx 0.118 \text{ V} \]

### Summary:
The standard cell potential \( E_{\text{cell}}^\Theta \) for the given reaction is approximately **0.118 V**.