The standard electrode potential for Daniel cell is `1.1V`. Calculate the standard Gibbs energy of the reaction (In KJ/mol)
`Zn_((s))+Cu_((aq))^(2+)rarrZn_((aq))^(2+)+Cu_((s))`

Options

A. `106.15`
B. `212.3`
C. `193`
D. `403`

Explanation

To calculate the standard Gibbs energy (ΔG°) of the reaction for the Daniel cell, we can use the relationship between Gibbs free energy and standard electrode potential:

\[ \Delta G° = -nFE° \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E° \) = standard electrode potential (in volts)

### Step-by-Step Solution:

1. **Identify the number of electrons transferred (n)**:
In the given reaction:
\[ \text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)} \]
Each zinc atom loses 2 electrons to become \( \text{Zn}^{2+} \) and each copper ion gains 2 electrons to become copper metal. Therefore, the number of electrons transferred \( n \) is 2.

**Hint**: Look for how many electrons are involved in the oxidation and reduction processes in the half-reactions.

2. **Use the given standard electrode potential (E°)**:
The standard electrode potential for the Daniel cell is given as \( E° = 1.1 \, V \).

**Hint**: Remember that the standard electrode potential is a measure of the driving force behind the redox reaction.

3. **Use Faraday's constant (F)**:
The value of Faraday's constant is \( F = 96500 \, C/mol \).

**Hint**: Faraday's constant represents the charge of one mole of electrons.

4. **Substitute the values into the Gibbs energy equation**:
Now we can substitute the values into the equation:
\[ \Delta G° = -nFE° \]
\[ \Delta G° = -2 \times 96500 \, C/mol \times 1.1 \, V \]

5. **Calculate ΔG°**:
\[ \Delta G° = -2 \times 96500 \times 1.1 = -212130 \, J/mol \]

6. **Convert to kilojoules**:
To convert joules to kilojoules, divide by 1000:
\[ \Delta G° = -212130 \, J/mol \div 1000 = -212.13 \, kJ/mol \]

### Final Answer:
\[ \Delta G° \approx -212.3 \, kJ/mol \]

### Summary:
The standard Gibbs energy of the reaction for the Daniel cell is approximately \(-212.3 \, kJ/mol\).