If the temperature coefficient of `EMF `if `-0.125V K^(-1), DeltaS` for the given cell at `25^(@)C` is `:`
`Fe|Fe^(2+)(aq)||Cd^(2+)(aq)|Cd`
1) `-26.125 k J K^(-1)`
2) `-24.125k J K^(-1)`
3) `-22.125 kJ K^(-1)`
4) `-20.125 k J K^(-1)`

Options

A. `-26.125 k J K^(-1)`
B. `-24.125k J K^(-1)`
C. `-22.125 kJ K^(-1)`
D. `-20.125 k J K^(-1)`

Explanation

To solve the problem, we need to calculate the change in entropy (ΔS) for the given electrochemical cell reaction using the provided temperature coefficient of EMF (ΔE/ΔT). Here’s the step-by-step solution:

### Step 1: Identify the cell reaction
From the cell representation \( \text{Fe} | \text{Fe}^{2+}(aq) || \text{Cd}^{2+}(aq) | \text{Cd} \), we can write the half-reactions:
- Anode: \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \)
- Cathode: \( \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \)

### Step 2: Write the overall cell reaction
The overall cell reaction can be written as:
\[ \text{Fe} + \text{Cd}^{2+} \rightarrow \text{Fe}^{2+} + \text{Cd} \]

### Step 3: Determine the number of electrons transferred (n)
From the half-reactions, we see that 2 electrons are transferred in the reaction. Therefore, \( n = 2 \).

### Step 4: Use the formula for ΔS
The relationship between the change in entropy (ΔS), the number of moles of electrons transferred (n), Faraday's constant (F), and the temperature coefficient of EMF (ΔE/ΔT) is given by:
\[ \Delta S = nF \left( \frac{\Delta E}{\Delta T} \right) \]

### Step 5: Substitute the known values
- \( n = 2 \)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( \frac{\Delta E}{\Delta T} = -0.125 \, \text{V/K} \)

Substituting these values into the formula:
\[ \Delta S = 2 \times 96500 \, \text{C/mol} \times (-0.125 \, \text{V/K}) \]

### Step 6: Calculate ΔS
Now, we perform the calculation:
\[ \Delta S = 2 \times 96500 \times (-0.125) \]
\[ \Delta S = 2 \times 96500 \times -0.125 = -24125 \, \text{J/K} \]
Converting this to kJ/K:
\[ \Delta S = -24.125 \, \text{kJ/K} \]

### Conclusion
Thus, the value of ΔS for the given cell at \( 25^\circ C \) is:
**Option 2: -24.125 kJ K^(-1)**

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