Problem 137 - A Complete Resource Book in Chemistry for JEE Main

Question

The half cell reaction for the corrosion \(2 \mathrm{H}^{+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}, E^{\circ}=1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Find the \(\Delta \mathrm{G}^{\circ}\) (in \(\mathrm{kJ}\) ) for the overall reaction. (a) \(-76\) (b) \(-322\) (c) \(-161\) (d) \(-152\)

Step-by-Step Solution

Answer

The Gibbs free energy change \(\Delta G^{\circ}\) is \(-322\) kJ.

Step 1: Identify the Reaction

Before combining the half-reactions, let's understand them: The first half-reaction involves the reduction of protons and oxygen to water with a standard reduction potential of 1.23 V. The second half-reaction involves the reduction of iron ions to solid iron with a standard reduction potential of -0.44 V.

Step 2: Overall Reaction Calculation

Now, combine the half-reactions. The overall reaction is obtained by reversing the first half-reaction (since iron oxidation occurs):\[ \text{Overall Reaction: } 2 \text{ Fe}^{2+} + 1/2 \text{ O}_2 + 2 \text{ H}_2\text{O} \rightarrow 2 \text{ Fe(s)} + 4 \text{ H}^+ \]

Step 3: Calculate Standard EMF

Calculate the standard electromotive force (EMF), \(E^{ ext{o}}_{cell}\), of the overall reaction by subtracting the reduction potential of the iron half-cell from the water half-cell:\[ E^{ ext{o}}_{cell} = 1.23 ext{ V} - (-0.44 ext{ V}) = 1.23 ext{ V} + 0.44 ext{ V} = 1.67 ext{ V} \]

Step 4: Calculate Gibbs Free Energy Change

Use the equation to find \(\Delta G^{\circ}\):\[ \Delta G^{\circ} = -nFE^{\circ}_{cell} \]where \(n\) is the number of moles of electrons transferred (4 moles), \(F\) is Faraday's constant (96485 C/mol). Thus,\[ \Delta G^{\circ} = -4 \times 96485 \times 1.67 \]\[ \Delta G^{\circ} = -322160 \] JoulesConvert to kilojoules:\[ \Delta G^{\circ} = -322 \text{ kJ} \]

Step 5: Select the Correct Answer

From the options provided, \(-322\) kJ matches \(\Delta G^{\circ}\). So, the correct answer is option (b): \(-322\).

Key Concepts

Gibbs free energy

Gibbs free energy, often abbreviated as \( \Delta G \), is a valuable concept in thermodynamics that allows us to predict whether a process will proceed spontaneously. It is a measure of the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. In electrochemistry, calculating \( \Delta G \) for a reaction helps determine the feasibility and extent to which an electrochemical reaction will occur without external influences.

When calculating \( \Delta G^{\circ} \) for an electrochemical cell, we use the formula: \[ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \]where:

• \( n \) is the number of moles of electrons transferred in the balanced equation,
• \( F \) is the Faraday constant (approximately 96485 C/mol),
• \( E^{\circ}_{\text{cell}} \) is the standard cell potential.

This formula highlights the relationship between the electrical energy of the cell, represented by \( E^{\circ}_{\text{cell}} \), and the amount of chemical work, represented by \( \Delta G^{\circ} \). A negative \( \Delta G^{\circ} \) signifies that the reaction can occur spontaneously, driven by a flow of electrons through the circuit.

half-reaction

The concept of a half-reaction is fundamental to understanding electrochemical processes. A half-reaction represents just one part of a redox (reduction-oxidation) reaction, specifically focusing on either the reduction or oxidation process separately.

In a redox reaction, oxidation refers to the loss of electrons, while reduction is the gain of electrons. Each half-reaction can be individually balanced and analyzed for its specific contribution to the overall process. For example:

• In the exercise given, the reduction half-reaction is represented by \( 2\text{ H}^{+} + 1/2 \text{ O}_{2} + 2 \text{ e}^{-} \rightarrow \text{ H}_{2}\text{O} \).
• Conversely, the oxidation half-reaction is \( \text{Fe}^{2+} + 2 \text{e}^{-} \rightarrow \text{Fe} (\text{s}) \).

Balancing half-reactions and then combining them is a critical step in understanding how electrons are transferred in the whole reaction. This approach allows us to deduce the overall redox reaction and helps in calculating related parameters such as \( E^{\circ}_{\text{cell}} \) and \( \Delta G^{\circ} \).

standard electrode potential

The standard electrode potential, denoted as \( E^{\circ} \), is an essential concept in electrochemistry. It refers to the voltage difference between an electrode and a reference electrode when the ion concentrations are at a standard state (1 M concentration, 298 K), and any gases involved are at 1 atm pressure.

The standard electrode potential is a measure of the driving force behind an electrochemical reaction. Positive \( E^{\circ} \) values indicate a strong tendency for a species to gain electrons and be reduced, whereas negative \( E^{\circ} \) values suggest a propensity for losing electrons and undergoing oxidation.

In our example:

• The reduction of \( \text{H}^{+} \) and \( \text{O}_{2} \) to form water has a standard potential of 1.23 V, demonstrating its stronger ability to attract electrons than the oxidation process.
• The oxidation of \( \text{Fe}^{2+} \) to \( \text{Fe} \) has a standard potential of -0.44 V, indicating its lesser tendency to gain electrons.

Combining individual electrode potentials according to the formula \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \) gives the overall cell potential, crucial for assessing the entire redox reaction's spontaneity.

electromotive force (EMF)

Electromotive force, or EMF, is a key term in electrochemistry that describes the potential difference (voltage) across the terminals of an electrochemical cell when no current is flowing. It indicates the ability of the cell to perform electrical work when connected in a closed circuit.

It's important to note that EMF, often represented as \( E_{\text{cell}} \), is directly linked to the energy changes involved in the redox reaction occurring within the electrochemical cell. The greater the EMF, the more potent the cell is in driving the reaction to completion.

For a standard electrochemical cell, the EMF is calculated from the standard reduction potentials for the reactions occurring at the cathode and anode:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
In the context of our exercise, the EMF of 1.67 V indicates a substantial driving force for the reaction, highlighting that considerable electrical energy can be harvested from this setup. Understanding EMF is essential for applications like battery design and evaluating the operational voltage that a chemical power source can provide.