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 \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2 t}\right)=0.771 \mathrm{~V}\) 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 is (a) \(1.212 \mathrm{~V}\).

Step 1: Write Half-Reactions

First, we need to identify the half-reactions involved and their standard electrode potentials. The reduction half-reactions are:1. \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) with \( E^{\circ} = 0.771 \mathrm{~V} \).2. \( \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe} \) with \( E^{\circ} = -0.441 \mathrm{~V} \).

Step 2: Reverse Reactions for Oxidation

Since the given reaction involves the oxidation of iron, we need to reverse the half-reaction involving \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} \):- Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^- \) with \( E^{\circ}_{\text{ox}} = 0.441 \mathrm{~V} \) (we change the sign because it's oxidation now).

Step 3: Balance Overall Reaction

The overall reaction given is \( \mathrm{Fe} + 2\mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+} \). This means:- One \( \mathrm{Fe} \) is oxidized to originate two electrons.- Two \( \mathrm{Fe}^{3+} \) will use these two electrons to reduce to \( \mathrm{Fe}^{2+} \).Thus, both processes are matched in electron transfer.

Step 4: Calculate Standard EMF

To find the standard emf of the reaction, we use the formula:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red}} + E^{\circ}_{\text{ox}} \]Substitute the values,\[ E^{\circ}_{\text{cell}} = 0.771 + 0.441 = 1.212 \mathrm{~V} \]

Step 5: Choose Correct Answer

The calculation gives us a standard emf of \( 1.212 \mathrm{~V} \). Comparing with the choices presented, the correct answer is:(a) \( 1.212 \mathrm{~V} \).

Key Concepts

Standard EMF Calculation

The standard electromotive force (emf) of a reaction is a key concept in electrochemistry. It helps us determine the voltage produced by an electrochemical cell. Calculating the standard emf involves using the electrode potentials for the half-reactions in the cell. The formula used is: \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red}} + E^{\circ}_{\text{ox}} \] In this expression, \( E^{\circ}_{\text{red}} \) is the standard reduction potential, and \( E^{\circ}_{\text{ox}} \) is the standard oxidation potential. Here's how to do it: - Identify the half-reactions involved.- Look up their standard electrode potentials.- Add the potentials, ensuring one is oxidation (reverse) and one is reduction. The result is the standard emf for the total cell reaction. In this problem, the cell voltages added up to 1.212 V, which means when the reaction occurs, it produces this voltage at standard conditions.

Redox Reactions

Redox reactions are processes where oxidation and reduction occur simultaneously. **Redox** is a combination of "reduction" and "oxidation," making it easier to remember. Here's how it works in simple terms: - **Reduction**: Gain of electrons. The species gets reduced.- **Oxidation**: Loss of electrons. The species gets oxidized. In the reaction \( \mathrm{Fe} + 2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+} \): - The iron (Fe) is oxidized; it loses electrons to become \( \mathrm{Fe}^{2+} \).- The \( \mathrm{Fe}^{3+} \) ions are reduced to \( \mathrm{Fe}^{2+} \) by gaining electrons. The overall reaction balances the number of electrons lost and gained, ensuring the redox process is complete.

Half-Reactions

Half-reactions are essential parts of redox processes. They represent the individual oxidation or reduction processes separately. For any redox reaction, the steps are: 1. **Identify half-reactions**: Assess the species being oxidized and reduced.2. **Write half-reactions**: Express them separately, showing the electron transfer. In this example: - Reduction: \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \), where \( E^{\circ} = 0.771 \mathrm{~V} \)- Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^- \), where \( E^{\circ}_{\text{ox}} = 0.441 \mathrm{~V} \) Reverse the signs for oxidation, when needed. Combine these balanced half-reactions to derive the overall redox equation. By recognizing each half, we ensure both processes are considered and the reaction is fully understood.