Question

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\) . b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\). c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

Short answer

a. No reaction occurs when Iodine crystals are added to a NaCl solution. b. For the reaction 2I⁻ + Cl₂ → I₂ + 2Cl⁻, we have \(\mathscr{E}^{\circ} = +0.82\,V\), \(\Delta G^{\circ} \approx -158\,kJ/mol\), and \(K \approx 1.5 × 10^{18}\). c. For the reaction 2Ag + Cu²⁺ → Cu + 2Ag⁺, we have \(\mathscr{E}^{\circ} = +0.46\,V\), \(\Delta G^{\circ} \approx -89.2\,kJ/mol\), and \(K \approx 4.5 × 10^{14}\). d. For the reaction 4Fe²⁺ + O₂ + 4H⁺ → 4Fe³⁺ + 2H₂O, we have \(\mathscr{E}^{\circ} = +2.00\,V\), \(\Delta G^{\circ} \approx -774\,kJ/mol\), and \(K \approx 2.2 × 10^{80}\).

Step-by-step solution

Determine if a reaction occurs

In this situation, both Iodine and Chlorine are halogens, which have a similar tendency to gain electrons. However, due to the lower electronegativity of Iodine compared to Chlorine, there is no redox reaction, and the substances do not react. b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\).

Write a balanced equation

In this situation, the \(\mathrm{Cl}_{2}\) gas will oxidize the iodide ions in the Nal solution: \[\mathrm{2I^- + Cl_2 → I_2 + 2Cl^-}\]

Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\)

From the reduction potential table, we find the standard reduction potentials: \(\mathrm{Cl_2}\) + 2e⁻ → 2\(\mathrm{Cl^-}\): \(E_{Cl}^{\circ} = +1.36\,V\) \(\mathrm{I_2}\) + 2e⁻ → 2\(\mathrm{I^-}\): \(E_{I}^{\circ} = +0.54\,V\) Now, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{Cl}^{\circ} - E_{I}^{\circ} = +1.36\,V - (+0.54\,V) = +0.82\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 2 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(2)(96485\,C/mol)(0.82\,V) \approx -158\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{158\,000\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 1.5 × 10^{18}\) c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\)

Write a balanced equation

In this situation, copper ions in the CuCl₂ solution will reduce and deposit on the silver wire: \[\mathrm{2Ag + Cu^{2+} → Cu + 2Ag^{+}}\]

Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\)

Using the standard reduction potential table, we find the reduction potentials: \(\mathrm{Cu^{2+}}\) + 2e⁻ → \(\mathrm{Cu}\): \(E_{Cu}^{\circ} = +0.34\,V\) \(\mathrm{Ag^{+}}\) + e⁻ → \(\mathrm{Ag}\): \(E_{Ag}^{\circ} = +0.80\,V\) Then, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{Ag}^{\circ} - E_{Cu}^{\circ} = +0.80\,V - (+0.34\,V) = +0.46\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 2 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(2)(96485\,C/mol)(0.46\,V) \approx -89.2\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{89\,200\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 4.5 × 10^{14}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air.

Write a balanced equation

In this situation, the iron ions in the FeSO₄ solution will oxidize in the air (exposed to O₂) in an acidic medium: \[\mathrm{4Fe^{2+} + O_2 + 4H^+ → 4Fe^{3+} + 2H_2O}\]

Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\)

From the standard reduction potential table, we find the reduction potentials: \(\mathrm{O_2}\) + 4e⁻ + 4H⁺ → 2\(\mathrm{H_2O}\): \(E_{O_2}^{\circ} = +1.23\,V\) \(\mathrm{Fe^{3+}}\) + e⁻ → \(\mathrm{Fe^{2+}}\): \(E_{Fe}^{\circ} = -0.77\,V\) Then, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{O_2}^{\circ} - E_{Fe}^{\circ} = +1.23\,V - (-0.77\,V) = +2.00\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 4 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(4)(96485\,C/mol)(2.00\,V) \approx -774\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{774\,000\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 2.2 × 10^{80}\)

