To calculate the heat of reaction (ΔH) taking place at 25°C for the given electrochemical cell, we can use the relationship between the cell potential (E), temperature (T), and the heat of reaction. The formula we will use is:
\[
\Delta H = nFE_{cell} - \frac{dE}{dT} \cdot T
\]
Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E_{cell} \) = cell potential at the given temperature
- \( \frac{dE}{dT} \) = change in cell potential with temperature
- \( T \) = temperature in Kelvin
### Step 1: Identify the values
- Given:
- \( E_{cell} \) at 25°C = 0.256 V
- \( E_{cell} \) at 35°C = 0.2595 V
- Convert temperatures to Kelvin:
- \( T_1 = 25°C = 298 \, K \)
- \( T_2 = 35°C = 308 \, K \)
### Step 2: Calculate \( \frac{dE}{dT} \)
Using the formula for the slope:
\[
\frac{dE}{dT} = \frac{E_{cell}(T_1) - E_{cell}(T_2)}{T_1 - T_2}
\]
Substituting the values:
\[
\frac{dE}{dT} = \frac{0.256 - 0.2595}{25 - 35} = \frac{-0.0035}{-10} = 0.00035 \, V/K = 3.5 \times 10^{-4} \, V/K
\]
### Step 3: Determine \( n \)
For the given cell \( Pt(H_2)|HCl(aq)||AgCl|Ag \), the number of electrons transferred \( n \) is 1.
### Step 4: Substitute values into the ΔH equation
Now we can substitute the values into the ΔH equation:
\[
\Delta H = nFE_{cell} - \frac{dE}{dT} \cdot T
\]
Substituting the known values:
\[
\Delta H = (1)(96500 \, C/mol)(0.256 \, V) - (3.5 \times 10^{-4} \, V/K)(298 \, K)
\]
Calculating each term:
1. \( nFE_{cell} = 96500 \times 0.256 = 24704 \, J \)
2. \( \frac{dE}{dT} \cdot T = (3.5 \times 10^{-4})(298) = 0.10493 \, V \)
Now, converting \( 0.10493 \, V \) to Joules (since \( 1 \, V = 1 \, J/C \)):
\[
0.10493 \, J
\]
### Step 5: Calculate ΔH
Now, we can calculate ΔH:
\[
\Delta H = 24704 \, J - 0.10493 \, J = 24703.89507 \, J
\]
Converting to kJ:
\[
\Delta H \approx 24.7039 \, kJ
\]
### Final Answer
Thus, the heat of reaction at 25°C is approximately:
\[
\Delta H \approx 24.70 \, kJ
\]
