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

Question

Given below are the half-cell reactions: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{0}=-1.18 \mathrm{~V}\) \(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2}+\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V}\) The \(\mathrm{E}^{0}\) 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

The answer is (c) \(-2.69 \mathrm{~V}\); the reaction will not occur.

Step 1: Understanding Half-cell Reactions

We have two half-cell reactions: \( \mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \); \( E^{0} = -1.18 \mathrm{~V} \) and \( 2(\mathrm{Mn}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}) \); \( E^{0} = +1.51 \mathrm{~V} \). To solve for the standard cell potential of the overall reaction, these half-cell reactions must be combined.

Step 2: Combine Half-cell Reactions

The target reaction is \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn} + 2 \mathrm{Mn}^{3+} \). Breaking this down: two \(\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+} \) reactions take place to form two \(\mathrm{Mn}^{3+} \), each requiring 1 electron, and one \(\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}\) reaction to form \(\mathrm{Mn}\), requiring 2 electrons.

Step 3: Determine the Overall Cell Reaction

First expand the given reactions:- For the oxidation reaction: \(2(\mathrm{Mn}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}) \), which actually means reversing it to \(\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+} + \mathrm{e}^{-}\). The potential becomes: \( E^{0} = -1.51 \mathrm{~V} \).- The reduction reaction remains as \(\mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\), \( E^{0} = -1.18 \mathrm{~V} \).

Step 4: Compute E° for the Target Reaction

For the overall cell reaction, add the potentials of the half-reactions. Since the reduction potential of \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\) is \(-1.18 \mathrm{~V}\) and the oxidation potential of \(\mathrm{Mn}^{3+}+ \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}\) is \(-1.51 \mathrm{~V} \), the total potential is:\[ E^{0} = (-1.18) + (-1.51) = -2.69 \mathrm{~V} \].

Step 5: Determine the Reaction Feasibility

The overall E° is negative at \(-2.69 \mathrm{~V}\), indicating the reaction is not feasible in a galvanic sense. A reaction with a negative cell potential is not spontaneous.

Key Concepts

Half-cell Reactions

A half-cell reaction is a chemical reaction that occurs at an electrode in an electrochemical cell. It involves the transfer of electrons either to or from a chemical species. In electrochemistry, the overall reaction occurring in a cell is the result of the combination of two half-cell reactions. Understanding half-cell reactions is essential because they are the fundamental processes that allow batteries, fuel cells, and other devices to convert chemical energy into electrical energy or vice versa.
In the given exercise, we have two half-cell reactions:

• First, the reduction of \( \mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \) with a standard electrode potential \( E^{0} = -1.18 \mathrm{~V} \).
• Second, the reduction \( \mathrm{Mn}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \), which is presented twice to maintain a balanced electron count, has a standard electrode potential \( E^{0} = +1.51 \mathrm{~V} \).

These reactions need to be correctly combined to determine the potential of the entire electrochemical cell.

Standard Electrode Potential

The standard electrode potential, denoted as \( E^{0} \), is a measure of the individual potential of a reversible electrode at standard state, which includes solutions at a concentration of 1 mol/L, gases at a pressure of 1 atm, and a temperature of 298 K (25 °C). This potential reflects the tendency of a species to be reduced; the more positive the potential, the greater the species' ability to gain electrons and be reduced.
Evaluating the reactions in the problem:

• The first reaction, \( \mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \), has a negative electrode potential of \(-1.18 \mathrm{~V} \), suggesting that \( \mathrm{Mn}^{2+} \) isn't inclined to pick up electrons under standard conditions.
• Conversely, the second reaction \( \mathrm{Mn}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \) has a positive \( E^{0} \) of \(+1.51 \mathrm{~V} \), indicating a strong propensity to be reduced.

The combination of these potentials allows us to determine the overall reaction potential, shedding light on whether the electrochemical reaction is spontaneous or requires an external force to proceed.

Redox Reactions

Redox reactions, or reduction-oxidation reactions, are chemical processes where electrons are transferred from one substance to another. These reactions involve two parts: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. They are foundational in many biochemical and industrial processes, as well as in the operation of electrochemical cells.
In the exercise, we combined two half-cell reactions to determine the overall cell potential by considering the following:

• The reaction \( \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+} + \mathrm{e}^{-} \), originally given as reduction, must be reversed for oxidation purposes, altering its \( E^{0} \) to \(-1.51 \mathrm{~V} \).
• The second reaction retains its reduction form as \( \mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \) with an \( E^{0} = -1.18 \mathrm{~V} \).

When added together, these values \( (-1.51) + (-1.18) \) result in an overall cell potential of \(-2.69 \mathrm{~V} \). A negative cell potential indicates a non-spontaneous, unfavorable reaction in the direction written without external energy input, illustrating a classic example of redox processes in electrochemistry.