The temperature coefficient of a cell whose operation is based on the reaction
`Pb(s)+HgCl_2(aq)toPbCl_2(aq)+Hg(l)` is :
`((dE)/(dT))_P=1.5xx10^(-4)VK^(-1)` at 298 K
The change in entropy (in J/k mol) during the operation is :

Options

A. 8627
B. 57.9
C. 28.95
D. 14.475

Answer

To find the change in entropy (ΔS) during the operation of the electrochemical cell based on the reaction:

\[ \text{Pb(s)} + \text{HgCl}_2(aq) \rightarrow \text{PbCl}_2(aq) + \text{Hg(l)} \]

we can use the relationship between the temperature coefficient of the cell (dE/dT), the number of electrons transferred (n), and Faraday's constant (F).

### Step-by-Step Solution:

1. **Identify the Reaction and Determine n:**
- The reaction involves the oxidation of lead (Pb) from an oxidation state of 0 to +2 (PbCl2) and the reduction of mercury (Hg) from +2 (HgCl2) to 0 (Hg).
- Each lead atom loses 2 electrons, and each mercury atom gains 2 electrons.
- Therefore, the number of electrons transferred in the reaction (n) is 2.

2. **Use the Given Temperature Coefficient:**
- The temperature coefficient of the cell is given as:
\[ \left( \frac{dE}{dT} \right)_P = 1.5 \times 10^{-4} \, \text{V/K} \]

3. **Use Faraday's Constant:**
- Faraday's constant (F) is approximately:
\[ F = 96500 \, \text{C/mol} \]

4. **Calculate the Change in Entropy (ΔS):**
- The relationship between ΔS, n, F, and dE/dT is given by:
\[ \Delta S = nF \left( \frac{dE}{dT} \right) \]
- Substituting the known values:
\[ \Delta S = 2 \times 96500 \, \text{C/mol} \times 1.5 \times 10^{-4} \, \text{V/K} \]

5. **Perform the Calculation:**
- Calculate the product:
\[ \Delta S = 2 \times 96500 \times 1.5 \times 10^{-4} \]
\[ \Delta S = 2 \times 96500 \times 0.00015 \]
\[ \Delta S = 28.95 \, \text{J/K/mol} \]

### Final Answer:
The change in entropy (ΔS) during the operation of the cell is:
\[ \Delta S = 28.95 \, \text{J/K/mol} \]