Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

For the electrochemical cell, \(\mathrm{M}\left|\mathrm{M}^{+} \| \mathrm{X}^{-}\right| \mathrm{X}\) \(E^{\circ} \mathrm{M}^{+} / \mathrm{M}=0.44 \mathrm{~V}\) and \(E^{\circ} \mathrm{X} / \mathrm{X}^{-}=0.33 \mathrm{~V}\). From these data, one can deduce that (a) \(\mathrm{M}+\mathrm{X} \longrightarrow \mathrm{M}^{+}+\mathrm{X}^{\text {- }}\) is the spontaneous reaction (b) \(\mathrm{M}^{+}+\mathrm{X} \longrightarrow \mathrm{M}+\mathrm{X}\) is spontaneous reaction (c) \(E_{c \text { cll }}=0.77 \mathrm{~V}\) (d) \(E_{\text {cell }}=-0.77 \mathrm{~V}\)

Step-by-Step Solution

Answer

Option (a) is the correct statement as the given reaction is spontaneous with a cell potential of 0.11 V.

Step 1: Understand the Problem Context

We have an electrochemical cell with reactions involving metals M and X. The standard electrode potentials are given for half-reactions: \(E^{\circ}_{\mathrm{M}^{+} / \mathrm{M}} = 0.44 \, \mathrm{V}\) and \(E^{\circ}_{\mathrm{X} / \mathrm{X}^{-}} = 0.33 \, \mathrm{V}\). We need to find out which statements about the cell are true, considering spontaneity and cell potential.

Step 2: Identify Anode and Cathode Reactions

In an electrochemical cell, the anode is where oxidation occurs and the cathode is where reduction occurs. The given values indicate that M is more positive and should act as the cathode, while X should act as the anode. The reactions can be written as follows: \( \text{Anode: } \mathrm{X}^{-} \rightarrow \mathrm{X} + e^{-}\) and \( \text{Cathode: } \mathrm{M}^{+} + e^{-} \rightarrow \mathrm{M} \).

Step 3: Calculate Cell Potential

The cell potential \(E_{\text{cell}}\) can be calculated using the formula: \[ E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \] Substituting the given values, we have: \[ E_{\text{cell}} = 0.44 \, \mathrm{V} - 0.33 \, \mathrm{V} = 0.11 \, \mathrm{V} \].

Step 4: Determine Reaction Spontaneity

A positive cell potential (\(E_{\text{cell}} > 0\)) indicates that the reaction is spontaneous. Since \(E_{\text{cell}} = 0.11 \, \mathrm{V}\), the reaction \(\mathrm{M}^{+} + \mathrm{X}^{-} \rightarrow \mathrm{M} + \mathrm{X}\) is spontaneous, equivalent to \(\mathrm{M} + \mathrm{X} \rightarrow \mathrm{M}^{+} + \mathrm{X}^{-}\) before electrode inversion.

Step 5: Analyze Given Options

Option (a) describes the spontaneous reaction correctly as shown by our calculation. Option (b) is incorrect because it describes the non-spontaneous reverse reaction. Option (c) presents an incorrect total cell potential based on our calculations, and option (d) is incorrect as well because it states a negative cell potential which contradicts the spontaneity.

Key Concepts

Standard Electrode Potential

Electrode potentials are a fundamental concept in electrochemistry that help us understand the behavior of electrochemical cells. The standard electrode potential, denoted as \(E^{\circ}\), is the voltage of a half-cell under standard conditions (1 M concentration, 25°C, and 1 atm pressure). This value is essential as it tells us the tendency of a species to gain or lose electrons.For any half-reaction, measuring its standard electrode potential gives us insight into whether the species will act as an oxidizing or reducing agent.
The more positive the \(E^{\circ}\) value, the greater the tendency to gain electrons (reduction), and conversely, the more negative, the stronger the tendency to lose electrons (oxidation).In the context of the given electrochemical cell, we have two half-cells:

• \( E^{\circ}_{\mathrm{M}^{+} / \mathrm{M}} = 0.44 \, \mathrm{V}\)
• \( E^{\circ}_{\mathrm{X} / \mathrm{X}^{-}} = 0.33 \, \mathrm{V}\)

These potentials show M, being more positive, will act as the cathode where reduction takes place, and X, being more negative, will act as the anode where oxidation occurs. This allows us to understand how electrons will flow in this electrochemical set-up.

Spontaneity of Reactions

A reaction is considered spontaneous if it occurs without needing to be driven by an additional energy source. In electrochemistry, spontaneity is closely linked to the cell potential (\(E_{\text{cell}}\)). If \(E_{\text{cell}}\) is positive, the reaction is spontaneous, meaning it will naturally proceed in the forward direction.For the exercise, we start by calculating the cell potential:\[ E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.44 \, \mathrm{V} - 0.33 \, \mathrm{V} = 0.11 \, \mathrm{V} \]The positive value indicates the cell reaction is spontaneous, which means that under standard conditions, M and X will react to form \(\mathrm{M}^{+}\) and \(\mathrm{X}^{-}\).To sum up:

• A positive \(E_{\text{cell}}\) corresponds to a spontaneous reaction.
• The reaction direction in a spontaneous case is from reactants to products, aligning with the positive cell potential calculation.

Cell Potential Calculation

The cell potential is the difference between the potentials of the cathode and the anode. It gives us a measure of the driving force behind the electrochemical reaction.Calculating the cell potential is straightforward:
Use the equation:\[ E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]Replacing with the known values:\[ E_{\text{cell}} = 0.44 \, \mathrm{V} - 0.33 \, \mathrm{V} = 0.11 \, \mathrm{V} \]This simple subtraction shows us the net voltage driving the reaction.
The positive \(E_{\text{cell}}\) tells us that the electrochemical cell can do work, as the reaction proceeds forward.To understand and calculate efficiently:

• Identify which species acts as the cathode and which as the anode based on their standard potentials.
• Perform the subtraction where the more positive potential is subtracted from the less positive potential.
• Analyze the result: A positive outcome indicates spontaneity and aligns with theoretical expectations based on half-cell potentials.