Problem 185 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard emf of a cell, involving one electron change is found to be \(0.591 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.\) \(\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(1.0 \times 10^{30}\) (b) \(1.0 \times 10^{1}\) (c) \(1.0 \times 10^{5}\) (d) \(1.0 \times 10^{10}\)

Step-by-Step Solution

Answer

(d) \( 1.0 \times 10^{10} \)

Step 1: Write the Nernst Equation and Solve for E°

The Nernst Equation is \( E = E^0 - \frac{RT}{nF} \ln K \), where \( E^0 \) is the standard emf of the cell, \( E \) is the cell potential at non-standard conditions, \( R \) is the gas constant \( 8.314 \mathrm{JK}^{-1}\mathrm{mol}^{-1} \), \( T \) is the temperature in Kelvin \( 298 \mathrm{K} \), \( n \) is the number of moles of electrons transferred in the cell reaction, \( F \) is Faraday's constant \( 96500 \mathrm{C}\mathrm{mol}^{-1} \), and \( K \) is the equilibrium constant. For standard conditions, \( E = 0 \) so we have \( E^0 = \frac{RT}{nF} \ln K \). We solve for \( E^0 \) which in this case is \( 0.591 \mathrm{~V} \).

Step 2: Rearrange to Solve for ln K

Rearrange the equation \( E^0 = \frac{RT}{nF} \ln K \) to solve for \( \ln K \): \[ \ln K = \frac{E^0 nF}{RT} \].

Step 3: Calculate ln K

Substitute the given values into the equation: \( \ln K = \frac{0.591 \times 1 \times 96500}{8.314 \times 298} \). Calculate the value: \[ \ln K = \frac{57043.5}{2477.572} \approx 23.015 \].

Step 4: Exponentiate to Find K

Since \( \ln K = 23.015 \), solve for \( K \) by exponentiating: \[ K = e^{23.015} \]. Using a calculator, find \( K \approx 1.0 \times 10^{10} \).

Step 5: Choose the Correct Option

The calculated equilibrium constant \( K \approx 1.0 \times 10^{10} \). Match this with the options provided: The correct choice is (d) \( 1.0 \times 10^{10} \).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \) or \( K_{eq} \), is a crucial concept in chemistry that quantifies the balance between Reactants and Products in a chemical reaction at equilibrium. It embodies the idea that reactions tend to reach a state where the concentrations of reactants and products no longer change over time.

In electrochemical cells, like the one mentioned in the exercise, the equilibrium constant can tell us how far a reaction will proceed before reaching equilibrium.

• If \( K \) is a large number (e.g., \( 1.0 \times 10^{10} \)), the reaction favors the formation of products.
• If \( K \) is small, it favors the reactants.

Understanding \( K \) allows scientists to predict how changes in conditions, like temperature or pressure, will affect the system.

Standard emf of a Cell

The term 'standard emf of a cell' refers to the cell potential when all reactants and products are at standard conditions. This includes concentrations of 1 M for aqueous solutions, pressures of 1 atm for gases, and a specified temperature (usually 25°C or 298K).

Mathematically, the standard emf (\( E^0 \)) is calculated using the Nernst Equation set to standard conditions:
\[ E^0 = \frac{RT}{nF} \ln K \]
This formula illustrates how the standard emf is directly related to the equilibrium constant of the reaction. If \( E^0 \) is positive, it indicates a spontaneous reaction under standard conditions, which aligns with the larger \( K \) values, suggesting product-favored equilibrium. Conversely, a negative \( E^0 \) would indicate the opposite.

Faraday's Constant

Faraday's constant, denoted as \( F \), is essential in electrochemistry, representing the charge of one mole of electrons. Numerically, it is approximately \( 96500 \) coulombs per mole.

In the context of the Nernst Equation, Faraday's constant is used to relate the moles of electrons transferred in a reaction to the potential energy generated or required by that reaction.

• It allows for converting chemical amount into charge, which is then used to compute cell potentials.
• Understanding \( F \) can help predict how changes in the number of electrons (\( n \)) affect the reaction's electromotive force.

This constant is integral in calculations that involve the transfers of electrons, such as those in batteries or electrolysis, making it a cornerstone in the study of electrochemical reactions.