The temperature coefficient of a given cell , `((delE)/(delT))_P` is `1.5xx10^(-4) VK^(-1)` at 300 K. The change in entropy of cell during the cource of reaction,
`Pb((s))+ HgCl_(2)((aq))rightarrowPbCl_(2) (aq)+ Hg(1)`

Options

A. (28.95J/K)
B. (14.47J/K)
C. (57.9J/K)
D. (27.70J/K)

Explanation

To solve the problem of finding the change in entropy of the cell during the course of the reaction, we will follow these steps:

### Step 1: Understand the relationship between temperature coefficient and entropy change
The temperature coefficient of the cell, given as \(\left(\frac{\Delta E}{\Delta T}\right)_P\), is related to the change in entropy (\(\Delta S\)) by the equation:
\[ \left(\frac{\Delta E}{\Delta T}\right)_P = -\frac{\Delta S}{nF} \]
where:
- \(n\) = number of moles of electrons transferred in the reaction
- \(F\) = Faraday's constant (\(96500 \, \text{C/mol}\))

### Step 2: Identify the number of electrons transferred
From the reaction:
\[ \text{Pb(s)} + \text{HgCl}_2(\text{aq}) \rightarrow \text{PbCl}_2(\text{aq}) + \text{Hg(l)} \]
- Lead (Pb) goes from an oxidation state of 0 to +2, losing 2 electrons.
- Mercury (Hg) goes from +2 in HgCl2 to 0 in elemental Hg, gaining 2 electrons.

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

### Step 3: Substitute the known values into the equation
Given:
- \(\left(\frac{\Delta E}{\Delta T}\right)_P = 1.5 \times 10^{-4} \, \text{V/K}\)
- \(F = 96500 \, \text{C/mol}\)
- \(n = 2\)

We can rearrange the equation to solve for \(\Delta S\):
\[ \Delta S = -nF \left(\frac{\Delta E}{\Delta T}\right)_P \]

### Step 4: Calculate the change in entropy
Substituting the values:
\[ \Delta S = -2 \times 96500 \, \text{C/mol} \times 1.5 \times 10^{-4} \, \text{V/K} \]
Calculating this:
\[ \Delta S = -2 \times 96500 \times 1.5 \times 10^{-4} = -28.95 \, \text{J/K} \]

### Step 5: Interpret the result
The negative sign indicates that the reaction leads to a decrease in entropy, which is consistent with the formation of a solid from ions in solution.

### Final Answer
The change in entropy of the cell during the course of the reaction is:
\[ \Delta S = 28.95 \, \text{J/K} \]

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