Question

At 300 K, the \( E^\ominus_{cell} \) of \[ \text{A(s)} + \text{B}^{2+}(\text{aq}) \rightleftharpoons \text{A}^{2+}(\text{aq}) + \text{B(s)} \] is 1.0 V. If \( \Delta_r S^\ominus \) of this reaction is 100 J K\(^{-1}\), what is \( \Delta_r H^\ominus \) (in kJ mol\(^{-1}\)) of this reaction?
(F = 96500 C mol\(^{-1}\))

• A. -163
• B. -223
• C. -193
• D. -163000

Hint:

Remember to convert all units to a consistent system (e.g., Joules and Kelvin) before performing calculations.

Answer 1

The Correct Option is A

We are given \( E^\ominus_{cell} = 1.0 \) V, \( \Delta_r S^\ominus = 100 \) J K\(^{
-1}\), and \( T = 300 \) K. We need to find \( \Delta_r H^\ominus \). First, we find \( \Delta_r G^\ominus \) using the formula: \[ \Delta_r G^\ominus =
-nFE^\ominus_{cell} \] where \( n \) is the number of electrons transferred, which is 2 in this case. \[ \Delta_r G^\ominus =
-2 \times 96500 \text{ C mol}^{
-1} \times 1.0 \text{ V} =
-193000 \text{ J mol}^{
-1} =
-193 \text{ kJ mol}^{
-1} \] Now, we use the Gibbs
-Helmholtz equation: \[ \Delta_r G^\ominus = \Delta_r H^\ominus
- T\Delta_r S^\ominus \] Rearranging for \( \Delta_r H^\ominus \): \[ \Delta_r H^\ominus = \Delta_r G^\ominus + T\Delta_r S^\ominus \] Substituting the given values: \[ \Delta_r H^\ominus =
-193000 \text{ J mol}^{
-1} + 300 \text{ K} \times 100 \text{ J K}^{
-1} \text{mol}^{
-1} \] \[ \Delta_r H^\ominus =
-193000 \text{ J mol}^{
-1} + 30000 \text{ J mol}^{
-1} =
-163000 \text{ J mol}^{
-1} =
-163 \text{ kJ mol}^{
-1} \]