Problem 47 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidised at the anode and oxygen at the cathode. If \(67.2\) litre of \(\mathrm{H}_{2}\) at STP react in 15 minutes, what is the average current produced? If the entire current is used for electro deposition of copper from copper (II) solution, how many grams of copper will be deposited? [1988 - 4 Marks] Anode reaction : \(\mathrm{H}_{2}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-}\) Cathode reaction : \(\frac{1}{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}\).

Step-by-Step Solution

Answer

The average current is 643.33 A, and 190.5 g of copper is deposited.

Step 1: Determine Moles of Hydrogen

Start by calculating the moles of hydrogen gas at STP using the volume given. Since volume of one mole of an ideal gas at STP is 22.4 L, use the formula: \[ \text{Moles of } \text{H}_2 = \frac{67.2 \text{ L}}{22.4 \text{ L/mol}} = 3 \text{ moles}\]

Step 2: Calculate Total Charge Transferred

Based on the anode reaction, the oxidation of one mole of \(\text{H}_2\) produces 2 moles of electrons. Therefore, the charge transferred is calculated via Faraday's constant (\(F = 96500 \text{ C/mol e}^-\)), expressed as:\[ \text{Total charge} = 3 \text{ moles} \times 2 \times 96500 \text{ C/mol} = 579000 \text{ C}\]

Step 3: Calculate Average Current

Using the relationship between current, charge, and time, compute the average current. Given the reaction duration is 15 minutes (or 900 seconds), use:\[ I = \frac{Q}{t} = \frac{579000 \text{ C}}{900 \text{ s}} = 643.33 \text{ A}\]

Step 4: Determine Copper Deposition

For copper deposition, the reaction \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \) requires 2 moles of electrons to deposit 1 mole of copper. Calculate the moles of copper using the total charge:\[ \text{Moles of Cu} = \frac{579000 \text{ C}}{2 \times 96500 \text{ C/mol}} = 3 \text{ moles}\]

Step 5: Convert Moles of Copper to Grams

Finally, convert moles to grams using the molar mass of copper (63.5 g/mol):\[ \text{Mass of Cu} = 3 \text{ moles} \times 63.5 \text{ g/mol} = 190.5 \text{ g}\]

Key Concepts

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is what we commonly understand as electricity. In electrochemical reactions, redox reactions take place. A redox reaction (short for reduction-oxidation) involves the transfer of electrons between two species.
- **Reduction** is the gain of electrons by a molecule, atom, or ion. For example, when oxygen gains electrons in a fuel cell, it undergoes reduction. - **Oxidation** is the loss of electrons, such as when hydrogen gas loses electrons at the anode.
In a fuel cell, this process occurs at electrodes and electrical energy is generated from chemical reactions. Understanding electrochemistry is crucial for designing and optimizing technologies like fuel cells and batteries.

Faraday's Law of Electrolysis

Faraday's law of electrolysis is a fundamental concept in electrochemistry. It states that the amount of a substance produced at each electrode during electrolysis is directly proportional to the total electric charge passed through the system.
Faraday’s laws can be summarized as:

• The mass of substances liberated or deposited at an electrode is proportional to the number of coulombs of electricity passed.
• The masses of different substances liberated or deposited by the passage of the same quantity of electricity are proportional to their chemical equivalent weights.

Using Faraday's constant (approximately 96500 C/mol of electrons), you can determine the total charge transferred in reactions, as seen in the exercise when calculating total charge based on moles of electrons produced. This helps in predicting the amount of metal deposited during electrolysis, like the copper in the example problem.

Anode Reaction

In the context of a fuel cell, the anode is where oxidation takes place. Specifically, in the exercise provided, the reaction at the anode is: \[\text{H}_2 + 2 \text{OH}^- \rightarrow 2 \text{H}_2\text{O} + 2 \text{e}^- \] Here, hydrogen gas (\(\text{H}_2\)) loses electrons, i.e., is oxidized, resulting in the creation of water and the release of electrons. These liberated electrons then travel through an external circuit to the cathode, creating an electric current. This conversion of hydrogen into water at the anode is a critical component of energy generation in hydrogen fuel cells. Understanding the anode's function is necessary for advancing fuel cell technology, where efficiency and fuel purity are knit closely with performance.

Cathode Reaction

At the cathode of a fuel cell, reduction occurs. The exercise involves the following cathode reaction: \[\frac{1}{2} \text{O}_2 + \text{H}_2\text{O} + 2 \text{e}^- \rightarrow 2 \text{OH}^- \] This reaction shows the reduction of oxygen, where oxygen molecules gain electrons and, in the presence of water, are transformed into hydroxide ions (\(\text{OH}^-\)). The attracted electrons flow from the anode side through an external circuit to the cathode, which creates the path for electric current. Reducing agents at the cathode, such as oxygen, are crucial since they facilitate the conversion of electrical energy in the cell. Understanding such reactions is significant for the effective designing of cathodes in fuel cells and realizing their potential in sustainable energy applications.