Problem 206 - A Complete Resource Book in Chemistry for JEE Main

Question

Given below are the half-cell reactions: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{\circ}=-1.18 \mathrm{~V}\) \(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2}+\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V}\) The \(\mathrm{E}^{\circ}\) for \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\) will be: (a) \(-0.33 \mathrm{~V}\); the reaction will not occur (b) \(-0.33 \mathrm{~V}\); the reaction will occur (c) \(-2.69 \mathrm{~V}\); the reaction will not occur (d) \(-2.69 \mathrm{~V}\); the reaction will occur

Step-by-Step Solution

Answer

(a) \(-0.33 \mathrm{~V}\); the reaction will not occur.

Step 1: Identify Given Half-Reactions

We are given two half-cell reactions: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-}\rightarrow \mathrm{Mn} \) with \(\mathrm{E}^{\circ}=-1.18 \mathrm{~V}\) and \( 2(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}) \) with \(\mathrm{E}^{\circ}=+1.51 \mathrm{~V}\). These transformations will be part of the overall reaction formation.

Step 2: Write Overall Reaction

For the reaction \(3\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\), we need to combine the two half-reactions. Adjust coefficients to balance electrons and ions so that the sum reflects the given overall reaction.

Step 3: Balance Electrons and Coefficients

The first half-reaction \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-}\rightarrow \mathrm{Mn}\) needs no adjustment. The second half-reaction is already multiplied by 2: \(2(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+})\). Both reactions balance to \(3\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\).

Step 4: Calculate Overall Cell Potential

Using the formula \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}}\), where \(-1.18 \mathrm{~V}\) is for reduction and \(+1.51 \mathrm{~V}\) is for oxidation. Thus, \(E^{\circ}_{\text{cell}} = +1.51\ \mathrm{V} - (-1.18\ \mathrm{V}) = -0.33\ \mathrm{V}\).

Step 5: Determine Feasibility of the Reaction

The overall cell potential \(-0.33 \mathrm{~V}\) is negative, indicating that the reaction is non-spontaneous under standard conditions.

Key Concepts

Half-cell reactions

In electrochemistry, a half-cell reaction is a fundamental concept that represents a single oxidation or reduction process. Every electrochemical cell comprises of two half-cells: one for oxidation (loss of electrons) and another for reduction (gain of electrons). These reactions take place at separate electrodes, called the anode and cathode.

When breaking down an overall chemical reaction into half-cell reactions, it allows us to look at each oxidation and reduction process separately. For instance, in the exercise provided, the reactions given:

• \( \text{Mn}^{2+} + 2 \text{e}^- \rightarrow \text{Mn} \)
• \( \text{Mn}^{3+} + \text{e}^- \rightarrow \text{Mn}^{2+} \)

Each of these represents either gaining or losing electrons, which is essential for balancing charge and verifying the flow of electrons. In practice, you must ensure that both half-reactions are balanced, especially regarding the number of electrons exchanged. This balance is crucial to combine them into a full redox equation accurately.

Standard electrode potential

The standard electrode potential, denoted as \( E^{\circ} \), is a vital measure in electrochemistry. It indicates the tendency of a chemical species to be reduced, essentially showing how likely it is to gain electrons. By convention, it’s measured at standard conditions: 25°C, 1M concentration, and 1 atm pressure.

In the exercise, we have two half-cell potentials:

• The first, \( E^{\circ} = -1.18 \text{ V} \), corresponds to \( \text{Mn}^{2+} + 2 \text{e}^- \rightarrow \text{Mn} \).
• The second potential, \( E^{\circ} = +1.51 \text{ V} \), is for \( \text{Mn}^{3+} + \text{e}^- \rightarrow \text{Mn}^{2+} \).

The values of these potentials are used to identify which substance will undergo reduction or oxidation under standard conditions.
The more positive the \( E^{\circ} \), the greater the substance’s affinity for electrons, favoring reduction. Conversely, a negative value indicates a tendency for oxidation when paired against a stronger oxidizing agent in complete redox reactions. Calculating the overall cell potential involves the difference between these values, guiding us on the reaction's spontaneity.

Redox reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes where the oxidation states of atoms are altered through the transfer of electrons. They happen in tandem, meaning when one species undergoes oxidation, another must be reduced.In the given exercise, we consider two separate redox processes through half-cell reactions:

• **Oxidation:** This involves the transformation \( \text{2 Mn}^{2+} \rightarrow \text{2 Mn}^{3+} + 2 \text{e}^- \), releasing electrons.
• **Reduction:** The corresponding conversion \( \text{Mn}^{2+} + 2 \text{e}^- \rightarrow \text{Mn} \), whereby electrons are gained.

Calculating the overall reaction potential is crucial to determine the reaction’s feasibility under standard conditions.The formula \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}} \) comes into play here. This difference in potentials enables us to assess if the reaction (\( -0.33 \text{ V} \) in this instance) is spontaneous or not. A negative value indicates non-spontaneity under standard conditions.

Thus, redox reactions are vital for understanding how energy is transferred and transformed in chemical processes, especially in fields like electrochemistry.