For the net cell reaction of the cell `Zn(s) |Xn^(2+) ||Cd^(2+) |Cd(s) DeltaG^(@)` in Kilojoules at `25^(@)C` is `(E_(cell)^(@) = 0.360 V)`:-

Options

A. `112.5`
B. `69.47`
C. `-34.73`
D. `-69.47`

Explanation

To solve the problem of calculating ΔG° for the given electrochemical cell reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Standard cell potential (E°_cell) = 0.360 V
- Faraday's constant (F) = 96500 C/mol
- Number of electrons transferred (n) = 2 (since Zn is oxidized from Zn to Zn²⁺ and Cd²⁺ is reduced to Cd)

2. **Use the Formula for ΔG°**:
The relationship between Gibbs free energy change (ΔG°) and cell potential (E°_cell) is given by the formula:
\[ ΔG° = -nFE°_{cell} \]

3. **Substitute the Values into the Formula**:
Substitute the known values into the formula:
\[ ΔG° = -2 \times 96500 \, \text{C/mol} \times 0.360 \, \text{V} \]

4. **Calculate ΔG° in Joules**:
Performing the multiplication:
\[ ΔG° = -2 \times 96500 \times 0.360 = -69348 \, \text{J} \]

5. **Convert ΔG° from Joules to Kilojoules**:
Since the answer is required in kilojoules, convert joules to kilojoules by dividing by 1000:
\[ ΔG° = -69348 \, \text{J} \div 1000 = -69.348 \, \text{kJ} \]

6. **Round the Answer**:
Rounding to two decimal places, we get:
\[ ΔG° \approx -69.35 \, \text{kJ} \]

7. **Final Answer**:
The final answer for ΔG° is approximately:
\[ ΔG° \approx -69.35 \, \text{kJ} \]