For a reaction `A(s)+2B^(o+) rarr A^(2+)+2B`
`K_(c)` has been found to be `10^(12)`. The `E^(c-)._(cell)` is

Options

A. `0.354V`
B. `0.708V`
C. `0.0098V`
D. `1.36V`

Explanation

To solve the problem, we need to calculate the standard cell potential \( E^\circ_{\text{cell}} \) for the given reaction using the relationship between the equilibrium constant \( K_c \) and the cell potential.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ A(s) + 2B^{+} \rightarrow A^{2+} + 2B \]

2. **Determine the Number of Electrons Transferred (n)**:
In this reaction, the oxidation state of \( A \) changes from 0 to +2, indicating that 2 electrons are lost. The \( B^{+} \) ions are reduced to \( B \), also involving 2 electrons. Therefore, the total number of electrons transferred \( n \) is:
\[ n = 2 \]

3. **Use the Nernst Equation**:
The relationship between the standard cell potential \( E^\circ_{\text{cell}} \) and the equilibrium constant \( K_c \) is given by:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c \]

4. **Substitute the Known Values**:
We know that \( K_c = 10^{12} \) and \( n = 2 \). Substituting these values into the equation gives:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \log(10^{12}) \]

5. **Calculate the Logarithm**:
The logarithm of \( 10^{12} \) is:
\[ \log(10^{12}) = 12 \]

6. **Substitute the Logarithm Value**:
Now substituting this back into the equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \times 12 \]

7. **Perform the Calculation**:
First, calculate \( \frac{0.0591}{2} \):
\[ \frac{0.0591}{2} = 0.02955 \]
Then multiply by 12:
\[ E^\circ_{\text{cell}} = 0.02955 \times 12 = 0.3546 \text{ V} \]

8. **Final Result**:
Rounding to three significant figures, we get:
\[ E^\circ_{\text{cell}} \approx 0.354 \text{ V} \]

### Conclusion:
The standard cell potential \( E^\circ_{\text{cell}} \) for the reaction is approximately **0.354 V**.