Problem 134 - A Complete Resource Book in Chemistry for JEE Main

Question

The emf of the following three galvanic cells (1) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (2) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (3) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.1 \mathrm{M})\right| \mathrm{Cu}\) are represented by \(E_{1}, E_{2}\) and \(E_{y}\). Which of the following statement is true? (a) \(E_{2}>E_{1}>E_{3}\) (b) \(E_{3}>E_{2}>E_{1}\) (c) \(E_{1}>E_{2}>E_{3}\) (d) \(E_{3}>E_{1}>E_{2}\)

Step-by-Step Solution

Answer

(a) \(E_{2}>E_{1}>E_{3}\).

Step 1: Understand the Components of the Cells

Each cell is composed of a zinc half-cell and a copper half-cell. The standard electrode potential for the half-reactions are as follows: 1. For Zn: \( \text{Zn}^{2+} + 2 \text{e}^- \rightarrow \text{Zn} \), \( E^0 = -0.76 \text{ V} \) 2. For Cu: \( \text{Cu}^{2+} + 2 \text{e}^- \rightarrow \text{Cu} \), \( E^0 = +0.34 \text{ V} \). The standard cell potential is given by the difference: \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.34 \text{ V} - (-0.76 \text{ V}) = 1.10 \text{ V} \).

Step 2: Apply Nernst Equation to Find EMF

The Nernst Equation is used to calculate the EMF under non-standard conditions: \[ E = E^0_{\text{cell}} - \frac{RT}{nF} \ln \left(\frac{[ ext{products}]}{[ ext{reactants}]}\right) \]. For the reactions, \( E = E^0_{\text{cell}} - \frac{0.059}{n} \log Q \), where \( Q \) is the reaction quotient. The reaction quotient \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \). Here \( n = 2 \), and \( E = 1.10 \text{ V} - 0.059 \frac{\log[\text{Zn}^{2+}]/[\text{Cu}^{2+}]}{2} \).

Step 3: Calculate EMF for Each Cell

For \(E_1\):Both \([\text{Zn}^{2+}]\) and \([\text{Cu}^{2+}]\) are 1 M, so \(Q = 1\) and \(E_1 = 1.10\). For \(E_2\):\([\text{Zn}^{2+}] = 0.1 \) and \([\text{Cu}^{2+}] = 1 \), so \(Q = \frac{0.1}{1}\). Plug into Nernst: \[ E_2 = 1.10 \text{ V} - \frac{0.059}{2} \log 0.1 = 1.10 + 0.0295 = 1.1295 \text{ V} \].For \(E_3\):\([\text{Zn}^{2+}] = 1 \) and \([\text{Cu}^{2+}] = 0.1\), so \(Q = \frac{1}{0.1}\). Plug into Nernst: \[ E_3 = 1.10 \text{ V} - \frac{0.059}{2} \log 10 = 1.10 - 0.0295 = 1.0705 \text{ V} \].

Step 4: Compare EMF Values

Compare the calculated EMF values:- \( E_2 = 1.1295 \text{ V} \)- \( E_1 = 1.10 \text{ V} \)- \( E_3 = 1.0705 \text{ V} \)Therefore, \( E_2 > E_1 > E_3 \).

Step 5: Select the Correct Answer

The computed EMF values show that \( E_2 > E_1 > E_3 \). Therefore, the correct statement is (a) \(E_{2}>E_{1}>E_{3}\).

Key Concepts

Nernst Equation

When we deal with electrochemical cells, the Nernst Equation is incredibly important. It lets us calculate the cell's electromotive force (EMF) under non-standard conditions. Knowing this, we can adjust for changes in concentration, temperature, and pressure.

The Nernst Equation is given by:

• \[E = E^0 - \frac{RT}{nF} \ln Q\]

Where:

• \(E\) is the EMF under non-standard conditions
• \(E^0\) is the standard electrode potential
• \(R\) is the gas constant (8.314 J/mol·K)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of moles of electrons exchanged in the reaction
• \(F\) is Faraday’s constant (96485 C/mol)
• \(Q\) is the reaction quotient, which represents the ratio of concentrations of the products to reactants

For practical calculations at room temperature, you often see a simplified form of the Nernst Equation:

• \[E = E^0 - \frac{0.059}{n} \log Q\]

This is used frequently because it makes calculations much easier.

Standard Electrode Potential

Standard electrode potential, denoted as \(E^0\), is a measure of the voltage created by a half-cell under standard conditions. These conditions include a concentration of 1 M for any solutions, 1 atm pressure for any gases, and a temperature of 25°C (298 K).

Standard electrode potential tells us how likely a chemical species is to be reduced. In this process, the more positive \(E^0\), the greater the species' ability to gain electrons and thus, the stronger its oxidizing power.
Here’s a breakdown of how it's used:

• The more positive the \(E^0\), the stronger the oxidizing agent.
• The more negative the \(E^0\), the stronger the reducing agent.

In electrochemical cells, we calculate the standard cell potential \(E^0_{cell}\) using the equation:

• \[E^0_{cell} = E^0_{cathode} - E^0_{anode}\]

It's the difference between the potential of the cathode and the anode. For example, in a Zn/Cu cell, the standard electrode potentials used are \(+0.34\) V for Cu and \(-0.76\) V for Zn.

Galvanic Cells

Galvanic cells, also known as voltaic cells, are devices that convert chemical energy into electrical energy through a spontaneous redox reaction. These batteries or cells operate using two different metals connected by a salt bridge and immersed in solutions containing metal ions.

Let's break down the key components:

• **Anode**: The electrode where oxidation occurs. This is where you lose electrons, and it has a negative charge.
• **Cathode**: The electrode where reduction occurs. This gains electrons, usually carrying a positive charge.
• **Salt Bridge**: This part allows the movement of ions to balance the charge in the cell and completes the circuit without mixing the different solutions.

In a zinc-copper galvanic cell, zinc acts as the anode. It gets oxidized, releasing electrons. Meanwhile, copper acts as the cathode and is where reduction occurs. Electrons naturally flow from the more negative electrode to the more positive electrode (anode to cathode).

• This flow of electrons from anode to cathode generates an electric current, which can power devices.