Problem 97 - Chemistry Principles and Practice

Question

The equilibrium constant at \(25^{\circ} \mathrm{C}\) is \(1.58 \times 10^{2}\) for $$\begin{aligned} 2 \mathrm{VO}^{2+}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \\\& 2 \mathrm{VO}_{2}^{2+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \end{aligned} $$ Calculate \(\Delta G^{\circ}\) and \(E^{\circ}\) for this reaction.

Step-by-Step Solution

Answer

\(\Delta G^{\circ} = -12.557 \text{ kJ/mol}, E^{\circ} = 0.065 \text{ V}\).

Step 1: Identify the Reaction Formula

The given reaction is: \(2 \text{VO}^{2+} (\text{aq}) + \text{Br}_2 (\ell) + 2 \text{H}_2\text{O} (\ell) \rightarrow 2 \text{VO}_2^{2+} (\text{aq}) + 2 \text{Br}^- (\text{aq}) + 4 \text{H}^+ (\text{aq})\). We need to calculate \(\Delta G^{\circ}\) and \(E^{\circ}\) for this reaction at \(25^{\circ}C\).

Step 2: Calculate \(\Delta G^{\circ}\)

The change in standard Gibbs free energy, \(\Delta G^{\circ}\), is calculated using the formula: \[ \Delta G^{\circ} = -RT \ln K \] where \(R = 8.314 \text{ J/(mol K)}\), \(T = 298 \text{ K}\), and \(K = 1.58 \times 10^{2}\). Substituting the values, we get: \[ \Delta G^{\circ} = -(8.314) \times (298) \times \ln(1.58 \times 10^2) \]. Calculate \(\ln(1.58 \times 10^2)\) and solve to find \(\Delta G^{\circ}\).

Step 3: Compute \(\ln(1.58 \times 10^2)\)

Calculate \(\ln(158)\): \[ \ln(158) \approx 5.0626 \]. Now, substitute this back into the expression to find \(\Delta G^{\circ}\): \[ \Delta G^{\circ} = -(8.314) \times (298) \times 5.0626 \approx -12557 \text{ J/mol} \] or \(-12.557 \text{ kJ/mol}\).

Step 4: Calculate Standard Cell Potential \(E^{\circ}\)

Use the relationship between \(\Delta G^{\circ}\) and \(E^{\circ}\): \[ \Delta G^{\circ} = -nFE^{\circ} \] where \(n = 2\) (since 2 moles of electrons are transferred per reaction), and \(F = 96485 \text{ C/mol}\) is Faraday's constant. Rearranging the formula gives: \[ E^{\circ} = -\frac{\Delta G^{\circ}}{nF} \]. Substitute \(\Delta G^{\circ} = -12557 \text{ J/mol}\), \(n = 2\), and \(F = 96485 \text{ C/mol}\): \[ E^{\circ} = -\frac{-12557}{2 \times 96485} \approx 0.065 \text{ V}\].

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K\), is a crucial variable in chemistry that describes the balance between products and reactants in a reversible chemical reaction at equilibrium. This constant is derived from the concentrations of substances involved, obeying the law of mass action. In our scenario, the provided equilibrium constant at \(25^{\circ}\)C is \(1.58 \times 10^2\), which indicates that at equilibrium, the reaction favors the formation of products over reactants.

To understand its impact, consider that if \(K\) is much greater than 1, the reaction strongly favors the products, while a \(K\) much less than 1 indicates a reaction that favors reactants. Thus, knowing \(K\) allows us to predict how far a reaction can proceed. This has implications for the amount of product we can expect from a reaction mixture at equilibrium, crucial in industrial and laboratory settings.

Furthermore, the equilibrium constant is utilized in calculating the Gibbs Free Energy change, \(\Delta G^{\circ}\), as in our exercise using the formula:

• \(\Delta G^{\circ} = -RT \ln K\)

This formula links thermodynamic stability and chemical potential difference, helping chemists understand whether a reaction is spontaneous.

Standard Cell Potential

The standard cell potential, denoted as \(E^{\circ}\), is a measure of the electrochemical potential or the voltage produced by a galvanic cell under standard conditions. It is linked closely with Gibbs Free Energy change, \(\Delta G^{\circ}\), which determines the maximum work obtainable from an electrochemical reaction.

The calculation of \(E^{\circ}\) from \(\Delta G^{\circ}\) is facilitated via the equation:

• \(\Delta G^{\circ} = -nFE^{\circ}\)

where \(n\) is the number of moles of electrons exchanged in the electrochemical process, and \(F\) is Faraday's constant (\(96485\) C/mol).

Rearranging the equation to solve for \(E^{\circ}\) gives:

• \(E^{\circ} = -\frac{\Delta G^{\circ}}{nF}\)

This relationship indicates that a negative \(\Delta G^{\circ}\) results in a positive \(E^{\circ}\), signifying a spontaneous process, as seen in our solution where \(E^{\circ}\) is approximately \(0.065\) Volts.

Understanding \(E^{\circ}\) is essential not only for assessing reaction spontaneity but also for predicting the feasibility of a reaction under specific conditions, making it a cornerstone in the study of electrochemistry.

Electrochemistry

Electrochemistry is a fascinating branch of chemistry that explores the connections between electrical forces and chemical changes. It covers a wide range of processes including battery function, electroplating, and electrolysis.

Central to electrochemistry is the idea of redox reactions, where electrons are transferred between substances. In our reaction, the transformation of \(\text{VO}^{2+}\) ions and \(\text{Br}_2\) is a classic redox scenario. Here, understanding the flow of electrons helps in determining the direction of the reaction and its energy requirements.

Electrochemistry not only explains how chemical energy is converted to electrical energy, as in batteries, but also utilizes electrical energy to bring about chemical changes, such as breaking down compounds in electrolysis.

Moreover, this field provides tools for quantitative descriptions of reaction tendencies and potentials. Utilizing concepts such as Gibbs Free Energy and standard cell potentials allows chemists to predict and manipulate reactions in diverse applications, from industrial synthesis to developing new energy storage technology.

Studying electrochemistry lays the groundwork for advancements in sustainable energy, like the development of more efficient batteries and fuel cells, proving its growing significance in today's tech-driven world.