Problem 179 - A Complete Resource Book in Chemistry for JEE Main

Question

For the redox reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})\) Taking place in a cell, \(E_{\text {cell }}^{\circ}\) is \(1.10\) volt. \(E_{\text {cell }}\) for the cell will be \((2.303 \mathrm{RT} / F=0.0591)\) (a) \(2.14 \mathrm{~V}\) (b) \(1.80 \mathrm{~V}\) (c) \(1.07 \mathrm{~V}\) (d) \(0.82 \mathrm{~V}\)

Step-by-Step Solution

Answer

The redox potential \( E \) for the cell is \( 1.07 \, \text{V} \).

Step 1: Write the Nernst Equation

The Nernst equation for a redox reaction is given as: \[ E = E^{ ext{cell}} - \frac{0.0591}{n} imes ext{log} \left( \frac{[ ext{ions in products}]}{[ ext{ions in reactants}]} \right) \] where \( E^{ ext{cell}} = 1.10 \, \text{V} \), \( n \) is the number of electrons transferred, and the bracket terms represent the concentrations of the ions. For the given reaction, \( n = 2 \) as two electrons are transferred.

Step 2: Identify the Ion Concentrations

For the given reaction, the concentration of \( \text{Cu}^{2+} \) is \( 0.1 \, \text{M} \), and the concentration of \( \text{Zn}^{2+} \) is \( 1 \, \text{M} \). The reaction quotient \( Q \) can be written as: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1}{0.1} = 10 \]

Step 3: Calculate the Redox Reaction Potential

Using the Nernst equation from Step 1, substitute the known values: \[ E = 1.10 - \frac{0.0591}{2} \times \text{log}(10) \] First, simplify the log term: \[ \text{log}(10) = 1 \] Then, calculate the change in potential: \[ \frac{0.0591}{2} \times 1 = 0.02955 \] Subtract this from the standard potential: \[ E = 1.10 - 0.02955 = 1.07045 \, \text{V} \] This can be approximated to the closest answer, \( 1.07 \, \text{V} \).

Key Concepts

Redox Reactions

Redox reactions are special chemical reactions characterized by the transfer of electrons between two substances. In these reactions, one substance undergoes oxidation, meaning it loses electrons, while another substance undergoes reduction and gains those electrons. This electron exchange is fundamental and can drive many chemical processes that are essential in both biological and industrial applications.
For instance, in the given exercise, zinc (Zn) is oxidized, losing two electrons to become Zn^{2+}, while copper ions (Cu^{2+}) are reduced, gaining those electrons to form solid copper (Cu). This transfer of electrons creates an electrochemical cell where the produced electrical energy can be harnessed for various purposes. Understanding redox reactions is crucial for comprehending how batteries work and how energy can be stored and utilized effectively. Moreover, these reactions are not just limited to chemistry labs but have real-world applications in technologies like batteries and metallurgy.

Electrochemistry Calculations

Electrochemistry calculations often involve understanding and applying the Nernst equation, which helps predict how the voltage of an electrochemical cell will change with varying concentrations of reactants and products. The Nernst equation is: \[ E = E^{\text{cell}} - \frac{0.0591}{n} \times \text{log} \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]This relationship is essential for calculating the cell potential under non-standard conditions.
In general, the cell potential is influenced by the concentrations of the reacting species — reactants and products. This is critical in predicting the feasibility and directionality of redox reactions under different conditions.
In the exercise, you calculated the cell potential using known concentrations and the number of electrons transferred ( = 2). Substituting these values into the equation allowed you to understand how the potential shifts from its standard value when conditions deviate from standard—0.0591 is a constant derived from the universal gas constant \( R \), the temperature in Kelvin, and the Faraday constant \( F \). Such calculations give a deeper insight into the practical working of cells and batteries in real-life applications.

Standard Electrode Potential

Standard electrode potential is the measure of the energy per charge available from a redox reaction under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It is denoted as \( E^{\circ}_{\text{cell}} \). The standard electrode potential of a reaction is a fundamental property and helps in determining how much potential a cell provides when it operates at standard-state conditions. A positive \( E^{\circ}_{\text{cell}} \) indicates a spontaneous redox reaction, contributing valuable insights into whether a given electrochemical cell can operate as a battery.
In the mentioned exercise, \( E^{\circ}_{\text{cell}} \) = 1.10 V serves as a baseline for calculating the actual cell potential under specific conditions using the Nernst equation. This helps in predicting how different concentrations of ionic species affect the overall potential, thereby allowing one to either maximize or optimize reaction conditions to achieve desired chemical processes efficiently. Understanding standard electrode potentials is crucial, not only for theoretical calculations but also for practical applications like designing and optimizing electrochemical cells and evaluating the feasibility of redox reactions in new chemical syntheses.