Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given that the standard potentials \(\left(\mathrm{E}^{0}\right)\) of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) and \(\mathrm{Cu}^{+} / \mathrm{Cu}\) are \(0.34\) \(\mathrm{V}\) and \(0.522 \mathrm{~V}\) respectively, the \(\mathrm{E}^{0}\) of \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\)is: (a) \(+0.182 \mathrm{~V}\) (b) \(+0.158 \mathrm{~V}\) (c) \(-0.182 \mathrm{~V}\) (d) \(-0.158 \mathrm{~V}\)

Step-by-Step Solution

Answer

The answer is (c) \\(-0.182 \, \\\mathrm{V}\\).

Step 1: Identify the Reactions

The two half-reactions given are: 1. \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^0 = 0.34 \, \text{V} \). 2. \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) with \( E^0 = 0.522 \, \text{V} \). We need to find \( E^0 \) for \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \).

Step 2: Use Nernst Equation for Each Reaction

Write the standard Nernst equation for each reaction:1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) is already given.2. \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \): We need to find this.3. \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) is already given.We will rearrange these reactions to make them consistent for calculation.

Step 3: Calculate the Missing Potential

The reaction \( \text{Cu}^{2+} \rightarrow \text{Cu}^+ + e^- \) can be derived by subtracting the second reaction from the first.Using the formula:\[E^0_{\text{cell}} = E^0_{\text{reduction}} - E^0_{\text{oxidation}}\]We set up:\[E^0_{\text{Cu}^{2+}/\text{Cu}} = E^0_{\text{Cu}^{2+}/\text{Cu}^+} + E^0_{\text{Cu}^+/\text{Cu}}\]\[0.34 = E^0_{\text{Cu}^{2+}/\text{Cu}^+} + 0.522\]Solving for \( E^0_{\text{Cu}^{2+}/\text{Cu}^+} \), we find:\[E^0_{\text{Cu}^{2+}/\text{Cu}^+} = 0.34 - 0.522 = -0.182 \, \text{V}\]

Step 4: Conclusion

Thus, the standard electrode potential \( E^0 \) of the reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) is \(-0.182 \, \text{V}\).

Key Concepts

Standard Electrode Potential

In electrochemistry, the concept of standard electrode potential helps us understand the energy levels associated with different half-reactions. Standard electrode potential, denoted as \( E^0 \), is the measure of the individual potential of a reversible electrode at standard state, which is 1 M concentration at 25°C and 1 atm pressure. The potential is measured in volts (V) relative to the standard hydrogen electrode (SHE), which is assigned a potential of 0 V.

Standard electrode potentials allow us to predict the direction of electron flow. The more positive the \( E^0 \) value, the greater the tendency for the species to gain electrons (i.e., be reduced), making it an oxidizing agent.

For the reactions given:

• \( ext{Cu}^{2+} + 2e^- \rightarrow ext{Cu} \) has \( E^0 = 0.34 ext{ V} \)
• \( ext{Cu}^+ + e^- \rightarrow ext{Cu} \) has \( E^0 = 0.522 ext{ V} \)

These values help us to derive the potential of other reactions by combining known half-cell reactions using the formula for cell potential differences.

Half-Cell Reactions

Half-cell reactions are the building blocks of electrochemistry. They represent either the oxidation or reduction reactions occurring in an electrochemical cell. In any electrochemical process, two half-reactions take place.

In our problem, we have two key reactions to consider:

• The first half-reaction: \( ext{Cu}^{2+} + 2e^- \rightarrow ext{Cu} \)
• The second half-reaction: \( ext{Cu}^+ + e^- \rightarrow ext{Cu} \)

The task here is to figure out the potential for the third half-reaction: \( ext{Cu}^{2+} + e^- \rightarrow ext{Cu}^+ \).

To solve this, understand that we can rearrange and combine these half-reactions strategically. Knowing the `standard electrode potential` of the known reactions, we use subtraction to find the unknown potential. Think of these as pieces in a puzzle, where combining them correctly reveals the missing value for \( E^0_{\text{Cu}^{2+}/\text{Cu}^+} \).

Nernst Equation

The Nernst Equation is a fundamental equation in electrochemistry that relates the cell potential of an electrochemical cell to the concentrations of the reactants and products. It allows us to calculate the cell potential under non-standard conditions. However, even under standard conditions, the logic used to derive new potentials can be applied, as seen in this exercise.

The general form of the Nernst equation is:
\[E = E^0 - \frac{RT}{nF} \ln Q\]
where:

• \( E \) is the cell potential under non-standard conditions.
• \( E^0 \) is the standard electrode potential.
• \( R \) is the universal gas constant (8.314 J/mol·K).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant (approximately 96,485 C/mol).
• \( Q \) is the reaction quotient.

In the exercise, we used the concept of potential differences to find the unknown standard potential. Although the Nernst equation wasn't explicitly utilized for concentration adjustments, it underpins the understanding of cell potential equilibrium and gradients.