`triangleH` for the reactio `Ag(s)+(1)/(2)Hg_(2)Cl_(2)(s)toAgCo(s)+Hg(l)` is `+1280` cal at `25^(@)C` this reaction can be conducted in a cell for which the emf`=0.0455` volt at this temperature. Calculate the temperature coefficient of the emf.

Answer

To solve the problem, we need to calculate the temperature coefficient of the emf (dE/dT) for the given reaction. We will use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and the emf of the cell.

### Step-by-Step Solution:

1. **Identify Given Values**:
- ΔH = +1280 cal (enthalpy change)
- E = 0.0455 V (emf of the cell)
- T = 25°C = 298 K (temperature in Kelvin)
- n = 2 (number of electrons transferred in the reaction, as inferred from the reaction)

2. **Convert ΔH to Joules**:
Since we need to work in SI units, we convert ΔH from calories to joules:
\[ \Delta H = 1280 \, \text{cal} \times 4.184 \, \text{J/cal} = 5350.72 \, \text{J} \]

3. **Use the Gibbs Free Energy Equation**:
The relationship between ΔG, ΔH, and the emf is given by:
\[ \Delta G = \Delta H - T \Delta S \]
And also,
\[ \Delta G = -nFE \]
By equating the two expressions for ΔG, we have:
\[ -nFE = \Delta H - T \Delta S \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ T \Delta S = \Delta H + nFE \]
Thus,
\[ \Delta S = \frac{\Delta H + nFE}{T} \]

5. **Substituting Values**:
Substitute the known values into the equation:
\[ \Delta S = \frac{5350.72 \, \text{J} + (2)(96500 \, \text{C/mol})(0.0455 \, \text{V})}{298 \, \text{K}} \]

6. **Calculate nFE**:
\[ nFE = 2 \times 96500 \times 0.0455 = 8775.1 \, \text{J} \]

7. **Calculate ΔS**:
\[ \Delta S = \frac{5350.72 + 8775.1}{298} = \frac{14125.82}{298} \approx 47.4 \, \text{J/K} \]

8. **Use the Temperature Coefficient Formula**:
The temperature coefficient of the emf is given by:
\[ \frac{dE}{dT} = -\frac{nF}{T^2} \Delta S \]
Substituting the values:
\[ \frac{dE}{dT} = -\frac{(2)(96500)}{(298)^2} \times 47.4 \]

9. **Calculate dE/dT**:
\[ \frac{dE}{dT} = -\frac{193000}{88804} \times 47.4 \approx -0.0455 \, \text{V/K} \]
This gives us:
\[ \frac{dE}{dT} \approx -3.389 \times 10^{-4} \, \text{V/°C} \]

### Final Answer:
The temperature coefficient of the emf is approximately:
\[ \frac{dE}{dT} \approx -3.389 \times 10^{-4} \, \text{V/°C} \]