Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.441 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}\) the standard emf of the reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+}\) will be (a) \(1.212 \mathrm{~V}\) (b) \(0.111 \mathrm{~V}\) (c) \(0.330 \mathrm{~V}\) (d) \(1.653 \mathrm{~V}\)

Step-by-Step Solution

Answer

The standard emf of the reaction is 1.212 V (option a).

Step 1: Understand the Half-Reactions

To determine the standard emf of the overall reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+}\), we need to consider the relevant half-reactions and their standard electrode potentials. The half-reactions given are: 1. \(\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}\) with \(E^{\circ} = -0.441 \, \mathrm{V}\).2. \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) with \(E^{\circ} = 0.771 \, \mathrm{V}\).

Step 2: Reverse the First Half-Reaction

Since the overall reaction has \(\mathrm{Fe}\) on the reactant side, we need to reverse the first half-reaction to match the overall process. This gives us: \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^-\) with \(E^{\circ} = +0.441 \, \mathrm{V}\), because reversing the reaction changes the sign of \(E^{\circ}\).

Step 3: Use the Given E° Value for the Second Half-Reaction

For the second half-reaction, \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\), we will use the given \(E^{\circ} = 0.771 \, \mathrm{V}\) without reversing it, as it already matches the product and reactant sides of the overall reaction step.

Step 4: Multiply the Second Reaction

To balance electrons for the overall reaction, multiply the second half-reaction by 2:\(2 \mathrm{Fe}^{3+} + 2e^- \rightarrow 2 \mathrm{Fe}^{2+}\), with \(E^{\circ} = 0.771 \, \mathrm{V}\). Note that when balancing half-reactions via multiplication, \(E^{\circ}\) remains unchanged.

Step 5: Calculate the Overall Emf

Now sum the two half-reaction potentials:\[E^{\circ}_{\text{overall}} = E^{\circ}_{\mathrm{Fe}/\mathrm{Fe}^{2+}} + E^{\circ}_{2\mathrm{Fe}^{3+}/\mathrm{2Fe}^{2+}} = 0.441 \, \mathrm{V} + 0.771 \, \mathrm{V} = 1.212 \, \mathrm{V}\]

Step 6: Identify the Correct Option

The calculated standard emf for the overall reaction is \(1.212 \, \mathrm{V}\), which corresponds to option (a).

Key Concepts

Standard Electrode Potential

The standard electrode potential, often denoted as \( E^{\circ} \), is a crucial component in electrochemistry. This value is measured in volts (V) and represents the potential difference between an electrode and its environment under standard conditions. It helps predict the tendency of a chemical species to be reduced or oxidized.

Standard electrode potentials are typically determined at a temperature of 25°C, with solutes at a concentration of 1 M and gases at 1 atmosphere of pressure. Positive \( E^{\circ} \) values indicate a strong tendency to gain electrons (get reduced), while negative values suggest a tendency to lose electrons (become oxidized).

Understanding \( E^{\circ} \) is essential for calculating the electromotive force in redox reactions, as shown in the exercise where the potentials for iron reactions are used to find the overall cell potential.

Half-Reactions

In electrochemistry, half-reactions are used to describe the individual oxidation and reduction processes that occur in a redox reaction. They simplify the analysis by breaking down the complex reaction into two parts: one that loses electrons (oxidation) and one that gains electrons (reduction).

For the given exercise involving iron, the half-reactions are:

• Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^-\)
• Reduction: \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \)

Each half-reaction is associated with its own standard electrode potential, which is critical for determining the overall energy change or electromotive force (EMF) of the cell or system. Remember, reversing a half-reaction flips the sign of \( E^{\circ} \), as seen with the first half-reaction for this exercise.

EMF (Electromotive Force)

The electromotive force (EMF) is the voltage developed by any source of electrical energy, such as a battery or a voltaic cell. In electrochemistry, it represents the energy per charge conveyed by the redox reaction within the cell.

To calculate the standard EMF of a cell, simply sum the standard electrode potentials of the oxidation and reduction half-reactions. This involves either keeping the original \( E^{\circ} \) or changing its sign based on the direction of the reaction.
The EMF in the exercise is calculated as:
\[ E^{\circ}_{\text{overall}} = E^{\circ}_{\mathrm{Fe}/\mathrm{Fe}^{2+}} + E^{\circ}_{2\mathrm{Fe}^{3+}/\mathrm{2Fe}^{2+}} = 0.441 \, \mathrm{V} + 0.771 \, \mathrm{V} = 1.212 \, \mathrm{V} \]
This value indicates the spontaneous direction of the reaction and its ability to perform electrical work.

Redox Reactions

Redox reactions are chemical reactions where electrons are transferred between species. These include both reduction (gain of electrons) and oxidation (loss of electrons). These reactions are fundamental in electrochemistry, as they drive the electrical circuits present in batteries and fuel cells.

For these reactions, it is important to identify which substances are oxidized and reduced, as seen in the exercise where iron transitions through multiple oxidation states. You can use half-reactions to facilitate this understanding by manually calculating the movement of electrons.

Key steps to remember in redox reactions:

• Determine the oxidation states of elements in compounds.
• Write the individual half-reactions for oxidation and reduction processes.
• Balance the electrons in each half-reaction before summing them to provide an overall balanced redox equation.

This structured approach simplifies complex reactions, making it easier to determine cell potentials and predict reaction spontaneity.