The standard Gibbs energy for the given cell reaction in kJ `mol^(-1)` at 298 K is
`Zn(s) + Cu^(2+) (aq) to Zn^(2+)(aq) + Cu(s), E^@ = 2V " at " 298 K `
(Faraday's constant, `F = 96000 C mol^(-1)`)

Options

A. `-192`
B. `192`
C. -384
D. 384

Explanation

To calculate the standard Gibbs energy (ΔG) for the given cell reaction at 298 K, we can use the following equation:

\[ \Delta G = -nFE^0 \]

Where:
- \( \Delta G \) = Gibbs free energy change (in joules)
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (in coulombs per mole)
- \( E^0 \) = standard cell potential (in volts)

### Step-by-step Solution:

1. **Identify the number of electrons transferred (n)**:
In the given reaction:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]
Zinc (Zn) is oxidized from 0 to +2, which means it loses 2 electrons. Copper (Cu) is reduced from +2 to 0, which means it gains 2 electrons. Therefore, \( n = 2 \).

2. **Identify the Faraday's constant (F)**:
The problem states that \( F = 96000 \, \text{C mol}^{-1} \).

3. **Identify the standard cell potential (E^0)**:
The standard cell potential is given as \( E^0 = 2 \, \text{V} \).

4. **Substitute the values into the Gibbs energy equation**:
Now we can substitute \( n \), \( F \), and \( E^0 \) into the equation:
\[ \Delta G = -nFE^0 = -2 \times 96000 \, \text{C mol}^{-1} \times 2 \, \text{V} \]

5. **Calculate ΔG**:
\[ \Delta G = -2 \times 96000 \times 2 = -384000 \, \text{J} \]

6. **Convert ΔG from joules to kilojoules**:
Since \( 1 \, \text{kJ} = 1000 \, \text{J} \):
\[ \Delta G = -384000 \, \text{J} \div 1000 = -384 \, \text{kJ} \]

### Final Answer:
The standard Gibbs energy for the given cell reaction at 298 K is:
\[ \Delta G = -384 \, \text{kJ mol}^{-1} \]