For a cell involving one electron `E_(cell)^(0)=0.59V` and 298K, the equilibrium constant for the cell reaction is:
[Given that `(2.303RT)/(F)=0.059V` at `T=298K`]

Options

A. `1.0xx10^(30)`
B. `1.0xx10^(2)`
C. `1.0xx10^(5)`
D. `1.0xx10^(10)`

Explanation

To find the equilibrium constant \( K_c \) for the cell reaction, we can use the Nernst equation and the relationship between the standard cell potential \( E^\circ_{cell} \) and the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Standard cell potential, \( E^\circ_{cell} = 0.59 \, V \)
- Number of electrons transferred, \( N = 1 \) (since the cell involves one electron)
- At 298 K, \( \frac{2.303RT}{F} = 0.059 \, V \)

2. **Use the Nernst Equation**:
The Nernst equation at equilibrium can be simplified to:
\[ E^\circ_{cell} = \frac{0.059}{N} \log K_c \]
Since we are at equilibrium, \( E_{cell} = 0 \).

3. **Substitute the Known Values**:
Substitute \( E^\circ_{cell} \) and \( N \) into the equation:
\[ 0.59 = \frac{0.059}{1} \log K_c \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ \log K_c = \frac{0.59}{0.059} \]

5. **Calculate the Logarithm**:
Now calculate \( \frac{0.59}{0.059} \):
\[ \log K_c = 10 \]

6. **Find \( K_c \) from the Logarithm**:
To find \( K_c \), we take the antilogarithm:
\[ K_c = 10^{10} \]

7. **Final Answer**:
Therefore, the equilibrium constant \( K_c \) for the cell reaction is:
\[ K_c = 10^{10} \]