Problem 18 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}\) \(\mathrm{Br}_{2}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br} ; \mathrm{E}^{\circ}=+1.09 \mathrm{~V}\) Considering the electrode potentials, which of the following represents the correct order of reducing power? [Main Online April 11, 2014] (a) \(\mathrm{Fe}^{2+}<\mathrm{Al}<\mathrm{Br}\) (b) \(\mathrm{Br}^{-<\mathrm{Fe}^{2+}<\mathrm{Al}}\) (c) \(\mathrm{Al}<\mathrm{Br}^{-}<\mathrm{Fe}^{2+}\) (d) \(\mathrm{Al}<\mathrm{Fe}^{2+}<\mathrm{Br}\)

Step-by-Step Solution

Answer

Option (b) "\(\mathrm{Br}^{-}<\mathrm{Fe}^{2+}<\mathrm{Al}\)" is correct.

Step 1: Understand Electrode Potentials

Reducing power is inversely related to electrode potential. Higher (more positive) reduction potential means lower reducing power and vice versa. We will use the provided electrode potentials to determine reducing power.

Step 2: Compare Electrode Potentials

Let's list the standard electrode potentials, \(E^\circ\), for the half-reactions:1. \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\); \(E^\circ = +0.77\, \mathrm{V}\)2. \(\mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}\); \(E^\circ = -1.66\, \mathrm{V}\)3. \(\mathrm{Br}_{2} + 2\mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}\); \(E^\circ = +1.09\, \mathrm{V}\)The more negative the potential, the greater the reducing power.

Step 3: Determine Reducing Power Order

Based on the electrode potentials:- \(\mathrm{Al}\) has the highest reducing power with \(E^\circ = -1.66\, \mathrm{V}\).- \(\mathrm{Fe}^{2+}\) follows with \(E^\circ = +0.77\, \mathrm{V}\).- \(\mathrm{Br}^{-}\) has the lowest reducing power with \(E^\circ = +1.09\, \mathrm{V}\).Thus, the order of reducing power is \(\mathrm{Al} > \mathrm{Fe}^{2+} > \mathrm{Br}^{-}\).

Step 4: Match the Order to Options

According to the calculated reducing powers, check each provided option:- Option (a) \(\mathrm{Fe}^{2+}<\mathrm{Al}<\mathrm{Br}\) is incorrect.- Option (b) \(\mathrm{Br}^{-}<\mathrm{Fe}^{2+}<\mathrm{Al}\) is correct.- Option (c) \(\mathrm{Al}<\mathrm{Br}^{-}<\mathrm{Fe}^{2+}\) is incorrect.- Option (d) \(\mathrm{Al}<\mathrm{Fe}^{2+}<\mathrm{Br}\) is incorrect.

Key Concepts

Reducing Power

Reducing power refers to the ability of a chemical species to donate electrons in a redox reaction. An effective reducing agent will have a high tendency to lose electrons and consequently, a **low reduction potential**.
Understanding reducing power is crucial. It allows chemists to predict how substances will behave in redox reactions.

• A **strong reducing agent** will easily give away electrons to another substance.
• Conversely, a substance with **low reducing power** resists losing electrons.

In this exercise, we decipher the reducing power by examining the electrode potentials of the species involved. Remember, the higher the negative value of the electrode potential, the greater the reducing power of that species. This inverse relationship is key when determining reducing capabilities for any given reaction.

Standard Electrode Potential

The standard electrode potential (\(E^{\circ}\)) measures the tendency of a chemical species to be reduced, i.e., to gain electrons.
It's like a scale that tells us how eager a molecule is to accept electrons.
When listed:

• **Positive values** suggest a species likes to gain electrons and is a strong oxidizing agent. This means it has lower reducing power.
• **Negative values** reveal a strong inclination to lose electrons, making the substance a strong reducing agent.

In practice, we use these values as reference points at standard conditions (1 M concentration, 1 atm pressure, and 25°C). While in this exercise, \(\mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}\) has an \(E^{\circ} = -1.66 \, \mathrm{V}\), highlighting aluminum’s high reducing power as it resists gaining electrons effectively.

Half-Reactions

In electrochemistry, reactions are often broken down into two half-reactions: oxidation and reduction. Each half-reaction shows part of the overall redox process.
Understanding half-reactions is crucial because they simplify the complexity of redox reactions.

• The **oxidation half** involves loss of electrons.
• The **reduction half** involves gaining electrons.

For instance, the half-reaction for \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\) involves the reduction of iron ions with \(E^{\circ} = +0.77 \, \mathrm{V}\).
By examining these half-reactions separately, chemists can predict how substances behave when combined and calculate the overall electrochemical cell potentials. This information is invaluable for balancing complex redox equations and designing efficient electrochemical cells.