Problem 57 - Chemistry: The Central Science in SI Units, Global Edition

Question

A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Step-by-Step Solution

Answer

For the given standard cell potential of +0.257 V and temperature of 298 K: (a) For n=1, the equilibrium constant \(K ≈ 9.33 \times 10^{10}\) (b) For n=2, the equilibrium constant \(K ≈ 8.69 \times 10^{21}\) (c) For n=3, the equilibrium constant \(K ≈ 7.67 \times 10^{32}\)

Step 1: Recap the Nernst Equation

The Nernst equation relates the standard cell potential (E°), the equilibrium constant (K), temperature (T), and the number of electron transferred (n) in a redox reaction: \[E° = \frac{RT}{nF} \ln K\] where R is the universal gas constant (R = 8.314 J/mol·K) and F is the Faraday constant (F = 96,485 C/mol).

Step 2: Calculate K for n = 1

Given E° = +0.257 V, T = 298 K, and n = 1, we can find the equilibrium constant K using the Nernst equation: \[K = e^{\frac{nFE°}{RT}}\] For n = 1: \[K = e^{\frac{1(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{25.277}\] \(K ≈ 9.33 \times 10^{10}\)

Step 3: Calculate K for n = 2

Now we will find the equilibrium constant K for n = 2. For n = 2: \[K = e^{\frac{2(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{50.555}\] \(K ≈ 8.69 \times 10^{21}\)

Step 4: Calculate K for n = 3

Finally, we will find the equilibrium constant K for n = 3. For n = 3: \[K = e^{\frac{3(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{75.832}\] \(K ≈ 7.67 \times 10^{32}\)

Step 5: Summarize the Results

For the given standard cell potential of +0.257 V and temperature of 298 K: (a) For n=1, the equilibrium constant K ≈ 9.33 × 10^{10} (b) For n=2, the equilibrium constant K ≈ 8.69 × 10^{21} (c) For n=3, the equilibrium constant K ≈ 7.67 × 10^{32}

Key Concepts

Standard Cell Potential

The concept of **standard cell potential** is foundational in understanding the behavior of electrochemical cells. It represents the voltage or electromotive force of a cell under standard conditions, which typically include a temperature of 298 K, a pressure of 1 atm, and solutions at 1 M concentration. These conditions ensure uniformity when comparing different cells.

The standard cell potential (denoted as \( E^\circ \)) is a measure of the cell's ability to drive an electric current. It is calculated by determining the difference in potential between the two half-cells involved in the redox reaction. Generally, a higher \( E^\circ \) value indicates a greater tendency for the reaction to occur spontaneously.

Understanding \( E^\circ \) allows scientists to predict the direction of electron flow, which is essential in designing batteries, electrolysis processes, and other applications in electrochemistry. For a given system, if \( E^\circ \) is positive, the redox reaction tends to proceed as written, generating electrical energy.

Equilibrium Constant

The **equilibrium constant** reflects the balance achieved in a chemical reaction where the rates of the forward and reverse reactions are equal. In the context of a redox reaction, it connects directly to the standard cell potential through the Nernst Equation. This equation enables us to determine the equilibrium constant (\( K \)) by recognizing the relationship between the cell's standard potential and the number of electrons transferred in the reaction.

According to the Nernst Equation:

• \( E^\circ = \frac{RT}{nF} \ln K \)

Where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of electrons transferred, and \( F \) is the Faraday constant.

From this equation, it's evident that if \( E^\circ \) is known, \( K \) can be calculated easily once you know \( n \). The equilibrium constant ultimately tells us how much of the reactants and products are favored in the equilibrium state. A large \( K \) value suggests a reaction heavily favoring products, while a small \( K \) value indicates reactants are favored.

Redox Reaction

A **redox reaction** is a process involving the transfer of electrons between two species. It consists of two parts: reduction and oxidation, which occur simultaneously. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.

In an electrochemical cell, these reactions are separated into two half-cells. One half-cell undergoes oxidation, losing electrons which then flow through a wire, as a current, to the other half-cell where reduction takes place. This flow creates an electrical potential measurable as cell potential.

The number of electrons transferred in a redox reaction, denoted by \( n \), significantly influences the cell potential and the resulting equilibrium constant. The understanding of redox reactions is crucial because they power many technologies we rely on, such as batteries and fuel cells. Moreover, they are essential to many biological processes, including cellular respiration.