Problem 52 - Chemistry The Central Science In Si Units

Question

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. \((\mathbf{c})\) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

Step-by-Step Solution

Answer

The balanced equation for (a) is: \( 2\mathrm{I}^{-} + \mathrm{Hg}_{2}^{2+} \rightarrow \mathrm{I}_{2} + 2\mathrm{Hg}^{+} \). Standard emf is \(+0.26 \, \text{V}\). \(\Delta G^{\circ} = -50172.2 \, \text{J/mol}\). \(K = 9.95 \times 10^8\).

Step 1: Write the Balanced Equation for Reaction (a)

For the reaction where aqueous iodide ion is oxidized to \( \mathrm{I}_{2}(s) \) by \( \mathrm{Hg}_{2}^{2+}(aq) \), we have: \( 2 \mathrm{I}^{-}(aq) + \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{I}_{2}(s) + 2 \mathrm{Hg}^{+}(aq) \). The iodide ions are oxidized, and the mercury(I) ions are reduced.

Step 2: Calculate the Standard EMF for Reaction (a)

The standard electrode potentials are: \( E^{\circ}(\mathrm{I}_{2}/\mathrm{I}^-) = +0.54 \, \text{V} \) and \( E^{\circ}(\mathrm{Hg}_2^{2+}/\mathrm{Hg}^+) = +0.80 \, \text{V} \). The standard emf is calculated by \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \), which equals \(+0.80 \, \text{V} - (+0.54 \, \text{V}) = +0.26 \, \text{V} \).

Step 3: Calculate \(\Delta G^{\circ}\) for Reaction (a)

Use the formula \( \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \), where \( n \) is the number of moles of electrons transferred (2 moles) and \( F \) is Faraday's constant (96485 C/mol). Therefore, \( \Delta G^{\circ} = -(2)(96485)(0.26) = -50172.2 \, \text{J/mol} \).

Step 4: Calculate the Equilibrium Constant \(K\) for Reaction (a)

Use the formula \( \Delta G^{\circ} = -RT \ln K \). Solve for \( K \): \( K = e^{\Delta G^{\circ}/-RT} \) where \( R = 8.314 \, \text{J/mol·K} \) and \( T = 298 \, \text{K} \). Substituting \( \Delta G^{\circ} = -50172.2 \) gives \( K = e^{50172.2/(8.314\times298)} \approx 9.95 \times 10^8 \).

Key Concepts

Standard Electrode Potential

The standard electrode potential, denoted as \( E^{\circ} \), is a measure of the intrinsic ability of a species to gain or lose electrons when connected to a standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It reveals if the substance is a good oxidizing or reducing agent. A higher \( E^{\circ} \) value indicates a stronger oxidizing agent, while a lower value signifies a more potent reducing agent.
In the context of our exercise, we looked at the reaction involving iodide ions being oxidized by mercurous ions to form solid iodine and mercury(I) ions. The given standard electrode potentials were \( E^{\circ}(\mathrm{I}_2/\mathrm{I}^-) = +0.54\, \text{V} \) and \( E^{\circ}(\mathrm{Hg}_2^{2+}/\mathrm{Hg}^+) = +0.80\, \text{V} \).

• The cell potential \( E^{\circ}_{\text{cell}} \) is calculated by taking the potential of the cathode and subtracting the potential of the anode.
• This resulted in a cell potential of \(+0.26\, \text{V}\).

A positive \( E^{\circ}_{\text{cell}} \) confirms the feasibility of the reaction under standard conditions. This understanding is foundational for predicting the direction of electron flow and the spontaneity of the reaction.

Gibbs Free Energy

Gibbs free energy change, represented as \( \Delta G^{\circ} \), determines whether a process will proceed spontaneously. A negative \( \Delta G^{\circ} \) indicates a spontaneous process. Its relation to electrochemistry is given by the equation:
\[ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \]
where:

• \( n \) is the number of moles of electrons transferred.
• \( F \) is Faraday's constant, approximately 96485 C/mol.
• \( E^{\circ}_{\text{cell}} \) is the standard electrode potential of the cell.

For our exercise's reaction \( a \), with two moles of electrons transferred and a cell potential of \(+0.26\, \text{V}\), the calculation yields \( \Delta G^{\circ} = -50172.2\, \text{J/mol}\). This negative value confirms the spontaneity.Understanding \( \Delta G^{\circ} \) values in electrochemistry helps determine if a particular redox reaction will occur on its own, making it crucial for practical applications like battery and fuel cell design.

Equilibrium Constant

The equilibrium constant, \( K \), is a measure of the position of equilibrium for a chemical reaction at a given temperature. It relates the concentrations of reactants and products at equilibrium. In terms of electrochemistry, \( K \) can be derived from the Gibbs free energy change using the formula:
\[ \Delta G^{\circ} = -RT \ln K \]
where:

• \( R \) is the ideal gas constant, 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin.

By rearranging this to solve for \( K \):
\[ K = e^{\Delta G^{\circ}/-RT} \]
For the exercise's example with \( \Delta G^{\circ} = -50172.2\, \text{J/mol} \), substituting \( R = 8.314\) J/mol·K and \( T = 298\, \text{K} \), we find \( K \approx 9.95 \times 10^8 \). Such a large \( K \) value indicates a reaction that heavily favors product formation at equilibrium.
Understanding how to calculate and interpret \( K \) is vital in predicting the extent of reactions, especially in designing processes and systems where controlled reaction conditions are necessary, such as environmental systems and industrial processes.