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) \(\mathrm{M}+\mathrm{X} \longrightarrow \mathrm{M}^{+}+\mathrm{X}^{-}\)is the spontaneous reaction (b) \(\mathrm{M}^{+}+\mathrm{X}^{-} \longrightarrow \mathrm{M}+\mathrm{X}\) is spontaneous reaction (c) \(E_{\text {cell }}=0.77 \mathrm{~V}\) (d) \(E_{\text {cell }}=-0.77 \mathrm{~V}\)

Step-by-Step Solution

Answer

(c) \(E_{\text{cell}}=0.77 \mathrm{~V}\) and (b) \(\mathrm{M}^{+} + \mathrm{X}^{-} \rightarrow \mathrm{M} + \mathrm{X}\) is spontaneous.

Step 1: Identify the Half-Reactions

The electrochemical cell is given as \(\mathrm{M}\left|\mathrm{M}^{+} \| \mathrm{X}^{-}\right| \mathrm{X}\). The standard electrode potentials provided are: \(E^{\circ} \mathrm{M}^{+} / \mathrm{M}=0.44 \mathrm{~V}\) and \(E^{\circ} \mathrm{X} / \mathrm{X}^{-}=0.33 \mathrm{~V}\). In this arrangement, \(\mathrm{M}^{+} + e^- \rightarrow \mathrm{M}\) is the reduction half-reaction with a potential of 0.44 V, and \(\mathrm{X} + e^- \rightarrow \mathrm{X}^{-}\) is the reduction half-reaction with a potential of 0.33 V.

Step 2: Write Reduction and Oxidation Potentials

For cell potential calculations, recall that the potential is obtained by subtracting the potential for the oxidation reaction from that of the reduction reaction. Here, the reduction reaction is \(\mathrm{M}^{+} + e^- \rightarrow \mathrm{M}\), and the oxidation reaction is \(\mathrm{X}^{-} \rightarrow \mathrm{X} + e^-\), since we take oxidation as the reverse of reduction.

Step 3: Calculate Cell Potential

The cell potential \(E_{\text {cell}}\) is given by:\[ E_{\text {cell}} = E^{\circ} (\text{reduction}) - E^{\circ} (\text{oxidation}) \]Substitute the given potentials:\[ E_{\text {cell}} = 0.44 \mathrm{~V} - (-0.33 \mathrm{~V}) = 0.44 \mathrm{~V} + 0.33 \mathrm{~V} = 0.77 \mathrm{~V} \]

Step 4: Determine the Spontaneous Reaction

The positive \(E_{\text {cell}} = 0.77 \mathrm{~V}\) indicates that the cell reaction is spontaneous as written: \(\mathrm{M}^{+} + \mathrm{X}^{-} \rightarrow \mathrm{M} + \mathrm{X}\). The negative sign of \(E_{\text {cell}}\) for the reverse indicates that it is non-spontaneous.

Key Concepts

Standard Electrode Potentials

Standard electrode potentials are a key concept in understanding electrochemical cells. These potentials are measured under standard conditions: 1 M concentration for each ion, a pressure of 1 atm for gases, and a temperature of 25°C. They are denoted by the symbol \( E^{\circ} \), which represents the standard potential for a given half-reaction.

These potentials indicate the tendency of a particular species to be reduced (gain electrons). A higher potential means a greater tendency to be reduced. For example, in our exercise, we have \( E^{\circ} \mathrm{M}^{+} / \mathrm{M}=0.44 \mathrm{~V} \) and \( E^{\circ} \mathrm{X} / \mathrm{X}^{-}=0.33 \mathrm{~V} \). This tells us that \( \mathrm{M}^{+} \) is more likely to undergo reduction than \( \mathrm{X} \).

Understanding these values allows us to predict the direction of electron flow in an electrochemical cell, as electrons flow from the electrode with lower potential to the one with higher potential.

Cell Potential Calculation

Cell potential, often denoted as \( E_{\text{cell}} \), is a measure of the electrical energy available from a redox reaction. To calculate it, we use the standard electrode potentials of the two half-reactions involved. These include a reduction potential and an oxidation potential, which is the opposite process to reduction.

To find the cell potential, we subtract the oxidation potential from the reduction potential. In the example given, the cell potential calculation looks like this:

• Reduction reaction (for \( \mathrm{M}^{+} \)): \( +0.44 \mathrm{~V} \)
• Oxidation reaction (reverse of \( \mathrm{X} \)): \( -0.33 \mathrm{~V} \)

Using the formula:\[E_{\text{cell}} = E^{\circ} (\text{reduction}) - E^{\circ} (\text{oxidation})\]we get:\[E_{\text{cell}} = 0.44 \mathrm{~V} - (-0.33 \mathrm{~V}) = 0.77 \mathrm{~V}\]This calculation confirms that the overall cell potential is 0.77 V.

Spontaneous Reaction

A reaction is considered spontaneous if it can occur without any external force. In electrochemistry, we determine this using the cell potential \( E_{\text{cell}} \). If \( E_{\text{cell}} \) is positive, the reaction proceeds spontaneously.

In the exercise, our calculated \( E_{\text{cell}} = 0.77 \mathrm{~V} \) is positive, indicating that the reaction \( \mathrm{M}^{+} + \mathrm{X}^{-} \rightarrow \mathrm{M} + \mathrm{X} \) occurs spontaneously. This means the chemical reaction can proceed without additional energy input, driven purely by the inherent energy difference.

It is important to remember that a non-spontaneous reaction would have a negative cell potential, which usually requires an external input of electrical energy to proceed.