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

Option (b) \(\mathrm{Fe}^{3+}\) decreases.

Step 1: Understanding Standard Electrode Potentials

Standard electrode potentials, denoted as \(E^{\circ}\), reflect the tendency of a component to be reduced. A more positive \(E^{\circ}\) value suggests a greater tendency for reduction.

Step 2: Identify Given Electrode Potentials

We are given two standard electrode potentials: \(E^{\circ}(\mathrm{Fe}^{2+} / \mathrm{Fe}) = -0.44 \text{ V}\) and \(E^{\circ}(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}) = 0.77 \text{ V}\). These potentials indicate iron's tendency to undergo redox reactions.

Step 3: Calculate Cell Potential

The overall cell reaction when mixing these species results in \(E^{\circ}_{\text{cell}} = E^{\circ}(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}) - E^{\circ}(\mathrm{Fe}^{2+} / \mathrm{Fe})\). This gives us \(E^{\circ}_{\text{cell}} = 0.77 - (-0.44) = 1.21 \text{ V}\).

Step 4: Determine the Direction of Reaction

Since \(E^{\circ}_{\text{cell}} = 1.21 \text{ V}\) is positive, the cell reaction is spontaneous. This implies \(\mathrm{Fe}^{3+}\) will be reduced to \(\mathrm{Fe}^{2+}\), causing \(\mathrm{Fe}^{3+}\) to decrease.

Step 5: Choose the Correct Option

Given that \(\mathrm{Fe}^{3+}\) decreases during the reaction due to spontaneous reduction to \(\mathrm{Fe}^{2+}\), option (b) is the correct choice.

Key Concepts

Standard Electrode Potentials

Standard electrode potentials, often represented as \(E^{\circ}\), depict how likely a chemical species is to gain electrons and be reduced. This is a significant aspect of electrochemistry, providing insight into which species will be reduced or oxidized in a redox reaction.

When you're evaluating electrode potentials, a positive \(E^{\circ}\) value suggests a higher tendency for reduction. This means the species is more likely to gain electrons rather than lose them. Conversely, a negative \(E^{\circ}\) value indicates the species will likely lose electrons, meaning it's more prone to oxidation.

In the exercise, we have two distinct potentials:

• \(E^{\circ}(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44 \text{ V}\) suggests that \(\mathrm{Fe}\) tends to oxidize to \(\mathrm{Fe}^{2+}\).
• \(E^{\circ}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = 0.77 \text{ V}\) suggests that \(\mathrm{Fe}^{3+}\) tends to reduce to \(\mathrm{Fe}^{2+}\).

Understanding these potentials can help predict what will occur when these species are mixed and which direction the redox reaction will take.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two species. They are an essential part of electrochemistry as they explain how electrons are traded between atoms, ions, or molecules. In a redox reaction, one species loses electrons, becoming oxidized, while another gains those electrons, becoming reduced.

When mixing \(\mathrm{Fe}^{2+}\), \(\mathrm{Fe}^{3+}\), and \(\mathrm{Fe}\), understanding which species gets reduced and which gets oxidized is crucial. Using the given standard electrode potentials:

• \(\mathrm{Fe}^{3+}\) gets reduced to \(\mathrm{Fe}^{2+}\) (as it has a higher positive \(E^{\circ}\)).
• \(\mathrm{Fe}^{2+}\) tends to oxidize to \(\mathrm{Fe}\) (as it has a negative \(E^{\circ}\)).

The cell potential equation, \(E^{\circ}_{\text{cell}} = E^{\circ}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) - E^{\circ}(\mathrm{Fe}^{2+}/\mathrm{Fe})\), helps us calculate the overall tendency for the reaction. In this case, it sums to \(1.21 \text{ V}\), indicating a spontaneous direction for the reaction: \(\mathrm{Fe}^{3+}\) is reduced to \(\mathrm{Fe}^{2+}\).

Spontaneity of Reactions

The spontaneity of a reaction tells us if a reaction will proceed on its own without any added energy. This is often determined by looking at the standard electrode potential, \(E^{\circ}_{\text{cell}}\).

In the exercise, the cell potential \(E^{\circ}_{\text{cell}} = 1.21 \text{ V}\) positively indicates a spontaneous reaction. Therefore, the process where \(\mathrm{Fe}^{3+}\) reduces to \(\mathrm{Fe}^{2+}\) will naturally occur without energy input. This is because:

• A positive \(E^{\circ}_{\text{cell}}\) signifies the reaction is thermodynamically favored.
• It means more energy is released by the reaction compared to the energy required to start it.

Spontaneous reactions release energy, often resulting in a decrease of \(\mathrm{Fe}^{3+}\) and an increase of \(\mathrm{Fe}^{2+}\) in the solution. This is why the correct answer to the exercise is that \(\mathrm{Fe}^{3+}\) decreases, as it naturally gets reduced during the reaction.