Question

Consider the cell reaction, at 300 K.
A\(^+ (aq)\) + B\(^3+\) (aq) \(\rightleftharpoons\) A\(^2+\) (aq) + B (g) Its \(E^\circ\) is 1.0 V. The \(\Delta_r H^\circ\) of the reaction is -163 kJ mol\(^{-1}\). What is \(\Delta_r S^\circ\) (in J K\(^{-1}\) mol\(^{-1}\)) of the reaction?
(F = 96500 C mol\(^{-1}\))

• A. 10
• B. 100
• C. 1000
• D. 10000

Hint:

To calculate the entropy change (\(\Delta_r S^\circ\)), use the relation between Gibbs free energy, enthalpy, and entropy: \(\Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ\).

Answer 1

The Correct Option is B

The relation between Gibbs free energy change (\(\Delta_r G^\circ\)) and the standard cell potential (\(E^\circ\)) is given by: \[ \Delta_r G^\circ = -n F E^\circ \] where: - \(n\) is the number of moles of electrons exchanged in the reaction,
- \(F\) is the Faraday constant (96500 C mol\(^{-1}\)),
- \(E^\circ\) is the standard electrode potential.
We also know that the Gibbs free energy change is related to the enthalpy change and entropy change as: \[ \Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ \] Equating both expressions for \(\Delta_r G^\circ\), we get: \[ -n F E^\circ = \Delta_r H^\circ - T \Delta_r S^\circ \] Rearranging to solve for \(\Delta_r S^\circ\): \[ \Delta_r S^\circ = \frac{\Delta_r H^\circ - (-n F E^\circ)}{T} \] Given: - \(\Delta_r H^\circ = -163 \, \text{kJ/mol} = -163000 \, \text{J/mol}\),
- \(E^\circ = 1.0 \, \text{V}\),
- \(n = 2\) (since 2 electrons are involved),
- \(F = 96500 \, \text{C/mol}\),
- \(T = 300 \, \text{K}\).
Substituting the values: \[ \Delta_r S^\circ = \frac{-163000 - (-2 \times 96500 \times 1.0)}{300} \] \[ \Delta_r S^\circ = \frac{-163000 + 193000}{300} = \frac{30000}{300} = 100 \, \text{J/K/mol} \]