Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider a \(70 \%\) efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and \(298 \mathrm{~K}\). Its cell reaction is $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ The work derived from the cell on the consumption of \(1.0 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) is used to compress \(1.00 \mathrm{~mol}\) of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in \(\mathrm{K}\) ) of the ideal gas? The standard reduction potentials for the two half-cells are given below. $$ \begin{aligned} &\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l), E^{0}=1.23 \mathrm{~V}, \\ &2 \mathrm{H}^{+}(a q)+2 e^{-} \rightarrow \mathrm{H}_{2}(g), E^{0}=0.00 \mathrm{~V} \end{aligned} $$ Use \(F=96500 \mathrm{C} \mathrm{mol}^{-1}, R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\).

Step-by-Step Solution

Answer

The change in temperature of the ideal gas is approximately -13.34 K.

Step 1: Calculate the Standard Cell Potential

The overall reaction is \( \mathrm{H}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2\mathrm{O}(l) \). To find the standard cell potential \(E^0_{cell}\), use the formula: \[E^0_{cell} = E^0_{cathode} - E^0_{anode}.\] Substitute the given values: \[E^0_{cell} = 1.23 \, \text{V} - 0.00 \, \text{V} = 1.23 \, \text{V}.\]

Step 2: Determine Maximum Theoretical Work

Use the formula for maximum theoretical work (Gibbs free energy change): \[\Delta G^0 = -nFE^0_{cell}.\] For this reaction, \(n = 2\) moles of electrons are transferred. Substitute \(F = 96500 \, \text{C/mol}\) and \(E^0_{cell} = 1.23 \, \text{V}\): \[\Delta G^0 = -2 \times 96500 \times 1.23 \approx -237390 \, \text{J/mol}.\]

Step 3: Compute Actual Work from Efficiency

The efficiency of the fuel cell is 70%, so the work done (\(W\)) is: \[W = \eta \cdot \Delta G^0.\] Substitute \(\eta = 0.70\) and \(\Delta G^0 \): \[W = 0.70 \times (-237390) \approx -166173 \, \text{J/mol}.\] For \(1.0 \times 10^{-3} \, \text{mol of H}_2\), the work is: \[W_{actual} = 1.0 \times 10^{-3} \times (-166173) = -166.173 \, \text{J}.\]

Step 4: Apply Work to Ideal Gas Compression

The work derived from the cell compresses \(1.00 \text{ mol}\) of a monoatomic ideal gas. The work \(W_{actual}\) is equal to the change in internal energy \(\Delta U\) because the process is adiabatic: \[\Delta U = W_{actual} = -166.173 \, \text{J}.\]

Step 5: Calculate Change in Temperature of Gas

For a monoatomic ideal gas, \(C_v = \frac{3}{2} R\). Use the formula: \[\Delta U = nC_v\Delta T.\] Substitute \(n = 1 \, \text{mol}\), \(C_v = \frac{3}{2} R = \frac{3}{2} \times 8.314\), and \(\Delta U = -166.173 \, \text{J}\): \[-166.173 = 1 \times \frac{3}{2} \times 8.314 \times \Delta T.\] Solve for \(\Delta T\): \[\Delta T = \frac{-166.173}{\frac{3}{2} \times 8.314} \approx -13.34 \, \text{K}.\]

Key Concepts

Fuel Cell Efficiency

Fuel cell efficiency is a measure of how effectively a fuel cell converts chemical energy into electrical energy. In this problem, we are dealing with a hydrogen-oxygen fuel cell. This type of fuel cell is known for its clean energy output, producing only water as a by-product. The efficiency of such a cell is determined by the ratio of the usable work (or energy output) to the total energy available from the fuel. Consider the standard efficiency equation:

• Efficiency (\(\eta\)) = \(\frac{\text{Usable Work}}{\Delta G^0}\)
• Usable work is the actual work performed by the system, affected by real-world inefficiencies.
• #Delta# \(G^0\) represents the change in Gibbs free energy, indicating the maximum possible work the reaction can perform at standard conditions.

In this case, the fuel cell efficiency is given as 70%. This means that 70% of the Gibbs free energy change is converted into useful work, while the remaining 30% is lost (often as heat). This efficiency rating is important for understanding the real-world performance of the fuel cell.

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) is a thermodynamic quantity that can predict the direction of chemical reactions and the maximum reversible work obtainable from a reaction at constant temperature and pressure.

The Role of Gibbs Free Energy in Fuel Cells

In the context of the fuel cell, \(\Delta G^0\) quantifies the maximum amount of work that can be extracted from the cell. This is used to determine how much energy is available to drive the reaction. The formula to find the change in Gibbs free energy for a standard cell reaction is:

• \(\Delta G^0 = -nFE^0_{cell}\)
• where \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant (96500 C/mol), and \(E^0_{cell}\) is the cell potential.

In this exercise, the value of \(\Delta G^0\) is calculated to be approximately -237390 J/mol, indicating a spontaneous reaction under standard conditions. Gibbs free energy fundamentally supports understanding how much usable energy can be derived from the reaction and the extent of efficiency loss due to inefficiencies in the fuel cell.

Ideal Gas Law

The ideal gas law describes the behavior of gases in terms of its pressure, volume, temperature, and number of moles. In this exercise, the work done by the fuel cell is harnessed to compress a monoatomic ideal gas.

Understanding the Compression using Ideal Gas Law

The ideal gas law is represented as:

• \(PV = nRT\)
• where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant (8.314 J/mol⋅K), and \(T\) is temperature.

In our scenario, the energy (work) derived from the fuel cell is used to compress the gas. Due to the adiabatic nature of the container (thermally insulated), no heat exchange occurs. The entire internal energy change (\(\Delta U\)) is caused by the work done. For a monoatomic ideal gas, the change in internal energy can be related to the change in temperature as follows:

• \(\Delta U = nC_v\Delta T\)
• with \(C_v\) being the molar heat capacity at constant volume (\(\frac{3}{2}R\) for monoatomic gases).

This relationship allows us to calculate the change in temperature, concluding that the temperature decreases by approximately 13.34 K as the gas is compressed adiabatically using the work derived from the fuel cell.