The temperature coefficient of the emf i.e. `(dE)/(dT)=0.00065"volt".deg^(-1)` for the cell `Cd|CdCl_(2)(1M)||AgCl(s)`
`|Ag` at `25^(@)C` calculate the entropy changes `triangleS_(298K)` for the cell reaction, `Cd+2AgCltCd^(++)+2Cl^(-)+2Ag`

Options

A. `-105.5JK^(-1)`
B. `-150.2JK^(-1)`
C. `-75.7JK^(-1)`
D. `-125.5JK^(-1)`

Answer

To calculate the entropy change (ΔS) for the cell reaction \( \text{Cd} + 2\text{AgCl} \rightarrow \text{Cd}^{2+} + 2\text{Cl}^- + 2\text{Ag} \) at 298 K, we can follow these steps:

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
We know that:
\[ \Delta G = \Delta H - T \Delta S \]
Also, from electrochemistry, we have:
\[ \Delta G = -nFE \]
where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E \) = cell potential (emf)

### Step 2: Relate the two equations
Setting the two expressions for ΔG equal to each other gives:
\[ -nFE = \Delta H - T \Delta S \]
Rearranging this equation, we find:
\[ T \Delta S = \Delta H + nFE \]

### Step 3: Use the temperature coefficient of emf
Given that the temperature coefficient of the emf is:
\[ \frac{dE}{dT} = 0.00065 \, \text{V/°C} \]
At \( T = 298 \, \text{K} \) (or \( 25 \, \text{°C} \)), we can express the change in emf with respect to temperature as:
\[ E = E_0 + \left(\frac{dE}{dT}\right)(T - T_0) \]
However, we can directly use the temperature coefficient for our calculations.

### Step 4: Calculate ΔS
From the relationship derived earlier, we can express ΔS as:
\[ \Delta S = \frac{\Delta H + nFE}{T} \]
But we will also use the temperature coefficient to find ΔS directly:
\[ \Delta S = nF \left(\frac{dE}{dT}\right) \]

### Step 5: Determine n and calculate ΔS
In our reaction, \( n = 2 \) (since 2 moles of electrons are transferred). Thus:
\[ \Delta S = nF \left(\frac{dE}{dT}\right) = 2 \times 96500 \, \text{C/mol} \times 0.00065 \, \text{V/°C} \]

Calculating this gives:
\[ \Delta S = 2 \times 96500 \times 0.00065 = 125.45 \, \text{J/K} \]

### Step 6: Consider the sign of ΔS
Since the reaction is spontaneous in the forward direction, we take the negative of the calculated value for ΔS:
\[ \Delta S = -125.45 \, \text{J/K} \]

### Final Answer
Thus, the entropy change \( \Delta S \) for the cell reaction at 298 K is:
\[ \Delta S = -125.45 \, \text{J/K} \]

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