Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the cell \(\mathrm{Zn}(\mathrm{s}) \mid \mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}\) (aq) \(\mid \mathrm{M}(\mathrm{s})\), different half cells and their standard electrode potentials are given below: (aq)/ (aq)/ \(\mathrm{Au}^{3+}(\mathrm{aq}) / \mathrm{Ag}^{+}(\mathrm{aq}) / \mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}^{2+}(\mathrm{a}\) \(\mathrm{Au}(\mathrm{s}) \quad \mathrm{Ag}(\mathrm{s}) \quad \mathrm{Fe}^{2+}(\mathrm{aq}) \mathrm{Fe}(\mathrm{s})\) \(\frac{\mathrm{M}^{\mathrm{x}+}(\mathrm{aq}) /}{\mathrm{M}(\mathrm{s})}\) (aq) 1 \(\mathrm{E}_{\mathrm{M}^{\mathrm{x}+} / \mathrm{M}}^{\circ} /(\mathrm{V}) \overline{1.40}\) 1 0 \(\begin{array}{lll}0.80 & 0.77 & -0.44\end{array}\) If \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\), which cathode will give a maximum value of \(\mathrm{E}_{\mathrm{cell}}^{\circ}\) per electron transferred? [Main Jan. \(11,2019(\mathrm{I})]\) (a) \(\mathrm{Ag}^{+} / \mathrm{Ag}\) (b) \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) (c) \(\mathrm{Au}^{3+} / \mathrm{Au}\) (d) \(\mathrm{Fe}^{2+} / \mathrm{Fe}\)

Step-by-Step Solution

Answer

The maximum \(E_{\text{cell}}^{\circ}\) is obtained with cathode (c) \(\mathrm{Au}^{3+} / \mathrm{Au}\).

Step 1: Understanding the Problem

We need to determine which half-cell reaction, when paired with the given half-cell reaction for Zn, will provide the highest electromotive force (emf), \( E_{\text{cell}}^{\circ} \), for the cell. The potential of \( \text{Zn}^{2+}/\text{Zn} \) is given as \(-0.76\, \text{V}\).

Step 2: Identifying Standard Electrode Potentials

We have four options for the cathode, each with differing standard electrode potentials. These are: \(\text{Au}^{3+}/\text{Au} = 1.40\, \text{V}\), \(\text{Ag}^{+}/\text{Ag} = 0.80\, \text{V}\), \(\text{Fe}^{3+}/\text{Fe}^{2+} = 0.77\, \text{V}\), and \(\text{Fe}^{2+}/\text{Fe} = -0.44\, \text{V}\).

Step 3: Electrochemical Cell Principle

An electrochemical cell's standard potential is given by \( E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \). Here, the anode reaction (\( \text{Zn}^{2+}/\text{Zn} \)) has \( E_{\text{anode}}^{\circ} = -0.76 \text{V} \).

Step 4: Calculate \(E_{\text{cell}}^{\circ}\) for Each Option

Calculate the emf for each proposed cathode:1. For \( \text{Au}^{3+}/\text{Au} \): \[ E_{\text{cell}}^{\circ} = 1.40 - (-0.76) = 2.16\, \text{V} \]2. For \( \text{Ag}^{+}/\text{Ag} \): \[ E_{\text{cell}}^{\circ} = 0.80 - (-0.76) = 1.56\, \text{V} \]3. For \( \text{Fe}^{3+}/\text{Fe}^{2+} \): \[ E_{\text{cell}}^{\circ} = 0.77 - (-0.76) = 1.53\, \text{V} \]4. For \( \text{Fe}^{2+}/\text{Fe} \): \[ E_{\text{cell}}^{\circ} = -0.44 - (-0.76) = 0.32\, \text{V}\]

Step 5: Determine the Maximum \(E_{\text{cell}}^{\circ}\)

The cathode that offers the maximum electromotive force is \( \text{Au}^{3+}/\text{Au} \) with \( E_{\text{cell}}^{\circ} = 2.16\, \text{V} \).

Key Concepts

Standard Electrode Potential

In electrochemistry, the standard electrode potential is a crucial concept. It represents the voltage difference between a given half-cell and a standard hydrogen electrode, which is set at zero volts. By measuring this potential, scientists can determine the tendency of a chemical species to gain electrons, meaning how easily it undergoes reduction.

For example, in the exercise provided, you have different metals like gold (\[ \mathrm{Au} \quad (E^{\circ} = 1.40 \, \mathrm{V}) \] ) with varying electrodes and their potentials. The value mentioned, such as 1.40 V for gold, indicates a strong ability to be reduced compared to others like iron (\[ \mathrm{Fe} \quad (E^{\circ} = -0.44 \, \mathrm{V}) \] ), which has a lower potential and hence is less likely to accept electrons.

Understanding these potentials helps in predicting which elements will serve as better cathodes or anodes in an electrochemical cell setup.

Electrochemical Cell

An electrochemical cell is a device capable of generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions. It's cleverly designed to separate oxidation and reduction reactions into separate compartments called half-cells.

Each half-cell contains an electrode and an electrolyte solution. One half-cell will undergo oxidation, where electron loss happens, and another will undergo reduction, gaining those electrons. For the cell with zinc and various cathodes in the exercise, zinc is the anode as it loses electrons, while the other metallic ions act as cathodes seeking those electrons.

In the exercise problem, the zinc half-cell is represented by the reaction:\[ \mathrm{Zn(s)} \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^{-} \] while potential cathodes include reactions like:\[ \mathrm{Au}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Au(s)} \] These cells create a flow of electrons which can be harnessed to do work, such as lighting a bulb or powering a device.

Electromotive Force

Electromotive force (emf) is not actually a force but rather a measure of the electrical energy produced or consumed by an electrochemical cell under standard conditions. It's crucial in determining the cell's ability to do work.

• Calculated as the difference between the standard electrode potentials of the cathode and the anode.
• Expressed when the cell reaction occurs spontaneously, resulting in a positive emf value.

In the problem, the emf can be calculated using the formula:\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \] This difference provides the potential energy available to move electrons through a circuit. For the zinc and gold cell, the calculation is:\[ E_{\text{cell}}^{\circ} = 1.40 \, \mathrm{V} - (-0.76 \, \mathrm{V}) = 2.16 \, \mathrm{V} \] A high emf like 2.16 V indicates a strong driving force for the flow of electrons, implying that the cell would be very efficient in converting chemical to electrical energy.