Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the following cell reaction: [2011] \(2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; E^{\circ}=1.67 \mathrm{~V}\) At \(\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1\) atm and \(\mathrm{pH}=3\), the cell potential at \(25^{\circ} \mathrm{C}\) is (a) \(1.47 \mathrm{~V}\) (b) \(1.77 \mathrm{~V}\) (c) \(1.87 \mathrm{~V}\) (d) \(1.57 \mathrm{~V}\)

Step-by-Step Solution

Answer

The cell potential is approximately 1.57 V, option (d).

Step 1: Understand the Nernst Equation

The Nernst equation relates the concentration of reactants and products with the cell potential. It is given by: \[ E = E^{ ext{°}} - \frac{RT}{nF} \ln Q \]Where: - \( E \) is the cell potential,- \( E^{ ext{°}} \) is the standard cell potential,- \( R \) is the universal gas constant \((8.314 \, \text{J/mol K})\),- \( T \) is the temperature in Kelvin,- \( n \) is the number of electrons transferred in the reaction,- \( F \) is the Faraday's constant \((96485 \, \text{C/mol})\),- \( Q \) is the reaction quotient.

Step 2: Calculate Reaction Quotient (Q)

The reaction quotient \( Q \) is calculated using the formula:\[ Q = \frac{[\text{Fe}^{2+}]^2}{P(\text{O}_2) [\text{H}^+]^4} \]Given: \([\text{Fe}^{2+}] = 10^{-3} \text{ M}, \quad P(\text{O}_2) = 0.1 \text{ atm}, \quad \text{pH} = 3\) implies \([\text{H}^+] = 10^{-3} \text{ M}\),\[ Q = \frac{(10^{-3})^2}{0.1 \times (10^{-3})^4} \]\[ Q = \frac{10^{-6}}{0.1 \times 10^{-12}} \]\[ Q = \frac{10^{-6}}{10^{-13}} = 10^7 \]

Step 3: Apply the Nernst Equation

Substitute the known values into the Nernst equation:\[ E = 1.67 \, \text{V} - \frac{0.0591}{n} \log Q \]Here, \( n = 4 \) as there are 4 electrons transferred.\[ E = 1.67 - \frac{0.0591}{4} \times \log(10^7) \]\[ E = 1.67 - \frac{0.0591}{4} \times 7 \]\[ E = 1.67 - 0.103425 \]\[ E = 1.566575 \, \text{V} \]

Step 4: Compare and Conclude

The calculated cell potential is approximately 1.57 V. Comparing this with the provided options, the closest match is option (d) 1.57 V.

Key Concepts

Nernst Equation

The Nernst Equation is an essential concept in electrochemistry that allows us to calculate the cell potential of an electrochemical cell under non-standard conditions. It connects the standard cell potential (\(E^{\circ}\)) with the cell potential (\(E\)) when the concentrations of reactants and products are not at standard state. The equation is expressed as:

• \( E = E^{\circ} - \frac{RT}{nF} \ln Q \)

Here, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, and \(F\) is the Faraday constant. \(Q\) denotes the reaction quotient, representing the ratio of concentrations of products to reactants.
The Nernst Equation is paramount when considering changes in concentration and pressure as it enables adjusting the cell potential value away from standard conditions to get a more accurate representation of the cell's natural environment.

Cell Potential

In electrochemistry, cell potential, also known as electromotive force (EMF), is the voltage produced by an electrochemical cell. It represents the driving force for the movement of electrons through an external circuit. Cell potential can be influenced by several factors such as concentration, temperature, and the nature of the reactants.
A positive cell potential indicates a spontaneous reaction, while a negative one suggests non-spontaneous activity. The standard cell potential (\(E^{\circ}\)) is measured when all reactants and products are in their standard states, typically 1 M concentration for solutions, 1 atm pressure for gases, and pure solids or liquids for their corresponding phases.
Adjustments from standard conditions using the Nernst equation allow us to establish the actual cell potential, taking into account the real concentrations and pressures present, hence providing a measured cell potential that accurately represents the cell’s spontaneous activities.

Reaction Quotient

The reaction quotient, \(Q\), is a crucial parameter in determining the direction of an electrochemical reaction at any given moment. It evaluates the relative amounts of products and reactants involved in a non-equilibrium situation.

• For the given half-reaction: \(2 \text{Fe}(\text{s}) + \text{O}_2(\text{g}) + 4 \text{H}^+(\text{aq}) \rightarrow 2 \text{Fe}^{2+}(\text{aq}) + 2 \text{H}_2\text{O}(\text{l})\), \(Q\) is derived as:
• \( Q = \frac{[\text{Fe}^{2+}]^2}{P(\text{O}_2) [\text{H}^+]^4} \)

This equation reflects the state of the reaction at any given point, allowing one to predict which direction the reaction will proceed to reach equilibrium. If \(Q < K_{eq}\), the reaction will move towards the forward direction (right), and vice versa for \(Q > K_{eq}\).
In clinical electrochemical cells, knowing the reaction quotient is essential for accurately applying the Nernst Equation to calculate the cell's potential.

pH

pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is a vital concept in chemistry and electrochemistry as it influences various reactions' intermediate and final outcomes.

• The formula for pH is: \( \text{pH} = -\log[\text{H}^+] \)

Where \([\text{H}^+]\) is the hydrogen ion concentration in the solution. A low pH indicates a high concentration of hydrogen ions, making the solution acidic, whereas a high pH means a low concentration and a basic solution.
In the given exercise, knowing the pH allows one to determine \([\text{H}^+]\), which is necessary for calculating the reaction quotient \(Q\). Thus, understanding the impact of pH on electrochemical reactions is crucial for accurate measurements and calculations. Maintaining the correct pH is imperative in applications ranging from industrial chemical processes to biological systems.