Question

At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.17 \mathrm{~V}\). The equilibrium constant for the reaction is \(5.5 \times 10^{5}\). What is the value of \(n\) for the reaction?

Short answer

The value of "n" for the reaction is approximately 2, meaning that 2 electrons are transferred in the redox reaction.

Step-by-step solution

List out given values

From the exercise, we have the following data: Standard cell potential, \(E^\circ = 0.17~V\) Equilibrium constant, \(K_c = 5.5 \times 10^5\) Temperature, \(T = 298~K\)

Write the relationship between equilibrium constant and standard cell potential

According to the relationship between Gibbs free energy and the equilibrium constant: \(\Delta G^\circ = -RT \ln K_c\) where \(\Delta G^\circ\) is standard Gibbs free energy change, \(R\) is the ideal gas constant, and \(T\) is the temperature. There's also a relationship between the cell potential and Gibbs free energy: \(\Delta G^\circ = -nFE^\circ\) where \(n\) is the number of electrons involved in the redox reaction, and \(F\) is the Faraday's constant (\(F = 96485 \, C/mol\)). Now, replace \(\Delta G^\circ\) in the first equation with that from the second equation: \[-nFE^\circ = -RT \ln K_c\]

Solve for "n" using the given values

Now, we will solve for n by rearranging the equation and using the given values: \[n = \frac{RT \ln K_c}{FE^\circ}\] Substitute the values to get: \[n = \frac{(8.314 \, J/(mol \cdot K))(298 \, K) \ln(5.5 \times 10^5)}{(96485 \, C/mol)(0.17 \, V)}\]

Calculate the value of "n"

Now, just do the calculation: \[n = \frac{(8.314)(298) \ln(5.5 \times 10^5)}{(96485)(0.17)}\] \[n \approx 1.795\] Since "n" represents the number of electrons transferred in the reaction, it should be rounded to the closest whole number since transferring partial electrons doesn't make physical sense: \[n \approx 2\] Therefore, the value of "n" for the reaction is 2, meaning that 2 electrons are transferred in the redox reaction.

Key concepts

Standard Cell Potential

Understanding the standard cell potential is essential in the realm of electrochemistry. This value, represented by the symbol \(E^\circ\), is the voltage or electrical potential difference of a cell when all components are in their standard states at a specified temperature, typically 25°C or 298K. A positive \(E^\circ\) indicates that the cell can perform work as a voltaic or galvanic cell, meaning it can spontaneously drive an electric current through an external circuit.

It serves as a measure of the driving force behind an electrochemical reaction. In the provided example, the standard cell potential of \(+0.17 \mathrm{~V}\) suggests that the reaction has a tendency to occur spontaneously. To enhance learner's understanding, an important piece of advice would be relating the standard cell potential to everyday batteries, where a higher \(E^\circ\) value would equate to a more 'powerful' battery.

Equilibrium Constant

The equilibrium constant, typically denoted as \(K_c\), is another pivotal concept in electrochemistry, representing the relative proportions of products to reactants at equilibrium. A larger value, such as \(5.5 \times 10^{5}\) seen in the example, indicates that at equilibrium, the reaction heavily favors the formation of products.

To comprehend the magnitude, one could picture a scale with reactants and products; a high equilibrium constant means the scale is heavily tipped towards the products. This understanding helps students predict the direction and extent of chemical reactions and correlate it to the battery's efficacy. The relationship between the standard cell potential and the equilibrium constant reveals how thermodynamically favorable a reaction is, with a higher \(K_c\) typically arising from a more positive \(E^\circ\).

Gibbs Free Energy

Gibbs free energy (\(\Delta G^\circ\)) is a thermodynamic function that helps predict the direction of chemical reactions. It determines whether a process will spontaneously proceed or not. A negative \(\Delta G^\circ\) indicates that a reaction is energetically favorable and can occur without added energy. Conversely, a positive \(\Delta G^\circ\) would mean that the reaction requires input of energy to proceed.

The relationship between standard cell potential and Gibbs free energy, \(\Delta G^\circ = -nFE^\circ\), connects electrical and chemical thermodynamics and underpins the bridge between electrochemistry and energy changes. For the exercise discussed, incorporating the Gibbs free energy into the equation allows for the determination of the number of moles of electrons transferred in the redox reaction, which is a core aspect of the process.

Faraday's Constant

In electrochemistry, Faraday's constant (\(F\)) is a crucial number, representing the charge of one mole of electrons, approximately \(96485 \, C/mol\). It serves as a bridge between the macroscopic properties that we can measure, such as current and time, and the microscopic events occurring on the atomic level during a redox reaction -- specifically the transfer of electrons.

Understanding Faraday's constant is vital for converting between amounts of substances and electric charge, which is key to solving many electrochemical problems, such as in the exercise. By applying Faraday's constant in calculations, students can relate abstract concepts to practical applications like computing the amount of material produced during electrolysis or the charge required to carry out a reduction or oxidation reaction. This eases the transition from theoretical knowledge to practical application.