Problem 66 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following four electrodes (1) \(\mathrm{Cu}^{2+}(0.0001 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s})\) (2) \(\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s})\) (3) \(\mathrm{Cu}^{2+}(0.01 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s})\) (4) \(\mathrm{Cu}^{2+}(0.001 \mathrm{M}) \mid \mathrm{Cu}(\mathrm{s})\) If the standard reduction potential of \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\) is \(+0.34\) \(\mathrm{V}\), the reduction potentials (in volts) of the above electrodes following the order (a) \(\mid>4>3>2\) (b) \(1>2>3>4\) (c) \(3>4>2>1\) (d) \(2>3>4>1\)

Step-by-Step Solution

Answer

The answer is (a): 1 > 4 > 3 > 2.

Step 1: Understand the Nernst Equation

To find the reduction potential at non-standard conditions, we use the Nernst equation: \[ E = E^0 - \frac{0.059}{n} \log \frac{[\text{{products}}]}{[\text{{reactants}}]} \] For the copper half-cell reaction: \[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)} \]The equation simplifies to: \[ E = E^0 - \frac{0.059}{2} \log [\mathrm{Cu^{2+}}] \] Here, \( E^0 \) is the standard potential (+0.34 V), \([\mathrm{Cu^{2+}}]\) is the concentration, and \( n = 2 \) is the number of moles of electrons exchanged.

Step 2: Calculate Reduction Potentials

Plug in the concentration values for each electrode into the Nernst equation:1. For the first electrode with [Cu2+] = 0.0001 M:\[ E_1 = 0.34 - \frac{0.059}{2} \log(0.0001) \]2. For the second electrode with [Cu2+] = 0.1 M:\[ E_2 = 0.34 - \frac{0.059}{2} \log(0.1) \]3. For the third electrode with [Cu2+] = 0.01 M:\[ E_3 = 0.34 - \frac{0.059}{2} \log(0.01) \]4. For the fourth electrode with [Cu2+] = 0.001 M:\[ E_4 = 0.34 - \frac{0.059}{2} \log(0.001) \]

Step 3: Simplify and Evaluate Each Potential

Calculating each term:1. For \( E_1 \):\[ E_1 = 0.34 - \frac{0.059}{2} \times (-4) = 0.34 + 0.118 = 0.458 \, \text{V} \]2. For \( E_2 \):\[ E_2 = 0.34 - \frac{0.059}{2} \times (-1) = 0.34 + 0.0295 = 0.3695 \, \text{V} \]3. For \( E_3 \):\[ E_3 = 0.34 - \frac{0.059}{2} \times (-2) = 0.34 + 0.059 = 0.399 \, \text{V} \]4. For \( E_4 \):\[ E_4 = 0.34 - \frac{0.059}{2} \times (-3) = 0.34 + 0.0885 = 0.4285 \, \text{V} \]

Step 4: Compare Potentials and Determine Order

From the results:- \( E_1 = 0.458 \, \text{V} \)- \( E_4 = 0.4285 \, \text{V} \)- \( E_3 = 0.399 \, \text{V} \)- \( E_2 = 0.3695 \, \text{V} \)The order of electrode potentials from high to low is: 1 > 4 > 3 > 2.

Key Concepts

Electrode potential

Electrode potential, sometimes referred to as electrode reduction potential, is essentially the ability of an electrode system to gain electrons. It represents how easily a chemical species can be reduced by gaining electrons. When you dip an electrode into a solution containing ions that can interact with it, a specific voltage, known as electrode potential, is established between the electrode and the solution. This potential can be measured and is essential in predicting the direction of electron flow in electrochemical cells.

• The greater the electrode potential, the greater the tendency of the metal to gain electrons and be reduced.
• This potential can be influenced by several factors, including the concentration of ions in the solution and the inherent properties of the materials involved.

Understanding electrode potential is crucial for determining how electrochemical cells function and for predicting redox reactions' spontaneity. In our example, we explore copper electrodes with varying copper ion concentrations to observe their corresponding electrode potentials.

Concentration effects on potential

Concentration directly impacts the electrode potential of an electrochemical cell due to the Nernst equation. As the concentration of ions in the solution changes, it alters the cell's electromotive force (emf). This can be thought of as adjusting the equilibrium position of the redox reaction at the electrode surface.
According to the Nernst equation:\[E = E^0 - \frac{0.059}{n} \log [\text{ion concentration}]\]

• When the concentration of the ion decreases, the value of the logarithmic term becomes more negative, which increases the calculated potential \(E\).
• Conversely, increasing the ion concentration makes the logarithm less negative, decreasing the electrode potential.

In the exercise, you can see how changing the concentration of \(\mathrm{Cu^{2+}}\) affects the reduction potential. Lower concentrations lead to higher electrode potentials because increased electron uptake is needed to balance the lowered ionic presence. It is crucial to consider these concentration effects when designing or analyzing electrochemical cells.

Standard reduction potential

The standard reduction potential is an essential concept in electrochemistry, representing the tendency of a chemical species to gain electrons under "standard conditions"—usually at a concentration of 1 M, a pressure of 1 atm, and a temperature of 25°C (298 K). The standard reduction potential is denoted by \(E^0\) and is measured in volts. This standard value serves as a reference point for comparing the intrinsic abilities of different electrodes to act as reducing agents.

• A positive \(E^0\) suggests that the species is more likely to gain electrons, meaning it is a better oxidizing agent.
• A negative \(E^0\) indicates a weaker tendency to be reduced.

In the problem, the standard reduction potential for the reaction \(\mathrm{Cu^{2+} \, + \, 2e^- \, \rightarrow Cu(s)}\) is given as +0.34 V. This indicates copper's relatively strong tendency to be reduced to its solid form under standard conditions. By comparing the measured potentials to this standard value, one can determine how concentration variations lead to potential changes using the Nernst equation. Understanding these concepts helps in calculating and predicting reaction behavior in electrochemical systems.