To find the change in enthalpy (ΔH) for the given electrochemical cell reaction, we can use the relationship between the cell potential (E), temperature (T), and the Gibbs free energy change (ΔG) as follows:
1. **Identify the Reaction**: The cell reaction can be represented as:
\[
\text{Cd}(s) + 2\text{AgCl}(s) \rightarrow \text{CdCl}_2(aq) + 2\text{Ag}(s)
\]
2. **Calculate the Change in Cell Potential (ΔE)**: We have two cell potentials given at two different temperatures:
- \( E_1 = 0.6915 \, V \) at \( T_1 = 0 \, °C \) (which is 273 K)
- \( E_2 = 0.6753 \, V \) at \( T_2 = 25 \, °C \) (which is 298 K)
The change in cell potential (ΔE) can be calculated as:
\[
\Delta E = E_2 - E_1 = 0.6753 \, V - 0.6915 \, V = -0.0162 \, V
\]
3. **Calculate the Change in Gibbs Free Energy (ΔG)**: The relationship between ΔG and E is given by:
\[
\Delta G = -nFE
\]
where:
- \( n \) = number of moles of electrons transferred (for this reaction, \( n = 2 \))
- \( F \) = Faraday's constant \( (96500 \, C/mol) \)
Thus, we can calculate ΔG at 298 K:
\[
\Delta G = -2 \times 96500 \, C/mol \times 0.6753 \, V = -130,000 \, J = -130 \, kJ
\]
4. **Use the Gibbs-Helmholtz Equation**: The relationship between ΔG, ΔH, and temperature is given by:
\[
\Delta G = \Delta H - T\Delta S
\]
Rearranging gives:
\[
\Delta H = \Delta G + T\Delta S
\]
Since we do not have ΔS directly, we can assume it to be constant over the temperature range. However, for simplicity, we can calculate ΔH using the average temperature and ΔE.
5. **Calculate ΔH**: Using the average temperature (T = 298 K):
\[
\Delta H \approx \Delta G + T\left(\frac{\Delta E}{\Delta T}\right)
\]
where \( \Delta T = T_2 - T_1 = 298 - 273 = 25 \, K \).
We can assume ΔS to be negligible for this calculation, thus:
\[
\Delta H \approx \Delta G = -130 \, kJ
\]
Thus, the change in enthalpy (ΔH) of the reaction at 25 °C is approximately **-130 kJ**.
