Question

The number of moles of hydrogen oxidized is:

• A. 0.33 moles
• B. 33.3 moles
• C. 3.0 moles
• D. 1.33 moles
• A. 2 moles
• B. 4 moles
• C. 1 mole
• D. 6 moles
• A. 96500 C
• B. 579000 C
• C. 193000 C
• D. 48250 C
• A. 324 g
• B. 648 g
• C. 108 g
• D. 216 g
• A. Lead storage battery
• B. A generator set
• C. Ni-Cd cells
• D. H₂-O₂ fuel cell

Answer 1

The Correct Option is C

To find the number of moles of hydrogen oxidized, we must determine the amount of hydrogen gas reacting using the provided volume and standard conditions. At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Given that 67.2 liters of H₂ is reacted, we can calculate the moles as follows:

Number of moles of H₂ = \(\frac{67.2 \ \text{L}}{22.4 \ \text{L/mol}}\)

Calculating this gives:

= \(\frac{67.2}{22.4} = 3.0 \ \text{moles}\)

Therefore, the number of moles of hydrogen oxidized is 3.0 moles. This matches with the provided correct answer.

Answer 2

To determine the number of moles of hydrogen oxidized, we first need to understand the reaction taking place at the anode in a fuel cell: 

H₂ + 2OH^- → 2H₂O + 2e^-

The volume of H₂ provided is 67.2 L at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies 22.4 L. We calculate the number of moles of H₂ using:

Number of moles = \(\frac{\text{Volume at STP}}{\text{Molar volume at STP}}\)

The calculation is:

Number of moles = \(\frac{67.2 \, \text{L}}{22.4 \, \text{L/mol}} = 3.0 \, \text{moles}\)

Hence, the number of moles of hydrogen oxidized is 3.0 moles.

Answer 3

The problem asks for the number of moles of electrons produced during the oxidation of hydrogen (H₂). Let's analyze the provided anode reaction in a fuel cell:

H₂ + 2OH^- → 2H₂O + 2e^-

From the equation, it is evident that 1 mole of H₂ produces 2 moles of electrons (2e^-).

Given in the passage, 67.2 L of H₂ at STP is consumed. At STP, 1 mole of any ideal gas occupies 22.4 L. Thus, we can find the moles of H₂:

Number of moles of H₂ = 67.2 L / 22.4 L/mol = 3 moles

Since 1 mole of H₂ produces 2 moles of electrons, 3 moles of H₂ will produce:

3 moles of H₂ × 2 moles of e^-/mole of H₂ = 6 moles of electrons

Thus, the number of moles of electrons produced in the oxidation process is 6 moles.

Answer 4

In the electrochemical reactions provided for a fuel cell utilizing H₂ and O₂, H₂ is oxidized at the anode. The anode reaction is:

H₂ + 2OH^- → 2H₂O + 2e^-

This implies that each mole of H₂ generates 2 moles of electrons (2e^-) during the oxidation process.

Now, the passage states 67.2 L of H₂ is consumed at Standard Temperature and Pressure (STP). At STP, 22.4 L of any gas is equivalent to 1 mole. Therefore, the moles of H₂ are:

67.2 L / 22.4 L/mol = 3 moles of H₂

Since each mole of H₂ produces 2 moles of electrons, 3 moles of H₂ will produce:

3 moles of H₂ × 2 moles of e^-/mole of H₂ = 6 moles of electrons

Thus, the number of moles of electrons produced in the oxidation of H₂ is 6 moles.

Answer 5

To find the quantity of electricity (in Coulombs) produced during the oxidation of 67.2 L of H₂, we follow these steps: 

1. First, determine the number of moles of H₂ gas at STP. At standard temperature and pressure (STP), one mole of any gas occupies 22.4 L. Thus, for 67.2 L of H₂:
   67.222.4=3.
2. According to the anode reaction, each mole of H₂ produces 2 moles of electrons (2e^-). Therefore, 3 moles of H₂ will produce 3 × 2 = 6 moles of electrons.
3. Use Faraday's constant to convert moles of electrons to Coulombs. Faraday's constant is approximately 96500 C/mol.
   Thus, the total charge (Q) is:
   6×96500=579000 C.
4. Therefore, the quantity of electricity produced is 579000 C.

Answer 6

To calculate the quantity of electricity produced from the oxidation of 67.2 L of H₂, we first refer to the chemical reactions provided: 

Anode reaction: H₂ + 2OH^- → 2H₂O + 2e^-

This indicates that 1 mole of H₂ produces 2 moles of electrons. The molar volume of an ideal gas at STP is 22.4 L/mol.

