Question

For the half-cell reaction \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{e}^{-} \rightarrow\) \(2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q), E^{\circ}=+0.27 \mathrm{V} .\) Using this result and \(\Delta G_{f}^{\circ}\left(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, s\right)=-210.7 \mathrm{kJ} \mathrm{mol}^{-1},\) determine \(\Delta G_{f}^{\circ}\left(\mathrm{Cl}^{-}, a q\right)\).

Short answer

The standard Gibbs free energy of formation for the chloride anion, \(\Delta G_f^\circ\left(\mathrm{Cl}^-\right)\), is approximately \(-131.36\,\mathrm{kJ/mol}\).

Step-by-step solution

Use the Nernst Equation to Relate \(E^\circ\) and \(\Delta G^\circ\)

The first step in solving this problem is to relate the standard electrode potential, \(E^\circ\), and the standard Gibbs free energy change, \(\Delta G^\circ\), using the Nernst Equation. The Nernst Equation can be written as: \[\Delta G^\circ = -nFE^\circ\] where \(\Delta G^\circ\) is the standard Gibbs free energy change, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant (96,485 C/mol), and \(E^\circ\) is the standard electrode potential.

Write the Balanced Chemical Equation

The balanced half-cell reaction is given as: \[\mathrm{Hg}_2\mathrm{Cl}_2(s) + 2\mathrm{e}^- \rightarrow 2\mathrm{Hg}(l) + 2\mathrm{Cl}^-(aq)\]

Calculate the Standard Gibbs Free Energy Change of the Reaction

To calculate the standard Gibbs free energy change of the reaction, we will use the given standard electrode potential, \(E^\circ = +0.27 V\). Since the balanced chemical equation involves the transfer of 2 electrons (n = 2), we can now apply the Nernst Equation to find the standard Gibbs free energy change: \[\Delta G^\circ = -nFE^\circ = -(2)(96485\,\mathrm{C/mol})(0.27\,\mathrm{V})\] \[\Delta G^\circ = -52010\,\mathrm{J/mol} = -52.01\,\mathrm{kJ/mol}\]

Use the Standard Gibbs Free Energy Change to Find \(\Delta G_f^\circ\left(\mathrm{Cl}^-\right)\)

Finally, we can use the standard Gibbs free energy change of the reaction and the standard Gibbs free energy of formation for Hg2Cl2 to determine the standard Gibbs free energy of formation for Cl-: \[\Delta G^\circ = \Delta G_f^\circ\left(\mathrm{Cl}^-\right) + \Delta G_f^\circ\left(\mathrm{Hg}\right) + 2\Delta G_f^\circ\left(\mathrm{Cl}^-\right) - \Delta G_f^\circ\left(\mathrm{Hg}_2\mathrm{Cl}_2\right) \] Since the standard free energy formation of elements in their standard states (Hg in this case) is zero we can rewrite the equation as: \[\Delta G^\circ = \Delta G_f^\circ\left(\mathrm{Cl}^-\right) + 2\Delta G_f^\circ\left(\mathrm{Cl}^-\right) - \Delta G_f^\circ\left(\mathrm{Hg}_2\mathrm{Cl}_2\right)\] Rearrange the equation to find the standard Gibbs free energy of formation for Cl-: \[\Delta G_f^\circ\left(\mathrm{Cl}^-\right)= \frac{1}{2} \Delta G^\circ + \frac{1}{2}\Delta G_f^\circ\left(\mathrm{Hg}_2\mathrm{Cl}_2\right)\] Now substitute the values from Steps 3 and the given value of \(\Delta G_f^\circ\left(\mathrm{Hg}_2\mathrm{Cl}_2\right)\): \[\Delta G_f^\circ\left(\mathrm{Cl}^-\right) = \frac{1}{2}(-52.01\,\mathrm{kJ/mol}) + \frac{1}{2}(-210.7\,\mathrm{kJ/mol})\] \[\Delta G_f^\circ\left(\mathrm{Cl}^-\right) = -26.005\,\mathrm{kJ/mol} - 105.35\,\mathrm{kJ/mol}\] \[\Delta G_f^\circ\left(\mathrm{Cl}^-\right) = -131.355\,\mathrm{kJ/mol}\] Therefore, the standard Gibbs free energy of formation for the chloride anion, \(\Delta G_f^\circ\left(\mathrm{Cl}^-\right)\), is approximately \(-131.36\,\mathrm{kJ/mol}\).

Key concepts

Nernst Equation

Understanding the Nernst Equation is essential for grasping the relationship between the electrical potential of a cell and the concentration of ions participating in the electrochemical reaction. In simple terms, the Nernst Equation enables us to calculate the electrode potential under non-standard conditions, which depends on the ion concentration.

When we look at the Nernst Equation, which is an expression of electrochemical equilibrium, it's given by the formula: