Problem 64 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Calculate the equilibrium constant for the reaction \(\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\) [1997 - 2 Marks] (given \(E_{\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}}^{\circ}=1.44 \mathrm{~V} ; E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.68 \mathrm{~V} ;\) )

Step-by-Step Solution

Answer

The equilibrium constant \( K \) for the reaction is approximately \( 2.29 \times 10^{13} \).

Step 1: Identify the Half-Reactions

The given reaction can be broken down into two half-reactions: \( \mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+} \) with a standard reduction potential \( E^\circ = 1.44 \text{ V} \) and \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with \( E^\circ = 0.68 \text{ V} \).

Step 2: Determine the Cell Potential

To find the cell potential for the full reaction, subtract the standard reduction potential of the iron half-reaction from the cerium half-reaction: \( E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}} - E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 1.44 \text{ V} - 0.68 \text{ V} = 0.76 \text{ V} \).

Step 3: Calculate the Equilibrium Constant using the Nernst Equation

Use the Nernst equation at standard conditions, which relates the cell potential to the equilibrium constant (\( K \)): \( E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K \). Here, \( n = 1 \) (since each half-reaction involves the transfer of one electron), \( R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \), \( T = 298 \text{ K} \), and \( F = 96485 \text{ C mol}^{-1} \). Rearranging gives \( \ln K = \frac{E^\circ_{\text{cell}} \times n \times F}{RT} \).

Step 4: Solve for the Equilibrium Constant

Substitute the known values into the rearranged equation: \( \ln K = \frac{0.76 \times 1 \times 96485}{8.314 \times 298} \approx 30.91 \). Thus, \( K \approx e^{30.91} \approx 2.29 \times 10^{13} \).

Key Concepts

Equilibrium Constant

In electrochemistry, the equilibrium constant, represented by \( K \), is a critical measure of the extent of a reaction at equilibrium. It tells us about the relative amounts of reactants and products at equilibrium. For a given electrochemical reaction, the equilibrium constant is determined from the standard cell potential \( E^\circ_{\text{cell}} \) using the Nernst equation. The relationship is given by the formula: \[ E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K \] where:

• \( R \) is the universal gas constant, usually 8.314 \( \mathrm{J \, mol^{-1} \, K^{-1}} \).
• \( T \) is the absolute temperature in Kelvin, often 298 K for standard conditions.
• \( n \) is the number of moles of electrons exchanged in the balanced equation; in this case, it's 1.
• \( F \) is Faraday's constant, approximately 96485 \( \mathrm{C \, mol^{-1}} \).

The larger the value of \( K \), the more the reaction favors products over reactants at equilibrium. In the given exercise, the calculated \( K \approx 2.29 \times 10^{13} \) indicates that the reaction favors product formation very strongly at equilibrium.

Nernst Equation

The Nernst Equation is a fundamental formula in electrochemistry that connects the cell potential of an electrochemical cell to its reaction quotient and standard potential. This equation is essential for understanding how the potential of an electrochemical cell depends on temperature, the number of moles of electrons transferred, and concentrations of reactants and products. The Nernst equation is written as:\[ E = E^\circ - \frac{RT}{nF} \ln Q \]where:

• \( E \) is the cell potential under non-standard conditions.
• \( E^\circ \) is the standard cell potential.
• \( Q \) is the reaction quotient, similar to the equilibrium constant but for non-equilibrium conditions.
• The rest are constants previously defined.

In this exercise, at equilibrium, \( E = 0 \), so \( E^\circ = \frac{RT}{nF} \ln K \) gives us an expression to find the equilibrium constant. This highlights how the cell potential can predict the direction and extent of reactions under certain conditions.

Standard Reduction Potential

The standard reduction potential, often denoted as \( E^\circ \), is a measure of the tendency of a chemical species to acquire electrons and undergo reduction. Measured in volts, it is a core concept in understanding electrochemical reactions. The more positive the standard reduction potential, the greater the species' tendency to be reduced. In the context of the previous exercise, we had two half-reactions:

• \( \mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+} \) with \( E^\circ = 1.44 \mathrm{V} \).
• \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with \( E^\circ = 0.68 \mathrm{V} \).

We use these values to determine the overall cell potential \( E^\circ_{\text{cell}} \). The formula is:

• \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)

This formula gives the electrochemical reaction \( E^\circ_{\text{cell}} = 0.76 \mathrm{V} \). The high cell potential indicates a strong driving force for the reaction to proceed as written, linking directly to a high equilibrium constant.