Problem 32 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An oxidation-reduction reaction in which 3 electrons are transferred has a \(\Delta \mathrm{G}^{0}\) of \(17.37 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). The value of \(\mathrm{E}_{\text {cell }}^{0}\) (in \(\mathrm{V}\) ) is \(\times 10^{-2}\) \(\left(1 \mathrm{~F}=96,500 \mathrm{C} \mathrm{mol}^{-1}\right)\)

Step-by-Step Solution

Answer

The value of \(E_{cell}^{0}\) is approximately \(-6.0 \times 10^{-2}\) V.

Step 1: Understand the given values

We have been given that 3 electrons are transferred, \(\Delta G^{0} = 17.37 \text{ kJ mol}^{-1}\), and \(T = 25^{\circ}\text{C}\). Additionally, we have \(1 \text{ F} = 96,500 \text{ C mol}^{-1}\). The temperature is not directly needed here since we are already given standard conditions.

Step 2: Convert Gibbs Free Energy to Joules

The given \(\Delta G^{0}\) is in kJ. Convert this to Joules by multiplying by 1,000:\(\Delta G^{0} = 17.37 \text{ kJ mol}^{-1} \times 1,000 = 17,370 \text{ J mol}^{-1}\).

Step 3: Apply the Nernst Equation for \(E_{cell}^{0}\)

The formula connecting \(\Delta G^{0}\) and \(E_{cell}^{0}\) is:\[ \Delta G^{0} = -nFE_{cell}^{0} \]where \(n = 3\) (the number of electrons transferred) and \(F = 96,500 \text{ C mol}^{-1}\).

Step 4: Substitute known values into the formula

Substitute \(\Delta G^{0} = 17,370 \text{ J mol}^{-1}\), \(n = 3\), and \(F = 96,500 \text{ C mol}^{-1}\) into the formula:\[ 17,370 = -3 \times 96,500 \times E_{cell}^{0} \].

Step 5: Solve for \(E_{cell}^{0}\)

Rearrange the equation to solve for \(E_{cell}^{0}\):\[ E_{cell}^{0} = -\frac{17,370}{3 \times 96,500} \approx -0.060 \text{ V} \].

Step 6: Express \(E_{cell}^{0}\) as \(\times 10^{-2}\)

Convert \(-0.060 \text{ V}\) to scientific notation as:\[ -6.0 \times 10^{-2} \text{ V}\].

Key Concepts

Gibbs Free Energy

In the realm of chemistry and physics, Gibbs Free Energy (\(\Delta G\)) is a pivotal concept. It gauges the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. Put simply, it's a measure of a system's spontaneity:

• A negative \(\Delta G\) signifies a spontaneous process.
• A positive \(\Delta G\) denotes a non-spontaneous process.

For the given exercise, we know \(\Delta G^{0} = 17.37 \text{ kJ mol}^{-1}\). Since this value is positive, it means the reaction isn't spontaneous under standard conditions.
This equation connects Gibbs Free Energy to other important variables: \[\Delta G^{0} = -nFE_{cell}^{0}\]This formula ties together the number of electrons \(n\), the Faraday constant \(F\) (charge per mole of electrons), and the standard cell potential \(E_{cell}^{0}\), bridging chemical energy and electrical potential.

Oxidation-Reduction Reaction

Oxidation-reduction reactions, often called "redox reactions," are a fundamental part of chemistry. They involve the transfer of electrons between chemical species. Here's what happens:

• Oxidation refers to the loss of electrons.
• Reduction involves gaining electrons.

Together, these processes help in transforming substances. In the exercise, we deal with the transfer of 3 electrons, represented by \(n = 3\). This indicates the reaction involves three pairs of oxidation-reduction events.
The significance of redox reactions extends beyond chemistry labs. They're crucial for biological processes like respiration and photosynthesis, and in industrial applications, such as batteries and corrosion prevention.

Standard Cell Potential

Standard cell potential (\(E_{cell}^{0}\)) is a key concept in electrochemistry. It represents the voltage or electrical potential difference of a cell under standard conditions (1 M concentration, 1 atm pressure, 25°C). \(E_{cell}^{0}\) is quantified in volts (V).With our exercise, calculating the standard cell potential involves using the Nernst equation to determine the electrical potential based on the Gibbs Free Energy change. We use the formula:\[\Delta G^{0} = -nFE_{cell}^{0}\]By rearranging this equation, we solve for \(E_{cell}^{0}\), discovering it to be approximately \(-0.060 \text{ V}\) or \(-6.0 \times 10^{-2} \text{ V}\).
Standard cell potentials help indicate how favorable a redox reaction is under standard conditions. Larger positive values show spontaneity towards the products, while negative values suggest non-spontaneity.