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) \(1>4>3>2\) (b) \(1>2>3>4\) (c) \(3>4>2>1\) (d) \(2>3>4>1\)

Step-by-Step Solution

Answer

Option (a) 1 > 4 > 3 > 2.

Step 1: Understand the Nernst Equation

The Nernst Equation is used to calculate the electrode potential at non-standard conditions. It is given by: \[E = E^0 - \frac{RT}{nF} \ln Q\]where:- \(E\) is the electrode potential.- \(E^0\) is the standard electrode potential (+0.34 V for \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\)).- \(R\) 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 (for copper, \(n = 2\)).- \(F\) is Faraday's constant (96485 C/mol).- \(Q\) is the reaction quotient, which for this system is the concentration of \(\mathrm{Cu}^{2+}\).

Step 2: Calculate Electrode Potentials Using the Nernst Equation

We assume the temperature \(T\) to be 298 K (room temperature). For each electrode, the potential \(E\) can be simplified to:\[E = E^0 - \frac{0.0592}{n} \log([\mathrm{Cu}^{2+}])\]Since \(n = 2\), this becomes:\[E = 0.34 - 0.0296 \times \log([\mathrm{Cu}^{2+}])\]

Step 3: Calculate Electrode Potential for Each Electrode

- **Electrode 1:** \([\mathrm{Cu}^{2+}] = 0.0001 \) M \[E_1 = 0.34 - 0.0296 \times \log(0.0001) = 0.34 - 0.0296 \times (-4) = 0.4584 \text{ V}\]- **Electrode 2:** \([\mathrm{Cu}^{2+}] = 0.1 \) M \[E_2 = 0.34 - 0.0296 \times \log(0.1) = 0.34 - 0.0296 \times (-1) = 0.3696 \text{ V}\]- **Electrode 3:** \([\mathrm{Cu}^{2+}] = 0.01 \) M \[E_3 = 0.34 - 0.0296 \times \log(0.01) = 0.34 - 0.0296 \times (-2) = 0.3992 \text{ V}\]- **Electrode 4:** \([\mathrm{Cu}^{2+}] = 0.001 \) M \[E_4 = 0.34 - 0.0296 \times \log(0.001) = 0.34 - 0.0296 \times (-3) = 0.4288 \text{ V}\]

Step 4: Compare the Calculated Potentials

The calculated potentials in decreasing order are:- \(E_1 = 0.4584 \text{ V}\)- \(E_4 = 0.4288 \text{ V}\)- \(E_3 = 0.3992 \text{ V}\)- \(E_2 = 0.3696 \text{ V}\)Thus, the decreasing order of electrode potentials is: 1 > 4 > 3 > 2.

Key Concepts

Nernst Equation

The Nernst Equation is a powerful tool used to determine the electrode potential when conditions vary from the standard state. Think of it as a bridge between the ideal world of standard conditions and the real-world where concentrations and temperatures might differ. The equation is given by:\[E = E^0 - \frac{RT}{nF} \ln Q\]Here's a quick breakdown:

• E is the electrode potential you want to find.
• E⁰ is the standard electrode potential, a known value for many reactions.
• R is the universal gas constant, equal to 8.314 J/mol·K.
• T is the absolute temperature in Kelvin.
• n is the number of moles of electrons transferred for the reaction.
• F is Faraday's constant, 96485 C/mol.
• Q represents the reaction quotient, calculated from ion concentrations.

In our example with copper electrodes, by assuming a constant temperature (usually 298 K), the equation simplifies considerably, making it easier to calculate potential based on the concentration of copper ions.

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. It's essentially like an eagerness meter for elements and ions wanting to gain electrons. The larger the reduction potential, the more likely the species will gain electrons. Each element or ion has a specific standard reduction potential measured in volts. For example, the standard reduction potential for the copper ion \(\mathrm{Cu}^{2+}\mid \mathrm{Cu}\)is +0.34 V. This is a baseline comparison to see how keen copper ions are to grab additional electrons compared to a hydrogen electrode, which is given a value of zero. This exercise looks at how potential changes with the concentration of the ion. By altering the concentration from standard conditions, we can observe how the eagerness to gain electrons shifts.

Concentration of Ions

Concentration of ions in a solution critically influences electrode potential. According to the Nernst Equation, the potential changes as a function of the ion concentration. The reaction quotient, or \(Q\) in the equation, depends directly on the concentration of ions, often calculated as the ratio of the concentrations of products to reactants. If the concentration of ions is high, the potential decreases, while a low concentration results in an increased potential. For example, if you have a \(\mathrm{Cu}^{2+}\)electrode with a concentration of 0.0001 M, the resulting potential is quite high at 0.4584 V. Compare this with a concentration of 0.1 M yielding a lower potential of 0.3696 V. This allows scientists to manipulate electrode potentials simply by tweaking ion concentrations, a crucial concept for practical applications in electrochemistry.

Standard Electrode Potential

The standard electrode potential is a fundamental reference point in electrochemistry. It's the measure of the potential difference between an electrode and a standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C temperature). Given as \(E^{0}\), this value is used to compare how various electrodes behave under the same conditions. The point of comparison is hydrogen's potential set at exactly 0.00 volts. Any electrode potential higher than this indicates a greater likelihood of reduction (gaining electrons). For copper (\(\mathrm{Cu}^{2+}\mid \mathrm{Cu}\)), the standard electrode potential is +0.34 V, showing it's more favorable to reduce copper ions as compared to hydrogen ions. This standard measure is used as the starting point in the Nernst Equation for calculating potentials under varying conditions. Thus, understanding \(E^{0}\) is integral to understanding more complex electrochemical behaviors.