1. Calculate the number of moles of H₂:
   Moles of H₂ = Volume / Molar volume = 67.2 L / 22.4 L/mol = 3 moles
2. Using the anode reaction, 1 mole of H₂ yields 2 moles of electrons, so 3 moles of H₂ will produce:
   3 moles of H₂ × 2 = 6 moles of electrons
3. Convert moles of electrons to charge (Faraday's constant is approximately 96500 C/mol):
   Total charge = 6 moles × 96500 C/mol = 579000 C

The quantity of electricity produced is thus 579000 C.

Answer 7

To determine the amount of silver deposited, we need to first understand the stoichiometry of the electrochemical reaction involved. The reduction of silver ions (Ag⁺) to silver metal (Ag) in an electrolytic process is given by the equation:

Ag⁺ + e^- → Ag

This indicates that one mole of electrons is required to deposit one mole of silver. The molar mass of silver is approximately 108 g/mol.

Step 1: Calculate the total charge used (Q) using the volume of hydrogen gas reacted.

The volume of hydrogen gas given is 67.2 L (at STP), which corresponds to 3 moles of H₂ because 1 mole of gas occupies 22.4 L at STP. Given that 2 moles of electrons are released per mole of H₂ (from the anode reaction), the total moles of electrons transferred is:

Total moles of electrons = 3 × 2 = 6

The charge (Q) passed can be calculated using Faraday's constant (F ≈ 96485 C/mol).

Q = 6 moles × 96485 C/mol = 578910 C

Step 2: Calculate the mass of silver deposited.

The number of moles of Ag deposited equals the moles of electrons, as each mole of electrons deposits one mole of silver.

Mass of Ag = moles of Ag × Molar mass of Ag = 6 moles × 108 g/mol = 648 g

Thus, the amount of silver deposited is 648 g.

Answer 8

To determine the amount of silver deposited, we first need to calculate the total charge transferred using the given volume of hydrogen gas reacted at standard temperature and pressure (STP).

Step 1: Calculation of moles of H₂
At STP, 1 mole of any gas occupies 22.4 L. Thus, the moles of H₂ reacted are:

\(n(H_2) = 67.2 \text{ L}<22.4 \text{ L/mol} = 3 \text{ mol}\)

Step 2: Determine total electrons involved
The anode reaction is: H₂ + 2OH^- → 2H₂O + 2e^-
From the reaction, 1 mole of H₂ gives 2 moles of electrons.

Total moles of electrons, n(e^-) = 3 mol H₂ × 2 mol e^-/mol H₂ = 6 mol e^-

Step 3: Calculate total charge (Q)
Using Faraday's constant (F = 96500 C/mol e^-), calculate the total charge:

Q = n(e^-) × F = 6 mol × 96500 C/mol = 579000 C

Step 4: Calculating mass of silver deposited
The reduction reaction for Ag⁺ is: Ag⁺ + e^- → Ag
1 mole of e^- deposits 1 mole of Ag. Thus, 6 moles of e^- deposit 6 moles of Ag.

Molar mass of silver (Ag) = 108 g/mol
Mass of Ag deposited = 6 mol × 108 g/mol = 648 g

Therefore, the amount of silver deposited is 648 g.

Answer 9

The source of electrical energy on the Apollo moon flight was the H₂-O₂ fuel cell. This choice is based on the principles of electrochemistry explained in the passage. In the context of an Apollo mission, fuel cells provided a reliable and efficient source of power. The hydrogen-oxygen (H₂-O₂) fuel cell is a type of galvanic cell where H₂ is oxidized at the anode and O₂ is reduced at the cathode, generating electricity through the following reactions:

Anode reaction: H₂ + 2OH^- → 2H₂O + 2e^-

Cathode reaction: O₂ + 2H₂O + 4e^- → 4OH^-

68.2 L of H₂ at standard temperature and pressure (STP) was utilized, sustaining power supply for the duration needed. Compared to other options like Lead storage batteries, generator sets, or Ni-Cd cells, fuel cells were preferred due to their higher energy density, ability to produce water as a byproduct, and suitability for space missions where weight and efficiency are critical.

Answer 10

The source of electrical energy on the Apollo moon flight was the H₂-O₂ fuel cell.

Fuel cells are a type of galvanic cell in which chemical energy from the combustion of fuels like hydrogen (H₂) is directly converted into electrical energy. In the case of H₂-O₂ fuel cells, hydrogen gas is oxidized at the anode, and oxygen is reduced at the cathode, producing electricity efficiently. This was ideal for space missions where reliability and efficiency are crucial.

Key Reactions:

Anode: H₂ + 2OH^- → 2H₂O + 2e^-

Cathode: O₂ + 2H₂O + 4e^- → 4OH^-

Through these reactions, a fuel cell efficiently combines hydrogen and oxygen to generate electricity, making it an excellent choice for applications in space, like the Apollo moon missions.