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 calculate the standard cell potential \( E^\circ_{\text{cell}} \) for the given reaction using the relationship between the equilibrium constant \( K_c \) and the standard cell potential. Here's a step-by-step breakdown of the solution:

### Step 1: Write the Nernst Equation
The Nernst equation relates the standard cell potential to the equilibrium constant:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K_c \]
At equilibrium, \( E_{\text{cell}} = 0 \). Therefore, we can set the equation to zero:
\[ 0 = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log K_c \]
Rearranging gives:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c \]

### Step 2: Identify the Reaction and Calculate \( n \)
The reaction given is:
\[ A(s) + 2B^+ \rightleftharpoons A^{2+} + 2B(s) \]
In this reaction, \( A \) is oxidized to \( A^{2+} \), and \( B^+ \) is reduced to \( B \).

To find \( n \), we need to determine the number of electrons transferred in the reaction:
- \( A \) loses 2 electrons to become \( A^{2+} \).
- \( 2B^+ \) gains 2 electrons to become \( 2B \).

Thus, the total number of electrons transferred \( n \) is 2.

### Step 3: Substitute Values into the Equation
Now we can substitute \( n \) and \( K_c \) into the equation. Given \( K_c = 10^{14} \):
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \log(10^{14}) \]

### Step 4: Calculate \( \log(10^{14}) \)
Using the properties of logarithms:
\[ \log(10^{14}) = 14 \]

### Step 5: Substitute and Calculate \( E^\circ_{\text{cell}} \)
Now substituting back into the equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{2} \times 14 \]
Calculating this gives:
\[ E^\circ_{\text{cell}} = 0.0591 \times 7 = 0.413 \text{ volts} \]

### Final Answer
Thus, the standard cell potential \( E^\circ_{\text{cell}} \) is:
\[ \boxed{0.413 \text{ volts}} \]