For a reaction `A(s) + 2B^(+) rarr A^(2+)+2B(s). K_(c)` has been found to be `10^(14)` The `E^(o) ""_(cell) ` is :

Options

A. `0.413V`
B. `0.709V`
C. `1.418`V
D. `2.127V`

Answer

To solve the problem, we need to find the standard cell potential \( E^\circ_{\text{cell}} \) for the reaction given the equilibrium constant \( K_c = 10^{14} \). We will use the Nernst equation and the relationship between the equilibrium constant and the standard cell potential.

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation relates the standard cell potential to the equilibrium constant:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c \]
where \( n \) is the number of moles of electrons transferred in the reaction.

2. **Identify the Reaction**:
The reaction given is:
\[ A(s) + 2B^+ \rightarrow A^{2+} + 2B(s) \]
From this reaction, we can see that 2 electrons are transferred (1 for each \( B^+ \) ion).

3. **Determine the Value of \( n \)**:
Since 2 moles of electrons are involved in the reaction, we have:
\[ n = 2 \]

4. **Substitute \( K_c \) into the Nernst Equation**:
We know \( K_c = 10^{14} \). Now we can substitute the values into the Nernst equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \log(10^{14}) \]

5. **Calculate \( \log(10^{14}) \)**:
Using the properties of logarithms:
\[ \log(10^{14}) = 14 \]

6. **Substitute \( \log(10^{14}) \) into the Equation**:
Now we can substitute this value back into the equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \times 14 \]

7. **Perform the Calculation**:
Calculate \( \frac{0.0591 \times 14}{2} \):
\[ E^\circ_{\text{cell}} = \frac{0.0591 \times 14}{2} = \frac{0.8274}{2} = 0.4137 \text{ V} \]

8. **Final Answer**:
Rounding to three significant figures, we find:
\[ E^\circ_{\text{cell}} \approx 0.414 \text{ V} \]

### Conclusion:
The standard cell potential \( E^\circ_{\text{cell}} \) is approximately \( 0.414 \text{ V} \).