Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard emf of a galvanic cell involving cell reaction with \(\mathrm{n}=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\left.F=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(2.0 \times 10^{11}\) (b) \(4.0 \times 10^{12}\) (c) \(1.0 \times 10^{2}\) (d) \(1.0 \times 10^{10}\)

Step-by-Step Solution

Answer

The equilibrium constant is \(1.0 \times 10^{10}\), option (d).

Step 1: Recall the Nernst Equation

The Nernst equation relates the emf of a galvanic cell to the standard emf and the reaction quotient. For a cell at equilibrium, the Nernst equation is: \[ E = E^0 - \frac{RT}{nF} \ln K \]where \( E \) is the cell emf at non-standard conditions, \( E^0 \) is the standard cell emf, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( K \) is the equilibrium constant.

Step 2: Set Emf to Zero at Equilibrium

At equilibrium, the cell emf (\( E \)) is zero. Substitute \( E = 0 \) into the Nernst equation:\[ 0 = E^0 - \frac{RT}{nF} \ln K \]

Step 3: Solve for the Equilibrium Constant

Rearrange the equation to solve for \( \ln K \):\[ \ln K = \frac{nFE^0}{RT} \]Substitute the known values into this rearranged formula: \( n = 2 \), \( F = 96500 \, \text{C/mol} \), \( E^0 = 0.295 \, \text{V} \), \( R = 8.314 \, \text{J/K mol} \), and \( T = 298 \, \text{K} \).

Step 4: Perform the Calculations

Substitute the numbers into the equation:\[ \ln K = \frac{2 \, \times \, 96500 \, \times \, 0.295}{8.314 \, \times \, 298} \]Calculate the value:\[ \ln K = \frac{56935}{2476.372} \approx 22.986 \]

Step 5: Convert to the Exponential Form

Find \( K \) by taking the exponential of \( \ln K \):\[ K = e^{22.986} \]Calculate the value:\[ K \approx 1.0 \times 10^{10} \]

Step 6: Identify the Correct Option

Compare the calculated \( K \) with the options given: (a) \(2.0 \times 10^{11}\), (b) \(4.0 \times 10^{12}\), (c) \(1.0 \times 10^{2}\), (d) \(1.0 \times 10^{10}\). The correct match is option (d) \(1.0 \times 10^{10}\).

Key Concepts

Standard emf

In the world of electrochemistry, the standard electromotive force (emf) is a crucial concept. It measures the potential difference between two electrodes of a galvanic cell under standard conditions: 1 M concentration for solutions, 1 atm pressure for gases, and 25°C temperature, which is equivalent to 298 K. The standard emf (denoted as \(E^0\)) is essentially the driving force behind the movement of electrons through the cell.

Standard emf tells us about the cell's ability to drive an electron current. A higher positive \(E^0\) means the cell reaction can be spontaneous and energetically favorable. Thus, understanding \(E^0\) is like gauging how willing a reaction is to proceed without external help.

• It is measured in volts (V).
• \(E^0\) provides insight into the cell reaction's tendency to occur.
• Used in calculating other thermodynamic quantities like Gibbs free energy and equilibrium constant.

In the given problem, the standard emf is \(0.295 \, \text{V}\), indicating a reaction with a specific spontaneity at standard conditions.

Equilibrium constant

The equilibrium constant, denoted by \(K\), offers insight into the balance between reactants and products in a chemical reaction at equilibrium. It acts like a telltale sign of how far a reaction proceeds before reaching a state where the rates of forward and reverse reactions are equal. When you apply the Nernst equation at equilibrium \( (E = 0) \), it helps in determining the equilibrium constant for the cell reaction.

The relationship is given by:\[ \ln K = \frac{nFE^0}{RT} \]
Here, **n** is the number of moles of electrons exchanged, **F** is Faraday’s constant, **R** is the gas constant, **T** is temperature in Kelvin, and **E^0** is the standard emf.

• A large \(K\) indicates a product-favored reaction.
• A small \(K\) implies a reactant-favored reaction.
• Helps predict the extent of a reaction.

For this exercise, the calculated \(K\) is approximately \(1.0 \times 10^{10}\), suggesting a strong tendency for product formation at equilibrium.

Gibbs free energy

Gibbs free energy, represented as \( \Delta G \), serves as a tool to predict the spontaneity of a reaction. A negative \( \Delta G \) hints at a spontaneous reaction under constant pressure and temperature, while a positive \( \Delta G \) implies non-spontaneity.

The connection between Gibbs free energy and the standard emf of a cell is crucial. They are linked by the equation:\[ \Delta G^0 = -nFE^0 \]

• \( \Delta G^0 \) represents the change in Gibbs free energy under standard conditions.
• A negative \( \Delta G^0 \) denotes a spontaneous reaction.
• This relationship helps reveal the total energy available from a cell reaction.

In essence, given the standard emf at 0.295 V, we could infer the standard Gibbs free energy change, which provides additional context to the equilibrium constant and likelihood of a reaction proceeding.

Electrochemistry

Electrochemistry is the study of chemical reactions involving the movement of electrons, or electricity. It bridges chemistry with electronics by exploring reactions where electricity is both a cause and an outcome.

In galvanic cells, which are a big focus in electrochemistry, chemical energy is converted into electrical energy via redox reactions (one species gets oxidized and the other reduced). Understanding electrochemical principles allows us to harness power from batteries and even fuel cells.

• Central concepts involve redox reactions, electrodes, and ionic conduction in electrolytes.
• Important terms include anode (oxidation site), cathode (reduction site), and cell potential (or voltage).
• Used in applications like electroplating, battery design, and electrolysis processes.

By mastering electrochemical principles, students can better understand the basis of how these reactions are applied in real-world energy solutions, such as in creating sustainable energy sources.