Question

At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.63 \mathrm{~V}\). The equilibrium constant for the reaction is \(3.8 \times 10^{10}\). What is the value of \(n\) for the reaction?

Short answer

The value of \(n\) for the reaction is approximately 2, which means the reaction transfers 2 electrons. This is calculated using the Nernst equation (\( Eº = \frac{RT}{nF} \ln K \)) and the given values of Eº, T, and K.

Step-by-step solution

Write down the Nernst equation

The Nernst equation is as follows: \( Eº = \frac{RT}{nF} \ln K \) Where: - Eº is the standard cell potential - R is the gas constant (8.314 J/mol·K) - T is the temperature in Kelvin - n is the number of electrons transferred in the reaction - F is the Faraday constant (96485 C/mol) - K is the equilibrium constant In this problem, we are given Eº, T, and K and are asked to find the value of n.

Plug in the given values

Substitute the given values into the Nernst equation. \( 0.63 \mathrm{~V} = \frac{(8.314 \frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}})(298 \mathrm{~K})}{n(96485 \frac{\mathrm{C}}{\mathrm{mol}})} \ln (3.8 \times 10^{10}) \)

Solve for n

Rearrange the equation to isolate n and then solve for its value. \( n = \frac{(8.314 \mathrm{\frac{J}{mol\cdot K}})(298 \mathrm{~K})(0.63 \mathrm{~V})}{(96485 \mathrm{\frac{C}{mol}}) \ln (3.8 \times 10^{10})} \) \( n \approx 1.97 \) Since the number of electrons transferred in the reaction must be an integer, we can round n to the nearest whole number.

Determine the value of n

Based on our calculation, the value of n is approximately 2. So, the reaction transfers 2 electrons.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K \), is a pivotal aspect of chemical reactions in equilibrium. It gives us a sense of the extent to which a reaction will proceed. To say it simply, the equilibrium constant describes the concentration ratio of products and reactants at equilibrium. When \( K \) is significantly greater than one, such as in the given example where \( K = 3.8 \times 10^{10} \), it suggests that the reaction heavily favors the formation of products.
In the context of electrochemistry, the Nernst equation relates the equilibrium constant to the standard cell potential \( (Eº) \). This interplay helps us determine other properties of the reaction, such as the number of electrons transferred \((n)\).
Understanding the equilibrium constant is important because it:

• Helps predict the direction of a chemical reaction.
• Assists in determining how far a reaction can go.
• Relates directly to the Nernst equation in determining cell potentials.

Standard Cell Potential

The standard cell potential, denoted as \( Eº \), is a measure of the driving force behind an electrochemical reaction. It indicates how much energy is released (or absorbed) when cell reactions occur. A positive \( Eº \) value, like the \( +0.63 \mathrm{~V} \) given in the exercise, signals a spontaneous reaction under standard conditions.
Standard conditions refer to:

• A temperature of 298 K (25 °C).
• Reactants and products are in their standard states.
• Solution concentrations of 1 M for dissolved substances.

The relevance of the standard cell potential is further amplified when used in the Nernst equation, as it allows for calculating the equilibrium constant and determining \( n \), the number of electrons exchanged in a reaction.
In summary, the standard cell potential:

• Indicates spontaneity of a reaction under standard conditions.
• Helps in calculating equilibrium constants using the Nernst equation.
• Plays a crucial role in understanding and predicting electrochemical behaviors.

Electrons Transferred

In electrochemical reactions, the number of electrons transferred is vital to understanding the full scope of the reaction. The symbol \( n \) denotes this number, representing how many electrons are exchanged between reactants as they are converted into products.
The standard procedure to find \( n \) involves using the Nernst equation, which in this exercise was used effectively. With the Nernst equation:\[ Eº = \frac{RT}{nF} \ln K \]We were able to rearrange and solve to find \( n \), ultimately determining that \( n = 2 \) for this specific reaction.
This means that each molecule of reactant interacted by transferring two electrons.
Key takeaways:

• \( n \) reveals the nature of the redox process within the reaction.
• It's critical to solving the Nernst equation for developing a deeper understanding of electrochemical reactions.
• The integer nature of \( n \) confirms whole electron transfers, crucial for determining the complete electrochemical equation.