Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

For a \(\mathrm{Ag}-\mathrm{Zn}\) button cell, net reaction is \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{ZnO}(\mathrm{s})+2 \mathrm{Ag}(\mathrm{s})\) \(\Delta \mathrm{G}_{f}^{\circ}\left(\mathrm{Ag}_{2} \mathrm{O}\right)=-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}(\mathrm{ZnO})=-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Hence \(E_{\text {cell }}^{\circ}\) of the button cell is (a) \(3.591 \mathrm{~V}\) (b) \(2.591 \mathrm{~V}\) (c) \(-1.591 \mathrm{~V}\) (d) \(1.591 \mathrm{~V}\)

Step-by-Step Solution

Answer

The correct answer is (d) 1.591 V.

Step 1: Calculate the Change in Gibbs Free Energy for the Reaction

First, we need to calculate the change in Gibbs free energy \( \Delta G^\circ \) for the entire reaction using the formation Gibbs free energy values provided. The net reaction is \( \mathrm{Zn}(\mathrm{s}) + \mathrm{Ag}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{ZnO}(\mathrm{s}) + 2 \mathrm{Ag}(\mathrm{s}) \). For this reaction, \( \Delta G^\circ = [\Delta G_{\mathrm{f}}^\circ(\mathrm{ZnO}) + 2 \cdot \Delta G_{\mathrm{f}}^\circ(\mathrm{Ag})] - [\Delta G_{\mathrm{f}}^\circ(\mathrm{Zn}) + \Delta G_{\mathrm{f}}^\circ(\mathrm{Ag}_2\mathrm{O})]\). Note that \(\Delta G_{\mathrm{f}}^\circ(\mathrm{Zn})\) and \(\Delta G_{\mathrm{f}}^\circ(\mathrm{Ag})\) are zero since they are in their standard states. Substitute \( \Delta G_{\mathrm{f}}^\circ(\mathrm{Ag}_2 \mathrm{O}) = -11.21 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \) and \( \Delta G_{\mathrm{f}}^\circ(\mathrm{ZnO}) = -318.3 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \).

Step 2: Substitute and Simplify to Find \( \Delta G^\circ \)

Substitute the formation Gibbs free energies into the \( \Delta G^\circ \) equation: \( \Delta G^\circ = [-318.3 \, \mathrm{kJ/mol} + 2 \cdot 0] - [0 - 11.21 \, \mathrm{kJ/mol}] \). Simplify this to get \( \Delta G^\circ = -318.3 \, \mathrm{kJ/mol} + 11.21 \, \mathrm{kJ/mol} = -307.09 \, \mathrm{kJ/mol} \).

Step 3: Convert \( \Delta G^\circ \) to Joules

Since the standard cell potential \( E^\circ_{cell} \) will be calculated in volts, we need \( \Delta G^\circ \) in Joules. Convert \(-307.09 \, \mathrm{kJ/mol} \) to Joules: \(-307.09 \, \mathrm{kJ/mol} \times 1000 \, \mathrm{J/kJ} = -307090 \, \mathrm{J/mol} \).

Step 4: Calculate Standard Cell Potential \( E^\circ_{cell} \)

Use the equation \( \Delta G^\circ = -nFE^\circ_{cell} \), where \( n \) is the number of moles of electrons transferred (\( n = 2 \), as 2 electrons are involved in converting \( \mathrm{Ag}_2 \mathrm{O} \) to \( 2 \mathrm{Ag} \)), \( F \) is Faraday’s constant (\( 96485 \, \mathrm{C/mol} \)). Rearranging gives \( E^\circ_{cell} = -\frac{\Delta G^\circ}{nF} \). Substitute the known values: \( E^\circ_{cell} = -\frac{-307090 \, \mathrm{J/mol}}{2 \times 96485 \, \mathrm{C/mol}} = 1.591 \, \mathrm{V} \).

Step 5: Identify the Correct Option

The calculated standard cell potential \( E^\circ_{cell} \) is 1.591 V. Therefore, the correct answer is (d) \( 1.591 \mathrm{~V} \).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a term used in chemistry to determine the spontaneity of a reaction. If \( \Delta G \) is negative, it indicates that a reaction will proceed spontaneously under constant temperature and pressure. Gibbs Free Energy combines enthalpy and entropy to give a comprehensive view of a reaction's energy profile. It is particularly useful in electrochemistry for understanding how much work can be obtained from an electrochemical cell.

To calculate the change in Gibbs Free Energy for a reaction, you combine the standard free energies of formation of the products and reactants. The formula is:

• \( \Delta G^\circ_{reaction} = \sum \Delta G^\circ_f \text{(products)} - \sum \Delta G^\circ_f \text{(reactants)} \)

In the context of the original exercise, it helped us ascertain the energetic capability of the \( \mathrm{Ag}-\mathrm{Zn} \) cell, leading to a negative \( \Delta G^\circ \) value, which signifies that the cell reaction can perform electrical work.

Standard Cell Potential

The Standard Cell Potential, \( E^\circ_{cell} \), is an essential concept in electrochemistry that quantifies the voltage or electrical potential difference across an electrochemical cell when all reactants and products are in their standard states. It is a measure of a cell's ability to produce electrical energy.

A positive \( E^\circ_{cell} \) indicates that the reaction is spontaneous and the cell can do work, while a negative \( E^\circ_{cell} \) implies the opposite.

• The relationship between \( \Delta G^\circ \) and \( E^\circ_{cell} \) is given by the equation \( \Delta G^\circ = -nFE^\circ_{cell} \).
• In this equation, \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant.

By rearranging the equation, we solve for \( E^\circ_{cell} \) to understand the potential electrical energy the cell can produce. In the original exercise, solving \( E^\circ_{cell} \) demonstrated that the \( \mathrm{Ag}-\mathrm{Zn} \) button cell produces a potential of 1.591 V.

Faraday's Constant

Faraday's Constant, often represented as \( F \), is a fundamental value in electrochemistry that represents the magnitude of electric charge per mole of electrons. The value of Faraday's Constant is approximately \( 96485 \) Coulombs per mole (C/mol). This constant is vital to converting between moles of electrons, electrical charge, and energy.

Faraday's Constant plays a crucial role in the equation \( \Delta G^\circ = -nFE^\circ_{cell} \), where it helps relate the change in Gibbs free energy to the standard cell potential. In simple terms, it helps bridge the gap between chemical equations and their electrical implications.

• It ensures the understanding of how much charge passes through the cell when a certain amount of electrical energy is generated.
• Observing its role in the original exercise highlighted how stoichiometry and electron transfer correlate with energy conversion in a practical context.

Understanding Faraday's Constant is crucial for anyone tackling problems related to electrochemical cells and reactions.