Problem 6 - A Complete Resource Book in Chemistry for JEE Main

Question

Given \(E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.44 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=\) \(0.77 \mathrm{~V}\) \(\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Fe}\) are placed together then (a) \(\mathrm{Fe}^{3+}\) increases (b) \(\mathrm{Fe}^{3+}\) decreases (c) \(\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{Fe}^{2+}\right]\) remains unchanged (d) \(\mathrm{Fe}^{2+}\) decreases

Step-by-Step Solution

Answer

The concentration of \( \text{Fe}^{3+} \) decreases, so option (b) is correct.

Step 1: Identify the Half-Reactions

First, identify the half-reactions involved based on the standard reduction potentials given:1. \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) with \( E^{\circ} = -0.44 \text{ V} \).2. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with \( E^{\circ} = 0.77 \text{ V} \).

Step 2: Determine the Cell Reaction

Join the half-reactions to form the full reaction keeping in mind the need to balance the electrons:1. Reverse the first reaction: \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \).2. Double the second reaction and add it up: \( 2\cdot (\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}) \) to \( 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \).Add these reactions to balance them:\[ \text{Fe} + 2\text{Fe}^{3+} \rightarrow 3\text{Fe}^{2+} \].

Step 3: Calculate the Overall Cell Potential

The overall cell potential \( E^{\circ}_{\text{cell}} \) is calculated by subtracting the potential of the oxidation reaction from the reduction reaction:\[ E^{\circ}_{\text{cell}} = E^{\circ}(\text{Fe}^{3+} / \text{Fe}^{2+}) - E^{\circ}(\text{Fe}^{2+} / \text{Fe}) = 0.77\text{ V} - (-0.44\text{ V}) = 1.21\text{ V} \].Since the potential is positive, the reaction is spontaneous as written.

Step 4: State the Reaction Direction and Outcome

The positive cell potential indicates spontaneous reaction in the written direction. In this reaction, \( \text{Fe}^{3+} \) is converted to \( \text{Fe}^{2+} \), meaning \( \text{Fe}^{3+} \) is consumed and decreases over time.

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is a fundamental concept used to predict the direction of redox reactions. The standard reduction potential, represented by \(E^{\circ}\), measures the tendency of a species to gain electrons and be reduced. This potential is measured in volts and allows us to compare the relative ease with which different species are reduced.

A reduction potential can be thought of as a kind of 'attraction' an element has for electrons; the higher the potential, the stronger the attraction. For example, in the reaction involving \( \text{Fe}^{2+} / \text{Fe} \), the standard reduction potential is \(-0.44 \text{ V}\). This value suggests that \( \text{Fe}^{2+} \) has a weaker attraction for electrons compared to \( \text{Fe}^{3+} / \text{Fe}^{2+}\), which has a potential of \(0.77 \text{ V}\).

By comparing these values, we can determine which reaction is more likely to occur and in what direction, ultimately guiding the prediction of the spontaneity of the reaction.

Half-Reaction Balancing

Balancing half-reactions is crucial in understanding the flow of electrons during redox reactions. A half-reaction shows either the reduction or oxidation process. To fully understand a redox system, each half-reaction must be balanced for mass and charge.

Consider the formation of the full reaction from the half-reactions of iron:

• First Half-Reaction: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) with a reduction potential \(-0.44 \text{ V}\).
• Second Half-Reaction: \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with a potential of \(0.77 \text{ V}\).
• To properly compose a full balanced redox reaction, these half-reactions must be manipulated to cancel their electron counts. The second reaction needs to be doubled to align the electron transfer, resulting in a balanced redox equation: \( \text{Fe} + 2\text{Fe}^{3+} \rightarrow 3\text{Fe}^{2+} \).

Balancing ensures the law of conservation of mass and charge in the reaction, providing a clear path for understanding how species interact in electrochemical processes.

Redox Reactions

Redox reactions are chemical processes in which oxidation and reduction occur simultaneously. These reactions play a vital role in a wide range of applications, from biological systems to industrial processes.

Understanding redox reactions requires grasping two simultaneous processes: oxidation, which involves the loss of electrons, and reduction, which involves the gain of electrons. For instance, in the exercise involving iron species, the iron metal (\( \text{Fe} \)) is oxidized to \( \text{Fe}^{2+} \), while \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \).

A key feature of redox reactions is the transfer of electrons from one species to another, leading to energy changes that can be harnessed in electrochemical cells. The cell potential, calculated by combining the respective standard reduction potentials of the half-reactions, indicates whether the overall reaction is spontaneous. A positive potential, as we calculated as 1.21 V, signifies that the reaction can happen on its own without external energy input.