Question

Given that: $$ \begin{array}{ll} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)$$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

Short answer

\(\Delta G^{\circ} = 13.51 \ \mathrm{kJ/mol}\) and \(K = 4.33 \times 10^{-3}\).

Step-by-step solution

Write the Balanced Equation for the Disproportionation Reaction

We need to determine the balanced equation for the given reaction. The given process is: \[ \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{Hg}^{2+}(aq) + \mathrm{Hg}(l) \]This implies that the mercury from \(\mathrm{Hg}_{2}^{2+}\) is both oxidized to \(\mathrm{Hg}^{2+}\) and reduced to \(\mathrm{Hg}(l)\). This is a balanced equation itself.

Determine the Standard Electrode Potential for the Overall Reaction

Use the provided standard electrode potentials to calculate the net standard potential for the reaction.Given:1. \[ 2\mathrm{Hg}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Hg}_{2}^{2+}(aq), \ E^{\circ} = 0.92 \mathrm{V} \]2. \[ \mathrm{Hg}_{2}^{2+}(aq) + 2e^{-} \rightarrow 2\mathrm{Hg}(l), \ E^{\circ} = 0.85 \mathrm{V} \]We reverse the first reaction to match the stoichiometry of the process:\[ \mathrm{Hg}_{2}^{2+}(aq) \rightarrow 2 \mathrm{Hg}^{2+}(aq) + 2e^{-}, \ E^{\circ} = -0.92 \mathrm{V} \]To find the \(E^{\circ}_{\text{net}}\), we add the reversed first reaction to the second:\[ E^{\circ}_{\text{net}} = -0.92 \mathrm{V} + 0.85 \mathrm{V} = -0.07 \mathrm{V} \]

Calculate Standard Free Energy Change (\(\Delta G^{\circ}\))

We use the relation between \(\Delta G^{\circ}\) and the standard cell potential:\[ \Delta G^{\circ} = -nFE^{\circ}_{\text{net}} \]where \(n\) is the number of moles of electrons exchanged (which is 2 in this reaction), \(F\) is the Faraday constant \((96485 \ \mathrm{C/mol})\), and \(E^{\circ}_{\text{net}} = -0.07 \ \mathrm{V}\).\[ \Delta G^{\circ} = -2 \times 96485 \ \mathrm{C/mol} \times -0.07 \ \mathrm{V} = 13507.9 \ \mathrm{J/mol} \approx 13.51 \ \mathrm{kJ/mol} \]

Calculate the Equilibrium Constant (\(K\))

We use the relation between \(\Delta G^{\circ}\) and the equilibrium constant \(K\):\[ \Delta G^{\circ} = -RT\ln K \]where \(R = 8.314 \ \mathrm{J/mol \cdot K}\) and \(T = 298 \ \mathrm{K}\).Rearranging and solving for \(K\):\[ K = e^{-\Delta G^{\circ}/RT} \]\[ K = e^{-13507.9 / (8.314 \times 298)} \approx e^{-5.44} \approx 4.33 \times 10^{-3} \]

Key concepts

Disproportionation Reaction

In the realm of electrochemistry, a disproportionation reaction is a type of redox reaction where a single element undergoes both oxidation and reduction in the same reaction. This means the element increases its oxidation state in one part of the reaction and decreases it in another.
For example, in the given problem, the reaction is \( \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{Hg}^{2+}(aq) + \mathrm{Hg}(l) \). Here, mercury in the \( \mathrm{Hg}_{2}^{2+} \) ion is both oxidized and reduced:

• Oxidation: \( \mathrm{Hg}_{2}^{2+} \) splits, with one mercury forming \( \mathrm{Hg}^{2+} \), increasing its oxidation state.
• Reduction: The other mercury forms \( \mathrm{Hg}(l) \), decreasing its oxidation state to zero.

This kind of reaction is significant because it reveals how elements can stabilize themselves through simultaneous electron gain and loss.

Standard Electrode Potential

The standard electrode potential \( (E^{\circ}) \) is a measure of the tendency of a chemical species to gain or lose electrons in an electrochemical cell. This value is determined under standard conditions, which include a concentration of 1 mol/L for ions, 1 atm pressure, and a temperature of 25°C (298 K).
In the given exercise, two standard electrode potentials are given:

• For \( \mathrm{Hg}_{2}^{2+} \) gaining electrons to form \( \mathrm{Hg}(l) \), the \( E^{\circ} = 0.85 \, \mathrm{V} \)
• For the reverse reaction, where \( \mathrm{Hg}_{2}^{2+} \) loses electrons to form \( 2\mathrm{Hg}^{2+} \), the \( E^{\circ} = -0.92 \, \mathrm{V} \)

To find the net electrode potential for the overall reaction, these values are combined, leading to a net potential\( (E^{\circ}_{\text{net}}) \) of \(-0.07 \, \mathrm{V} \). This value indicates the non-spontaneity of the disproportionation reaction since a negative potential implies a non-spontaneous reaction under standard conditions.

Standard Free Energy Change

The standard free energy change \( (\Delta G^{\circ}) \) of a reaction is directly related to the standard electrode potential by the equation \( \Delta G^{\circ} = -nFE^{\circ} \). Here,

• \( n \) is the number of moles of electrons exchanged in the reaction,
• \( F \) is Faraday's constant (\( 96485 \, \mathrm{C/mol} \)), and
• \( E^{\circ} \) is the standard electrode potential.

For the process \( \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{Hg}^{2+}(aq) + \mathrm{Hg}(l) \), \( n = 2 \). Plugging in the values gives:
\[ \Delta G^{\circ} = -2 \times 96485 \, \mathrm{C/mol} \times -0.07 \, \mathrm{V} = 13507.9 \, \mathrm{J/mol} \approx 13.51 \, \mathrm{kJ/mol} \]
This positive \( \Delta G^{\circ} \) value indicates that the disproportionation reaction is non-spontaneous under standard conditions.

Equilibrium Constant

The equilibrium constant \( (K) \) of a reaction is closely tied to the standard free energy change through the equation \( \Delta G^{\circ} = -RT\ln K \). In this equation:

• \( R \) is the universal gas constant (8.314 J/mol·K),
• \( T \) is the temperature (in Kelvin), and
• \( \ln K \) is the natural logarithm of the equilibrium constant.

To find \( K \) for the reaction \( \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{Hg}^{2+}(aq) + \mathrm{Hg}(l) \), we rearrange the equation:
\[ K = e^{-\Delta G^{\circ}/RT} \]
Substituting our values gives:
\[ K = e^{-13507.9 / (8.314 \times 298)} \approx e^{-5.44} \approx 4.33 \times 10^{-3} \]
This small \( K \) value suggests that the reaction heavily favors the reactants, consistent with the positive \( \Delta G^{\circ} \) and negative \( E^{\circ}_{\text{net}} \), reinforcing the non-spontaneity of the reaction at equilibrium.