Key concepts

Standard Cell Potential

The standard cell potential, denoted as \(\mathscr{E}^{\circ}\), is a crucial part of electrochemistry. It represents the voltage or electric potential difference between two half-cells in a galvanic cell under standard conditions. Standard conditions typically mean 1 M concentrations for all aqueous solutions, 1 atm pressure for gases, and a temperature of 25°C (298 K). These conditions allow us to compare cell potentials consistently. The standard cell potential can be determined using reduction potential values from a standard potential table. We take the difference between the reduction potentials of the cathode and anode reactions. The formula is:\[\mathscr{E}_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\]A positive \(\mathscr{E}^{\circ}\) indicates a spontaneous reaction, implying that the energy released can drive an electric current through an external circuit. Without a positive standard cell potential, a reaction under standard conditions would not proceed as a galvanic cell.

Gibbs Free Energy

The concept of Gibbs free energy (\(\Delta G^{\circ}\)) is important for understanding the spontaneity of a chemical reaction. It links the standard cell potential \(\mathscr{E}^{\circ}\) with thermodynamics.The Gibbs free energy change tells us whether a reaction will occur spontaneously. When \(\Delta G^{\circ}\) is negative, the reaction occurs spontaneously, releasing free energy. We calculate \(\Delta G^{\circ}\) using the formula:\[\Delta G^{\circ} = -nF\mathscr{E}_{cell}^{\circ}\]Where:

• \(n\) = number of moles of electrons exchanged
• \(F\) = Faraday's constant, approximately 96485 C/mol
• \(E_{cell}^{\circ}\) = standard cell potential in volts (V)

This relationship helps us understand how energy changes in chemical processes are related to electrical energy.

Equilibrium Constant

The equilibrium constant (\(K\)) is a powerful tool in understanding the extent of a chemical reaction. It connects thermodynamics and electrochemistry and is related to both \(\Delta G^{\circ}\) and \(\mathscr{E}^{\circ}\).We calculate \(K\) using the following equation:\[K = e^{-\frac{\Delta G^{\circ}}{RT}}\]Where:

• \(\Delta G^{\circ}\) = standard Gibbs free energy change
• \(R\) = gas constant, 8.314 J/mol·K
• \(T\) = temperature in Kelvin

A large \(K\) value (much greater than 1) suggests that the products are heavily favored at equilibrium, signifying that the reaction proceeds almost to completion. Conversely, a small \(K\) value means the reactants are favored, and the reaction does not proceed far toward products. Understanding \(K\) helps predict the final composition of the reaction mix.

Reduction Potential

Reduction potential, another key concept in electrochemistry, is a measure of the tendency of a chemical species to acquire electrons and thus be reduced. Every chemical reaction of this nature is assigned a standard reduction potential, often found in tables, facilitating comparison across different reactions. This potential is measured in volts and determines whether a particular species will gain electrons in a redox reaction.The reduction potential of a species can be explained using half-reactions. For instance, the potential for a reaction where chlorine gas is reduced to chloride ions is:\[\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-: \quad E_{Cl}^{\circ} = +1.36\,V\]The higher the reduction potential, the greater the species' tendency to undergo reduction. Using these potentials, we can predict and calculate the electron transfer tendencies in electrochemical cells.

Electrochemistry

Electrochemistry is the branch of chemistry dealing with reactions that involve electrons moving between atoms. These are broadly called redox (reduction-oxidation) reactions. In electrochemistry, we study the movement of electrons in both spontaneous chemical reactions, such as those in galvanic cells, and non-spontaneous reactions, like those in electrolysis. Galvanic cells, in particular, use spontaneous reactions to generate electrical energy. In these systems, oxidation occurs at the anode, where electrons are lost, while reduction happens at the cathode, where electrons are gained. Key topics in electrochemistry include:

• Standard cell potentials and how they characterize the ability of cells to produce energy.
• Balancing redox equations to understand the electron transfers between species.
• The practical applications of electrochemical principles in batteries, corrosion, and electroplating.

Through understanding electrochemistry, we gain insight into powerful and versatile industrial applications, as well as fundamental scientific principles.