Question

For the cell reaction, \[ Hg_2Cl_2 (s) + H_2 (1 \, {atm}) \rightarrow 2Hg (l) + 2H^+ (a=1) + 2Cl^- (a=1) \] The standard cell potential is \( \mathcal{E}^0 = 0.2676 \) V, and \( \left(\frac{\partial \mathcal{E}^0}{\partial T}\right)_P = -3.19 \times 10^{-4} \) V K\(^{-1}\). The standard enthalpy change of the reaction (\( \Delta_r H^0 \)) at 298 K is \( -x \) kJ mol\(^{-1}\). The value of \( x \) is ___________ (rounded off to two decimal places). [Given: Faraday constant \( F = 96500 \) C mol\(^{-1}\)]

Hint:

The Nernst equation relates the cell potential to the standard cell potential and the activities of the reactants and products. The temperature dependence of the cell potential is related to the entropy change of the reaction. The Gibbs-Helmholtz equation relates the enthalpy change to the Gibbs free energy change and the entropy change.

Answer 1

The relationship between the standard Gibbs free energy change (\( \Delta_r G^0 \)), the standard cell potential (\( \mathcal{E}^0 \)), and the number of electrons transferred (\( n \)) is: \[ \Delta_r G^0 = -n F \mathcal{E}^0 \] The number of electrons transferred in the given cell reaction can be determined by looking at the oxidation and reduction half-reactions: Oxidation: \( H_2 (g) \rightarrow 2H^+ (aq) + 2e^- \) (\(n=2\)) Reduction: \( Hg_2Cl_2 (s) + 2e^- \rightarrow 2Hg (l) + 2Cl^- (aq) \) (\(n=2\)) So, the number of electrons transferred is \( n = 2 \). Now, calculate the standard Gibbs free energy change at 298 K: \[ \Delta_r G^0 = -(2 \, \text{mol e}^-) \times (96500 \, \text{C mol}^{-1}) \times (0.2676 \, \text{V}) \] \[ \Delta_r G^0 = -51658.8 \, \text{J mol}^{-1} = -51.6588 \, \text{kJ mol}^{-1} \] The temperature dependence of the standard Gibbs free energy change is given by: \[ \left(\frac{\partial \Delta_r G^0}{\partial T}\right)_P = -\Delta_r S^0 \] We also know that: \[ \left(\frac{\partial \mathcal{E}^0}{\partial T}\right)_P = \frac{\Delta_r S^0}{nF} \] So, the standard entropy change (\( \Delta_r S^0 \)) is: \[ \Delta_r S^0 = nF \left(\frac{\partial \mathcal{E}^0}{\partial T}\right)_P = (2 \, \text{mol e}^-) \times (96500 \, \text{C mol}^{-1}) \times (-3.19 \times 10^{-4} \, \text{V K}^{-1}) \] \[ \Delta_r S^0 = -61.607 \, \text{J mol}^{-1} \text{K}^{-1} = -0.061607 \, \text{kJ mol}^{-1} \text{K}^{-1} \] The standard enthalpy change (\( \Delta_r H^0 \)) is related to \( \Delta_r G^0 \) and \( \Delta_r S^0 \) by: \[ \Delta_r H^0 = \Delta_r G^0 + T \Delta_r S^0 \] At 298 K: \[ \Delta_r H^0 = -51.6588 \, \text{kJ mol}^{-1} + (298 \, \text{K}) \times (-0.061607 \, \text{kJ mol}^{-1} \text{K}^{-1}) \] \[ \Delta_r H^0 = -51.6588 \, \text{kJ mol}^{-1} - 18.358886 \, \text{kJ mol}^{-1} \] \[ \Delta_r H^0 = -70.017686 \, \text{kJ mol}^{-1} \] Given that \( \Delta_r H^0 = -x \) kJ mol\(^{-1}\), the value of \( x \) is 70.017686. Rounding off to two decimal places, \( x = 70.02 \). This falls within the given range of 69.00 to 71.00.