Problem 104 - Chemistry The Molecular Science

Question

Consider these unbalanced equations for two reactions: \(\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Hg}(\ell) \longrightarrow \underset{\mathrm{Hg}_{2}^{2+}}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) $$ \mathrm{Hg}^{2+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Hg}_{2}^{2+}(\mathrm{aq})+\mathrm{Br}_{2}(\ell) $$ For each reaction: (a) balance the equation. (b) draw a cell diagram. (c) calculate the standard cell potential. (d) calculate the \(\Delta_{t} G^{\circ} .\) (e) determine whether each reaction is product-favored. Explain your reason.

Step-by-Step Solution

Answer

Both reactions are balanced, product-favored, and have positive cell potentials.

Step 1: Balance the Equations

For the first reaction: 1. Balance the elements that appear in one compound on each side, starting with Hg and NO: \( 2 \text{NO}_3^- + 8 \text{H}^+ + 3 \text{Hg} \rightarrow \text{Hg}_2^{2+} + 2 \text{NO} + 4 \text{H}_2\text{O} \)2. For the second reaction, balance using: \( 2 \text{Hg}^{2+} + 2 \text{Br}^- \rightarrow \text{Hg}_2^{2+} + \text{Br}_2 \).

Step 2: Draw Cell Diagrams

For each balanced reaction, the cell diagram is as follows:1. For the first reaction: \( \text{Hg}(\ell) \mid \text{Hg}_2^{2+}(\text{aq}) \mid \text{NO}(\text{g}) , \text{NO}_3^-(\text{aq}) , \text{H}^+(\text{aq}) \)2. For the second reaction: \( \text{Br}_2(\ell) \mid \text{Br}^-(\text{aq}) \mid \text{Hg}_2^{2+}(\text{aq}) \mid \text{Hg}^{2+}(\text{aq}) \).

Step 3: Calculate Standard Cell Potentials

1. Look up standard reduction potentials from tables: - \( \text{Hg}_2^{2+} + 2\text{e}^- \rightarrow 2\text{Hg} \): +0.85 V - \( \text{NO}_3^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{NO} + 2\text{H}_2\text{O} \): +0.96 V2. Standard cell potential for 1st reaction: \[ E^\circ_{\text{cell, 1st}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.96 - 0.85 = 0.11 \text{ V} \]3. For the second reaction: - \( \text{Br}_2 + 2\text{e}^- \rightarrow 2\text{Br}^- \): +1.07 V4. Standard cell potential for 2nd reaction: \[ E^\circ_{\text{cell, 2nd}} = 1.07 - 0.85 = 0.22 \text{ V} \].

Step 4: Calculate Gibbs Free Energy Change

Using \( \Delta G^\circ = -nFE^\circ \):1. For the first reaction, \( n = 6 \text{ moles of electrons} \): \[ \Delta G_1^\circ = -(6)(96485)(0.11) = -63618 \text{ J/mol} \]2. For the second reaction, \( n = 2 \text{ moles of electrons} \): \[ \Delta G_2^\circ = -(2)(96485)(0.22) = -42454 \text{ J/mol} \].

Step 5: Determine Reaction Favorability

1. Both reactions are product-favored if \( \Delta G^\circ < 0 \) and/or \( E^\circ_{\text{cell}} > 0 \).2. Since \( \Delta G_1^\circ = -63618 \text{ J/mol} \) and \( E^\circ_{\text{cell, 1st}} = 0.11 \text{ V} \), the first reaction is product-favored.3. Since \( \Delta G_2^\circ = -42454 \text{ J/mol} \) and \( E^\circ_{\text{cell, 2nd}} = 0.22 \text{ V} \), the second reaction is also product-favored.

Key Concepts

Cell Potential

In electrochemistry, the cell potential, often termed as electromotive force (EMF), is a measure of the potential difference between two electrodes in an electrochemical cell. This potential difference drives the redox reactions that occur in these cells. It is measured in volts (V). To compute the standard cell potential, one must find the difference between the reduction potential at the cathode and the reduction potential at the anode.

This standard cell potential is determined using a table of standard reduction potentials. The formula to calculate it is given by:

• For the first reaction: \( E^\circ_{\text{cell, 1st}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.96 - 0.85 = 0.11 \text{ V} \)
• For the second reaction: \( E^\circ_{\text{cell, 2nd}} = 1.07 - 0.85 = 0.22 \text{ V} \)

The positive value of the cell potential indicates that the reaction is spontaneous under standard conditions. When undertaking electrochemical calculations, the choice of the cathode and anode is crucial, as the direction of electron flow dictates the reaction process.Breaking this down further, a positive cell potential signifies that electrons will flow from the anode (where oxidation occurs) to the cathode (where reduction takes place) spontaneously. A cell diagram can help visually represent the electrochemical cell and track the flow of electrons and ions in the system, emphasizing the half-reactions that occur in the cell compartments.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a thermodynamic quantity that helps predict whether a process or a reaction is spontaneous. Simply put, if the change in Gibbs Free Energy (\(\Delta G\)) is negative, the reaction is spontaneous and product-favored under constant temperature and pressure.

For an electrochemical cell, \(\Delta G\) is closely related to the cell potential. It can be calculated by using the equation:

• \( \Delta G^\circ = -nFE^\circ \)

Where:

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

For instance, considering the examples provided:

• For the first reaction, using \(n = 6 ext{ moles of electrons}\): \[ \Delta G_1^\circ = -(6)(96485)(0.11) = -63618 \text{ J/mol} \]
• For the second reaction with \(n = 2 ext{ moles of electrons}\):\[ \Delta G_2^\circ = -(2)(96485)(0.22) = -42454 \text{ J/mol} \]

A negative \(\Delta G^\circ\), as calculated above, confirms that both reactions are product-favored, meaning the reactions proceed on their own under standard conditions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes in which electrons are transferred between two substances. During these reactions, one substance gains electrons (reduction) while the other loses electrons (oxidation).

Oxidation refers to the loss of electrons. Typically, metals tend to lose electrons and become oxidized, as seen with \(\text{Hg}\) being oxidized to \(\text{Hg}_2^{2+}\) in the provided reactions. On the other hand, reduction is the gain of electrons, which can often be observed in nonmetals.

To correctly balance these redox reactions, it is essential to ensure that the number of electrons lost equals the number of electrons gained. This principle is vital in maintaining charge conservation within the reaction:

• For example, in the first reaction, \((2 ext{NO}_3^- + 8 ext{H}^+ + 3 ext{Hg} ightarrow ext{Hg}_2^{2+} + 2 ext{NO} + 4 ext{H}_2 ext{O}) \)
• And the second reaction, \((2 ext{Hg}^{2+} + 2 ext{Br}^- ightarrow ext{Hg}_2^{2+} + ext{Br}_2)\)

These equations highlight the transfer of electrons. Proper balancing involves ensuring that both mass and charge are conserved across the chemical equation.

Chemists often visualize these reactions using cell diagrams that show the materials involved and the electron movement through the cell, providing a framework to understand how different materials interact within electrochemical applications.