The standard electrode potential E° and its temperature coefficient $$\left(\frac{dE}{dT}\right)$$ for a cell are 2 V and $$-5 \times 10^{-4}$$ V K$$^{-1}$$ at 300 K, respectively. The reaction is Zn(s) + Cu$$^{2+}$$(aq) $$\to$$ Zn$$^{2+}$$(aq) + Cu(s). The standard reaction enthalpy $$(\Delta_r H^-)$$ at 300 K in mol$$^{-1}$$ is [Use R = 8 J K$$^{-1}$$ mol$$^{-1}$$ and F = 96,500 C mol$$^{-1}$$]

Given Data from the Problem

• Standard cell potential (E) = 2 V
• Temperature coefficient (dE/dT) = -5 * 10^-4 V K^-1
• Temperature (T) = 300 K
• Faraday's constant (F) = 96,500 C mol^-1
• Number of electrons transferred (n) = 2 (since Zn changes to Zn2+ and Cu2+ changes to Cu)

2. Relevant Formulas

To find the standard reaction enthalpy (H), we use the thermodynamic relationship for electrochemical cells:

H = G + T * S

Where:

• G = -n * F * E (Standard Gibbs free energy change)
• S = n * F * (dE/dT) (Standard entropy change)

Substituting G and S into the main equation gives: H = -n * F * E + T * [n * F * (dE/dT)]

Factoring out (-n * F) simplifies the equation to: H = -n * F * [E - T * (dE/dT)]

3. Step-by-Step Calculation

Step 3A: Calculate the term inside the brackets [E - T * (dE/dT)]

• T * (dE/dT) = 300 * (-5 * 10^-4)
• T * (dE/dT) = -1500 * 10^-4
• T * (dE/dT) = -0.15 V

Now substitute this back into the bracket:

• [E - T * (dE/dT)] = 2 - (-0.15)
• [E - T * (dE/dT)] = 2 + 0.15 = 2.15 V

Step 3B: Calculate the total Enthalpy (H)

• H = -n * F * 2.15
• H = -2 * 96,500 * 2.15
• H = -193,000 * 2.15
• H = -414,950 J mol^-1

Step 3C: Convert Joules to Kilojoules (kJ mol^-1)

• H = -414,950 / 1000
• H = -414.95 kJ mol^-1