The `E_(cell)` for a given cell is `1.2346` and `1.2340V` at `300K ` and `310K`, respectively. Calculate the change in entropy during the cell reaction if the redox change involves three electrons.

Options

A. `-17.37 J K^(-1)`
B. `+17.37 J K ^(-1)`
C. `173.7J K^(-1)`
D. `5.79 J K^(-1)`

Explanation

To solve the problem of calculating the change in entropy during the cell reaction, we can follow these steps:

### Step 1: Identify the given values
- \( E_{cell} \) at 300K = 1.2346 V
- \( E_{cell} \) at 310K = 1.2340 V
- Number of electrons transferred, \( n = 3 \)
- Faraday's constant, \( F = 96500 \, \text{C/mol} \)

### Step 2: Calculate the change in cell potential (\( \Delta E \))
\[ \Delta E = E_{cell} (310K) - E_{cell} (300K) = 1.2340 \, \text{V} - 1.2346 \, \text{V} = -0.0006 \, \text{V} \]

### Step 3: Calculate the change in temperature (\( \Delta T \))
\[ \Delta T = 310K - 300K = 10K \]

### Step 4: Use the formula for change in entropy (\( \Delta S \))
The formula relating change in entropy to change in cell potential and temperature is:
\[ \Delta S = -\frac{nF \Delta E}{\Delta T} \]

### Step 5: Substitute the values into the formula
\[ \Delta S = -\frac{3 \times 96500 \, \text{C/mol} \times (-0.0006 \, \text{V})}{10 \, \text{K}} \]

### Step 6: Calculate \( \Delta S \)
\[ \Delta S = -\frac{3 \times 96500 \times -0.0006}{10} \]
\[ \Delta S = -\frac{-173.7}{10} = 17.37 \, \text{J/K} \]

### Conclusion
The change in entropy during the cell reaction is:
\[ \Delta S = 17.37 \, \text{J/K} \]