Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the electrochemical cell, \(M\left|M^{+} \| X^{-}\right| X, E^{\circ}\left(M^{+} / M\right)=0.44 \mathrm{~V}\) and \(E^{\circ}\left(X / X^{-}\right)=0.33 \mathrm{~V}\) From this data one can deduce that [2000S] (a) \(M+X \rightarrow M^{+}+X^{-}\)is the spontaneous reaction (b) \(M^{+}+X^{-} \rightarrow M+X\) is the spontaneous reaction (c) \(\mathrm{E}_{\text {cell }}=0.77 \mathrm{~V}\) (d) \(\mathrm{E}_{\mathrm{ccll}}=-0.77 \mathrm{~V}\)

Step-by-Step Solution

Answer

(b) \(M^{+}+X^{-} \rightarrow M+X\) is the spontaneous reaction.

Step 1: Understand the Cell Diagram

The cell diagram given is \( M | M^+ || X^- | X \). This indicates that \( M \) is the anode and \( X \) is the cathode. Therefore, the anode reaction is \( M \rightarrow M^+ + e^- \) and the cathode reaction is \( X + e^- \rightarrow X^- \).

Step 2: Recall the Standard Electrode Potential Convention

For a cell represented as anode | anode ion || cathode ion | cathode, the cell potential \( E_{cell} \) is calculated as \( E_{cell} = E^{ ext{cathode}} - E^{ ext{anode}} \). The more positive the electrode potential, the stronger the oxidizing agent presents at the cathode.

Step 3: Calculate the Standard Cell Potential

Using the given standard reduction potentials, \( E^{ ext{cathode}} = E^ ext{o}(X/X^-) = 0.33 \text{ V} \) and \( E^{ ext{anode}} = E^ ext{o}(M^+/M) = 0.44 \text{ V} \), calculate the cell potential: \[ E_{cell} = E^{ ext{cathode}} - E^{ ext{anode}} = 0.33 \text{ V} - 0.44 \text{ V} = -0.11 \text{ V} \]

Step 4: Interpret the Cell Potential

A negative \( E_{cell} \) means the cell reaction, as written (\( M \rightarrow M^+ \) and \( X \rightarrow X^- \)), is non-spontaneous. To make it spontaneous, the reverse direction (\( M^+ + X^- \rightarrow M + X \)) would need to be considered as it would make \( E_{cell} \) positive.

Step 5: Select the Correct Option

With \( E_{cell} = -0.11 \text{ V} \), the original reaction is not spontaneous \((M+X\rightarrow M^++X^-)\). Thus, option (b), the reverse reaction \((M^+ + X^- \rightarrow M + X)\), is the spontaneous one. Also, options (c) and (d) are incorrect because they incorrectly calculate \( E_{cell} \) as \( 0.77 \text{ V} \) or \( -0.77 \text{ V} \).

Key Concepts

Standard Electrode Potential

The concept of Standard Electrode Potential is central to understanding electrochemical cells. This refers to the voltage, or potential difference, measured under standard conditions for a specific electrochemical reaction at the electrode. Essentially, it's a measure of how readily a chemical species will gain or lose electrons.

Each substance has a standard reduction potential value, often represented as \( E^{\circ} \). This value indicates how likely an element is to undergo reduction. More positive values suggest a greater tendency to gain electrons and act as an oxidizing agent. Conversely, more negative values indicate a preference to lose electrons and act as a reducing agent.

In the given electrochemical cell, the standard electrode potential for the cathode \( E^{\circ}(X/X^{-}) \) is 0.33 V, and for the anode \( E^{\circ}(M^{+}/M) \) is 0.44 V. This allows us to predict the direction of electron flow and determine the cell's reaction potential, which is crucial in assessing whether a reaction will happen spontaneously.

Cell Potential

In an electrochemical cell, the cell potential, often referred to as the cell voltage, is the measure of the electrochemical reaction’s driving force. This is the difference in potential energy between the two electrodes connected in a cell, calculated by subtracting the anodic potential from the cathodic potential.

For the electrochemical cell, the cell potential \( E_{cell} \) is calculated using the formula: \[E_{cell} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
Using the values provided, we have:

• \( E^{\circ}_{\text{cathode}} = 0.33 \text{ V} \)
• \( E^{\circ}_{\text{anode}} = 0.44 \text{ V} \)

Thus, the cell potential is: \[E_{cell} = 0.33 \text{ V} - 0.44 \text{ V} = -0.11 \text{ V} \]
A negative cell potential means that, as-written, the cell reaction will not proceed spontaneously. For a reaction to be spontaneous, the cell potential should be positive, indicating a thermodynamically favorable reaction.

Spontaneous Reaction

Spontaneity in electrochemistry refers to a reaction's natural tendency to occur without external intervention. For a reaction in an electrochemical cell to be spontaneous, the cell potential \( E_{cell} \) must be positive.

The given exercise involves determining the spontaneous reaction between substances \( M \) and \( X \) in a cell with potentials calculated to be \( -0.11 \text{ V} \). This indicates that the forward reaction, as described \( (M + X \rightarrow M^+ + X^-) \), is not spontaneous on its own.

Instead, the spontaneous process occurs when the reaction direction is reversed: \( (M^+ + X^- \rightarrow M + X) \). This reversal changes the outlook of the cell potential, making the reaction positive and indicating spontaneous behavior. Understanding these factors is vital, as it influences which reactions can occur naturally and which require additional energy.