Problem 52 - Chemistry The Central Science

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. (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

For Part (a), the balanced redox equation is \(\mathrm{Hg}_{2}^{2+} + 2I^- \rightarrow \mathrm{I}_{2} + 2Hg\), the standard emf is \(+0.25~V\), the standard Gibbs free energy change at \(298~K\) is \(-48242.5~\text{J}~\text{mol}^{-1}\), and the equilibrium constant at \(298~K\) is approximately \(7.8 \times 10^{10}\).

Step 1: Write half-cell reactions

First, we need to write the corresponding half-cell reactions for both the oxidation and reduction processes: Oxidation half-reaction: \(\mathrm{2I}^{-} \rightarrow \mathrm{I}_{2} + 2e^-\) Reduction half-reaction: \(\mathrm{Hg}_{2}^{2+} + 2e^- \rightarrow \mathrm{2Hg}\)

Step 2: Write the balanced redox equation

Now, we combine the two half-cell equations to form a balanced redox equation: \(\mathrm{Hg}_{2}^{2+} + 2I^- \rightarrow \mathrm{I}_{2} + 2Hg\)

Step 3: Calculate the standard emf

Using the standard reduction potentials for the given half-cell reactions (which can be found in a reference table), we can calculate the standard emf as follows: \(E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation}\) For this reaction: \(E^{\circ}_{cell} = (+0.79~V) - (+0.54~V) = +0.25~V\)

Step 4: Calculate \(\Delta G^{\circ}\) at \(298~K\)

Now, we can calculate the standard Gibbs free energy change using the following relation: \(\Delta G^{\circ} = -nFE^{\circ}_{cell}\) Where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday's constant (\(96485~C~mol^{-1}\)), and \(E^{\circ}_{cell}\) is the standard emf. For this reaction: \(n=2\) (2 moles of electrons are transferred) \(\Delta G^{\circ} = -2 \times 96485~C~mol^{-1}\times 0.25~V = -48242.5~\text{J}~\text{mol}^{-1} \)

Step 5: Calculate the equilibrium constant \(K\) at \(298~K\)

Lastly, we can find the equilibrium constant \(K\) at \(298~K\) using the standard Gibbs free energy change and the following equation: \(\Delta G^{\circ} = -RT \ln{K}\) Where \(R\) is the gas constant (\(8.314~\text{J}~\text{K}^{-1}\text{mol}^{-1}\)) and \(T\) is the temperature (in Kelvin). Solving for \(K\): \( K = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{\frac{48242.5}{8.314\times298}} \approx 7.8 \times 10^{10}\) So, for Part (a), we have a balanced redox equation, a standard emf of \(0.25~V\), a standard Gibbs free energy change of \(-48242.5~\text{J}~\text{mol}^{-1}\), and an equilibrium constant of about \(7.8 \times 10^{10}\) at \(298~K\). Now, we'll proceed to Part (b) and (c) similarly. ## Part (b) ## (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. ## Part (c) ## (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\).

Key Concepts

Standard Electrode Potential

Understanding standard electrode potential is key to grasping redox reactions. It gives us insight into how likely a substance is to gain or lose electrons, which is fundamental in determining the direction of a redox reaction.

The standard electrode potential, represented as \(E^{ ext{°}}\), is measured under standard conditions: 1 M concentration, 1 atm pressure, and 25°C (298 K). The value is recorded in volts (V) and is measured relative to the standard hydrogen electrode, which is assigned a potential of 0 V.

• When \(E^{ ext{°}}\) is positive, the species is a strong oxidizing agent and has a greater tendency to gain electrons.
• When \(E^{ ext{°}}\) is negative, the species is a strong reducing agent and prefers to lose electrons.

In the reaction provided in the exercise, iodide ions are oxidized by mercury ions, and the calculated standard emf (electromotive force) of 0.25 V indicates the net electron flow in the reaction. This positive emf confirms that the reaction will proceed spontaneously under standard conditions. Calculating \(E^{ ext{°}}_{ ext{cell}}\) involves subtracting the \(E^{ ext{°}}\) of the oxidation half-reaction from the \(E^{ ext{°}}\) of the reduction half-reaction.

Gibbs Free Energy

Gibbs free energy relates closely to the spontaneity of a chemical process. This concept helps understand whether a reaction will occur without external energy input. The change in Gibbs free energy (\(\Delta G^{\text{°}}\)) is linked to the standard emf of a cell reaction.

The formula \(\Delta G^{\text{°}} = -nFE^{\text{°}}_{\text{cell}}\) calculates this energy change, where \(n\) is the number of moles of electrons exchanged, \(F\) is Faraday's constant (about \(96485\, \text{C}\, \text{mol}^{-1}\)), and \(E^{\text{°}}_{\text{cell}}\) is the cell potential.

• If \(\Delta G^{\text{°}}\) is negative, the reaction is spontaneous.
• If it's positive, the reaction needs energy to proceed.

In the given reaction, the \(\Delta G^{\text{°}}\) is calculated to be \(-48242.5 \text{ J} \text{mol}^{-1}\), indicating that the process is indeed spontaneous. This negative value aligns with the positive emf calculated earlier, further confirming that the reaction can occur on its own under standard conditions.

Equilibrium Constant

The equilibrium constant \(K\) provides vital information about the position of equilibrium in a reaction. It's a measure of the ratio of products to reactants when a reaction ceases to change over time. At equilibrium, forward and reverse reaction rates are equal, and \(K\) values can signal which side of the reaction is favored.

The relationship between \(\Delta G^{\text{°}}\) and \(K\) is given by the equation \(\Delta G^{\text{°}} = -RT\ln K\), where \(R\) is the gas constant (8.314 \(\text{J}\, \text{K}^{-1}\, \text{mol}^{-1}\)), and \(T\) is the temperature in Kelvin.

• A \(K\) much greater than 1 indicates products are heavily favored, as is the case with \(K \approx 7.8 \times 10^{10}\) in the exercise.
• Conversely, a \(K\) much smaller than 1 shows that reactants predominate.

This large \(K\) value suggests that at equilibrium, the products (\(\mathrm{I}_{2}\) and \(\mathrm{Hg}\)) are significantly more prevalent than the reactants, underlining the reaction's tendency to move towards completion.