The temperature coefficient, of the emf, i.e., `(dE)/(dt) = -0.00065` Volt `deg^(-)` for the cell, `Cd| CdCl_(2)(1 M) ||AgCl(s)| Ag` at `25^(@)`.
Calculate the entropy change `DeltaS_(298K)` for the cell reaction,
`Cd + 2AgCl rightarrow Cd^(2+) + 2Cl^(-) + 2Ag`

Options

A. `-105.5JK^(-1)`
B. `-105.2JK^(-1)`
C. `-75.7JK^(-1)`
D. `-125.5JK^(-1)`

Answer

To solve the problem, we need to calculate the entropy change (ΔS) for the cell reaction given the temperature coefficient of the emf (dE/dT) and the cell reaction. Here’s the step-by-step solution:

### Step 1: Identify the given values
- **dE/dT** = -0.00065 V/°C
- **Temperature (T)** = 298 K
- **Cell reaction**: \( \text{Cd} + 2\text{AgCl} \rightarrow \text{Cd}^{2+} + 2\text{Cl}^{-} + 2\text{Ag} \)

### Step 2: Determine the number of moles of electrons transferred (n)
From the cell reaction, we can see that:
- 1 mole of Cd loses 2 electrons to form \( \text{Cd}^{2+} \).
Thus, \( n = 2 \).

### Step 3: Use the relationship between ΔS, dE/dT, and n
Using the formula derived from the Gibbs-Helmholtz equation:
\[ \Delta S = nF \left( \frac{dE}{dT} \right) \]
Where:
- \( F \) = Faraday's constant = 96500 C/mol

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ \Delta S = 2 \times 96500 \, \text{C/mol} \times \left( -0.00065 \, \text{V/°C} \right) \]

### Step 5: Calculate ΔS
Calculating the above expression:
\[ \Delta S = 2 \times 96500 \times -0.00065 \]
\[ \Delta S = 2 \times 96500 \times -0.00065 = -125.5 \, \text{J/K} \]

### Final Result
Thus, the entropy change \( \Delta S \) at 298 K for the cell reaction is:
\[ \Delta S = -125.5 \, \text{J/K} \]

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