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. \(E_{\text {otl }}\) for the cell will be \((2.303 \mathrm{RT} / F=0.0591)\) [2003] (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 cell potential is 1.07 V, option (c).

Step 1: Write the Nernst equation

The Nernst equation modifies the standard cell potential to account for non-standard conditions. It is given by: \[ E_{cell} = E_{cell}^{ ext{°}} - \frac{0.0591}{n} \log \frac{[ ext{Zn}^{2+}]}{[ ext{Cu}^{2+}]} \]where \( E_{cell}^{ ext{°}} = 1.10 \) V is the standard cell potential, \( n = 2 \) is the number of moles of electrons transferred, and \([\text{Zn}^{2+}]=1\) M, \([\text{Cu}^{2+}]=0.1\) M are the concentrations of the zinc and copper ions, respectively.

Step 2: Calculate the reaction quotient

The reaction quotient \( Q \) is calculated as the ratio of the concentrations of the products to the reactants: \[ Q = \frac{[ ext{Zn}^{2+}]}{[ ext{Cu}^{2+}]} = \frac{1}{0.1} = 10 \]

Step 3: Substitute values into the Nernst equation

Substitute the values into the Nernst equation. \[ E_{cell} = 1.10 - \frac{0.0591}{2} \log(10) \]Calculate \( \frac{0.0591}{2} \log(10) \) as follows: \[ \frac{0.0591}{2} = 0.02955 \]\[ \log(10) = 1 \]

Step 4: Perform the calculation

Now compute the value: \[ E_{cell} = 1.10 - (0.02955 \cdot 1) \] \[ E_{cell} = 1.10 - 0.02955 = 1.07045 \] V

Step 5: Compare and conclude

The calculated cell potential \(E_{cell} = 1.07045\) V rounds to \(1.07\) V. Thus, the correct answer is \(\text{option (c)}\).

Key Concepts

Redox Reactions

Redox reactions often seem complex at first, but when we break them down, they reveal a fascinating dance of electrons between reactants. At their core, redox reactions are chemical reactions wherein oxidation and reduction occur simultaneously.

• Oxidation: This is the loss of electrons by a molecule, atom, or ion.
• Reduction: This is the gain of electrons by a molecule, atom, or ion.

In our example of the reaction between zinc and copper ions, - Zinc is oxidized, as it loses electrons. - Copper ions are reduced, as they gain those electrons and form solid copper. It's much like a baton being passed in a relay race, with electrons being the baton exchanged between two runners (oxidizing agent and reducing agent). This transfer is the foundational basis for many processes, particularly in electrochemistry, where these electron transfers drive the flow of electric current.

Electrochemical Cells

Electrochemical cells harness the energy from redox reactions to produce electricity. This device is an ingenious way of transforming chemical energy into electrical energy. They work due to differing potentials of the electrode materials, which gives rise to electron flow. There are two key components in an electrochemical cell:

• Anode: The electrode where oxidation occurs. In our example, zinc serves as the anode as it loses electrons.
• Cathode: The electrode where reduction happens. Copper acts as the cathode in this scenario, receiving electrons.

The overall electric circuit is completed through a salt bridge or a porous membrane that separates the two half-cells and allows ions to flow without allowing the solutions to mix. This maintains charge balance as electrons move through the external circuit, powering devices or creating measurable current. Electrochemical cells are essential in batteries, electroplating, and corrosion prevention, showcasing their wide array of practical applications.

Cell Potential Calculation

The cell potential (electromotive force or emf) of an electrochemical cell is key to understanding its capability to perform work. Calculating the cell potential under non-standard conditions is where the Nernst Equation comes into play. The Nernst Equation is expressed as:\[ E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q \]Where, - \( E_{cell}^{\circ} \) is the standard cell potential,- \( R \) is the universal gas constant,- \( T \) is temperature in Kelvin,- \( n \) is the number of moles of electrons exchanged,- \( F \) is Faraday's constant, and- \( Q \) is the reaction quotient, calculated from concentrations of the products and reactants.For simplicity in most standard conditions at room temperature, like in our problem, the expression simplifies to:\[ E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q \]In the provided exercise, substituting the known values:- Standard potential \( E_{cell}^{\circ} = 1.10 \text{ V} \),- Reaction quotient \( Q = 10 \), - Number of electrons \( n = 2 \),we calculated the cell potential \( E_{cell} \), yielding a value of \( 1.07 \text{ V} \). This tells us the potential difference that drives electrons through the circuit when the cell operates under these non-standard conditions.