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 :

Options

A. `-4.2xx10^(-4)VK^(-1)`
B. `-3..81xx10^(-4)VK^(-1)`
C. `0.11VK^(-1)`
D. `7.62xx10^(-4)VK^(-1)`

Answer

To solve the problem, we need to find the temperature coefficient of the electromotive force (e.m.f) of the cell, denoted as \((\frac{dE}{dT})_p\). We will use the relationship between Gibbs free energy, enthalpy, and entropy for the reaction given.

### Step-by-Step Solution:

1. **Identify the given values:**
- \(\Delta H = -218 \, \text{kJ/mol} = -218 \times 10^3 \, \text{J/mol}\)
- \(E_{\text{cell}} = 1.015 \, \text{V}\)
- Temperature \(T = 300 \, \text{K}\)
- Number of electrons transferred \(n = 2\) (from the reaction stoichiometry)

2. **Convert \(\Delta H\) to Joules:**
\[ \Delta H = -218 \times 10^3 \, \text{J/mol} \]

3. **Use the relationship between Gibbs free energy, enthalpy, and entropy:**
\[ \Delta G = \Delta H - T \Delta S \]
and also,
\[ \Delta G = -nFE_{\text{cell}} \]
where \(F\) is Faraday's constant (\(F = 96500 \, \text{C/mol}\)).

4. **Set the two expressions for \(\Delta G\) equal to each other:**
\[ -nFE_{\text{cell}} = \Delta H - T \Delta S \]

5. **Rearranging for \(\Delta S\):**
\[ T \Delta S = \Delta H + nFE_{\text{cell}} \]
\[ \Delta S = \frac{\Delta H + nFE_{\text{cell}}}{T} \]

6. **Substituting the known values:**
\[ \Delta S = \frac{-218 \times 10^3 + 2 \times 96500 \times 1.015}{300} \]

7. **Calculate \(nFE_{\text{cell}}\):**
\[ nFE_{\text{cell}} = 2 \times 96500 \times 1.015 = 195,370 \, \text{J/mol} \]

8. **Substituting back into the equation for \(\Delta S\):**
\[ \Delta S = \frac{-218000 + 195370}{300} \]
\[ \Delta S = \frac{-22630}{300} = -75.4333 \, \text{J/(mol K)} \]

9. **Now, we can find \((\frac{dE}{dT})_p\):**
\[ \left(\frac{dE}{dT}\right)_p = \frac{\Delta S}{nF} \]
\[ \left(\frac{dE}{dT}\right)_p = \frac{-75.4333}{2 \times 96500} \]
\[ \left(\frac{dE}{dT}\right)_p = \frac{-75.4333}{193000} \approx -0.000390 \, \text{V/K} \]

10. **Final answer:**
\[ \left(\frac{dE}{dT}\right)_p \approx -3.9 \times 10^{-4} \, \text{V/K} \]