Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard reduction potentials of \(\mathrm{Ag}, \mathrm{Cu}, \mathrm{Co}\) and \(\mathrm{Zn}\) are \(0.799,0.337,-0.277\) and \(-0.762 \mathrm{~V}\) respectively. Which of the following cells will have maximum cell emf? (a) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{IM}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (b) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{lM})\right| \mathrm{Ag}\) (c) \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{IM})\right| \mathrm{Ag}\) (d) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Co}^{2+}(\mathrm{IM})\right| \mathrm{Co}\)

Step-by-Step Solution

Answer

Option (b) has the maximum cell emf of 1.561 V.

Step 1: Identify the Half Reactions

We need to identify the half-reactions for each option, where the more positive standard reduction potential will be the reduction reaction (cathode) and the less positive will be the oxidation reaction (anode).

Step 2: Calculate EMF for Option (a) using Zn and Cu

For the cell \( \text{Zn} \mid \text{Zn}^{2+} (\text{1M}) \parallel \text{Cu}^{2+} (\text{1M}) \mid \text{Cu} \): Zinc is oxidized and copper is reduced. The EMF is calculated as:\[E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.337 - (-0.762) = 1.099 \, \text{V}\]

Step 3: Calculate EMF for Option (b) using Zn and Ag

For the cell \( \text{Zn} \mid \text{Zn}^{2+} (\text{1M}) \parallel \text{Ag}^{+}(\text{1M}) \mid \text{Ag} \): Zinc is oxidized and silver is reduced. The EMF is:\[E_{\text{cell}} = 0.799 - (-0.762) = 1.561 \, \text{V}\]

Step 4: Calculate EMF for Option (c) using Cu and Ag

For the cell \( \text{Cu} \mid \text{Cu}^{2+} (\text{1M}) \parallel \text{Ag}^{+} (\text{1M}) \mid \text{Ag} \): Copper is oxidized and silver is reduced. The EMF is:\[E_{\text{cell}} = 0.799 - 0.337 = 0.462 \, \text{V}\]

Step 5: Calculate EMF for Option (d) using Zn and Co

For the cell \( \text{Zn} \mid \text{Zn}^{2+} (\text{1M}) \parallel \text{Co}^{2+} (\text{1M}) \mid \text{Co} \): Zinc is oxidized and cobalt is reduced. The EMF is:\[E_{\text{cell}} = -0.277 - (-0.762) = 0.485 \, \text{V}\]

Step 6: Compare EMFs and Determine Maximum

Compare the calculated EMFs: - Option (a): 1.099 V - Option (b): 1.561 V - Option (c): 0.462 V - Option (d): 0.485 V The maximum EMF comes from Option (b) with 1.561 V.

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is a critical concept that indicates the tendency of a chemical species to acquire electrons and be reduced. The higher the standard reduction potential, the more likely the species is to gain electrons. These potentials are typically measured under standard conditions, which include 1 M concentrations for all aqueous species, a temperature of 25°C, and a pressure of 1 atm for gases.

The standard reduction potential is measured in volts (V) and is often denoted as \(E^0\). This value determines how easily an element or compound can be reduced in an electrochemical cell. For example, in the exercise, the standard reduction potentials for different metals like \(\text{Ag}\), \(\text{Cu}\), \(\text{Co}\), and \(\text{Zn}\) are given as 0.799 V, 0.337 V, -0.277 V, and -0.762 V respectively.

Knowing these values allows us to determine the order of reactivity and predict which metals will serve as good reducing agents and which as oxidizing agents. A more positive \(E^0\) value suggests a better oxidizing agent, while a more negative one suggests a better reducing agent.

Cell Potential Calculation

When constructing an electrochemical cell, calculating the cell potential (EMF) is crucial to understanding the cell's efficiency and feasibility. The cell potential, \(E_{\text{cell}}\), is the difference between the reduction potentials of the cathode and the anode. It can be calculated using the formula:

\[E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\]

For example, in one of the exercise's options, zinc (Zn) acts as the anode and copper (Cu) as the cathode. The calculation for \(E_{\text{cell}}\) is performed as follows:

• Identify the reduction potential for both half-reactions.
• Subtract the anode potential from the cathode potential.

The subtraction gives the overall cell potential or EMF. For the Zn-Cu cell in the exercise:

\[E_{\text{cell}} = 0.337 \, \text{V} - (-0.762 \, \text{V}) = 1.099 \, \text{V}\]
This cell potential indicates how much work the cell can perform, with a higher EMF suggesting a more effective cell.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are processes in which electrons are transferred between species. In an electrochemical cell, one species undergoes oxidation (loses electrons), and another undergoes reduction (gains electrons).

In the context of the given exercise, it's important to identify which metal serves as the reducing agent (anode) and which as the oxidizing agent (cathode). This identification is based on their standard reduction potentials:

• The species with a lower (or more negative) standard reduction potential tends to lose electrons, hence undergoes oxidation.
• The species with a higher (or more positive) standard reduction potential tends to gain electrons, hence undergoes reduction.

Let's consider the Zn-Ag cell from the exercise:

• \(\text{Zn}\) has a standard reduction potential of -0.762 V, indicating it will be oxidized and act as the anode.
• \(\text{Ag}\) has a standard reduction potential of 0.799 V, indicating it will be reduced and act as the cathode.

This reaction yields an EMF of 1.561 V, showing a strong tendency for electrons to transfer from Zn to Ag, efficiently powering the electrochemical cell.