At 300 k, `/_\ H`for the reaction
`Zn(s)+AgCl(s)toZnCl_2(aq)+2Ag(s)` is
`-218 KJ/mol `while the e.m.f of the cell was 1.015V. `((dE)/(dT))_p` of the cell is `-0.000381 VK^(-1)`. Calculate `/_\S` for the given cell reaction.

Options

A. -73.53J/Kmol
B. 83.53J/Kmol
C. 100J/Kmol
D. None of these

Answer

To solve the problem, we need to calculate the change in entropy (ΔS) for the given cell reaction using the provided information. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the Reaction and Given Data
The reaction is:
\[ \text{Zn(s)} + \text{AgCl(s)} \rightarrow \text{ZnCl}_2(aq) + 2\text{Ag(s)} \]

Given data:
- ΔH = -218 kJ/mol
- e.m.f (E) = 1.015 V
- \(\left(\frac{dE}{dT}\right)_p = -0.000381 \, \text{V/K}\)

### Step 2: Determine the Number of Electrons Transferred (n)
In the reaction:
- Zinc (Zn) is oxidized from 0 to +2 (loses 2 electrons).
- Silver (Ag) is reduced from +1 to 0 (gains 1 electron per atom, and there are 2 Ag atoms).

Thus, the total number of electrons transferred (n) in the reaction is:
\[ n = 2 \]

### Step 3: Relate ΔS to \(\left(\frac{dE}{dT}\right)_p\)
The relationship between the change in entropy (ΔS) and the temperature coefficient of the cell (\(\left(\frac{dE}{dT}\right)_p\)) is given by:
\[ \left(\frac{dE}{dT}\right)_p = \frac{\Delta S}{nF} \]

Where:
- F = Faraday's constant = 96500 C/mol

### Step 4: Rearranging the Equation to Solve for ΔS
Rearranging the equation gives:
\[ \Delta S = \left(\frac{dE}{dT}\right)_p \cdot nF \]

### Step 5: Substitute the Values
Substituting the known values:
\[ \Delta S = \left(-0.000381 \, \text{V/K}\right) \cdot (2) \cdot (96500 \, \text{C/mol}) \]

### Step 6: Calculate ΔS
Calculating the value:
\[ \Delta S = -0.000381 \cdot 2 \cdot 96500 \]
\[ \Delta S = -0.000381 \cdot 193000 \]
\[ \Delta S = -73.53 \, \text{J/K/mol} \]

### Final Answer
Thus, the change in entropy (ΔS) for the given cell reaction is:
\[ \Delta S = -73.53 \, \text{J/K/mol} \]

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