The equilibrium constant, K, is `2.0 xx 10^(19)` for the cell
`Ni(s) || Ni^(2+)(aq) || Hg_(2)^(2+)(aq) | Hg(l)`
The value of `E^(@)` at `25^(@)`C for this cell is closest to:

Options

A. `-1.14V`
B. `-0.57V`
C. `0.57V`
D. `1.14V`

Answer

To find the standard cell potential \( E^\circ \) for the given electrochemical cell, we can use the relationship between the equilibrium constant \( K \) and the standard cell potential. The equation we will use is:

\[ E^\circ = \frac{RT}{nF} \ln K \]

### Step-by-Step Solution:

1. **Identify the values needed for the equation**:
- The equilibrium constant \( K = 2.0 \times 10^{19} \)
- The temperature \( T = 25^\circ C = 298 \, K \)
- The universal gas constant \( R = 8.314 \, \text{J/(mol K)} \)
- The Faraday constant \( F = 96500 \, \text{C/mol} \)
- The number of electrons transferred \( n = 2 \) (from the reaction involving nickel and mercury).

2. **Substitute the known values into the equation**:
\[ E^\circ = \frac{(8.314 \, \text{J/(mol K)})(298 \, K)}{(2)(96500 \, \text{C/mol})} \ln(2.0 \times 10^{19}) \]

3. **Calculate \( \ln(2.0 \times 10^{19}) \)**:
\[ \ln(2.0 \times 10^{19}) = \ln(2.0) + \ln(10^{19}) = 0.693 + 19 \times 2.303 \approx 0.693 + 43.877 = 44.570 \]

4. **Calculate the numerator**:
\[ (8.314)(298) \approx 2477.572 \, \text{J/mol} \]

5. **Calculate the denominator**:
\[ (2)(96500) = 193000 \, \text{C/mol} \]

6. **Combine the results**:
\[ E^\circ = \frac{2477.572}{193000} \times 44.570 \]
\[ E^\circ \approx 0.01283 \times 44.570 \approx 0.572 \, \text{V} \]

7. **Final result**:
The value of \( E^\circ \) at \( 25^\circ C \) for this cell is approximately \( 0.57 \, \text{V} \).

### Answer:
The value of \( E^\circ \) at \( 25^\circ C \) for this cell is closest to **0.57 V**.