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Chapter 18: Problem 100

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

Short Answer

Expert verified
The volumes of hydrogen and oxygen gas produced during the electrolysis of water by a current of 2.50 A in 15.0 min are \(0.262 \mathrm{~L}\) H鈧� and \(0.131 \mathrm{~L}\) O鈧�, respectively.

Step by step solution

01

Calculate the charge passed through the circuit

We are given that the current (I) is 2.50 A and the electrolysis takes place over 15.0 minutes. With this information, we can calculate the charge (Q) passed through the circuit using the formula: Q = I 脳 t where t is the time in seconds. First, convert the minutes to seconds: t = 15.0 min 脳 (60 s/min) = 900 s Now, calculate the charge: Q = 2.50 A 脳 900 s = 2250 C (coulombs)
02

Determine the moles of electrons passed through the circuit

Next, we need to find the moles of electrons that passed through the circuit in this process. We know that 1 mole of electrons has a charge of 96485 C (Faraday's constant). So, we can find the number of moles (n) of electrons using the formula: n = Q / Faraday's constant n = 2250 C / 96485 C/mol = 0.0233 moles of electrons
03

Determine the stoichiometry of the electrolysis reaction

Now we need to find how many moles of gases are produced by the given moles of electrons in the electrolysis of water. The balanced electrolysis reaction is: 2H鈧侽 (l) 鈫� 2H鈧� (g) + O鈧� (g) From the stoichiometry of the reaction, we find that 4 moles of electrons are required to produce 2 moles of hydrogen gas and 1 mole of oxygen gas: 4 e鈦� + 2H鈧侽 (l) 鈫� 2H鈧� (g) + O鈧� (g)
04

Calculate the moles of gases produced

Now, we can calculate the moles of gases produced using the stoichiometry from the balanced reaction: Moles of H鈧� = (0.0233 moles e鈦� / 4 moles e鈦�) 脳 2 moles H鈧� = 0.0117 moles H鈧� Moles of O鈧� = (0.0233 moles e鈦� / 4 moles e鈦�) 脳 1 moles O鈧� = 0.00583 moles O鈧�
05

Convert moles of gases to volumes at STP

Finally, we can convert the moles of gases to volumes at STP (Standard Temperature and Pressure: 273 K and 1 atm). Using the molar gas volume at STP, which is 22.4 L/mol, we can find the volumes of hydrogen and oxygen gas: Volume of H鈧� = 0.0117 moles H鈧� 脳 22.4 L/mol = 0.262 L H鈧� Volume of O鈧� = 0.00583 moles O鈧� 脳 22.4 L/mol = 0.131 L O鈧� The volumes of hydrogen and oxygen gas produced during the electrolysis of water by a current of 2.50 A in 15.0 min are 0.262 L and 0.131 L, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charge Calculation
To find out how much "charge" has passed through a circuit during electrolysis, you need to know the current and the time. The formula for charge is \(Q = I \times t\). Here, the current \(I\) is 2.50 A (amperes), and the time \(t\) needs to be in seconds. You convert 15.0 minutes into seconds by multiplying by 60, because there are 60 seconds in a minute, giving you 900 seconds.
Plugging these values into the formula gives you \(Q = 2.50 \times 900 = 2250\) C, where C stands for coulombs, the unit of charge.
This number represents the total charge that flowed through the circuit during the 15 minutes of electrolysis.
Moles of Electrons
Once you have the total charge, you can calculate how many "moles of electrons" passed through the circuit. This is done using Faraday's constant, which is 96485 C/mol. This constant tells you how much charge one mole of electrons carries.
You use the formula \(n = \frac{Q}{\text{Faraday's constant}}\), where \(n\) stands for the moles of electrons. By applying our charge value, the equation becomes \(n = \frac{2250}{96485} \approx 0.0233\) moles of electrons.
This result means that 0.0233 moles of electrons participated in the electrolysis process.
Stoichiometry
"Stoichiometry" involves using a balanced chemical equation to see how substances react and in what proportions. In the electrolysis of water, the balanced reaction is:
\[2\text{H}_2\text{O} (l) \rightarrow 2\text{H}_2 (g) + \text{O}_2 (g)\]
This reaction tells us that 4 moles of electrons are needed to make 2 moles of hydrogen gas and 1 mole of oxygen gas. From the earlier calculation, we found 0.0233 moles of electrons are available, so we relate them to the stoichiometry.
  • Moles of \(\text{H}_2\) = \(\left(\frac{0.0233}{4}\right) \times 2 = 0.0117\) moles
  • Moles of \(\text{O}_2\) = \(\left(\frac{0.0233}{4}\right) \times 1 = 0.00583\) moles
This step helps us find out how many moles of hydrogen and oxygen gases are produced in the reaction.
Molar Volume at STP
"Molar volume" at STP (Standard Temperature and Pressure) allows us to convert the moles of gas calculated previously into a physical volume in liters. At STP, one mole of any gas occupies 22.4 L. This is very useful for understanding the actual size of the gas produced.
Using this conversion factor, the volumes are calculated as follows:
  • Volume of \(\text{H}_2\) = \(0.0117 \hspace{0.1cm} \text{moles} \times 22.4 \hspace{0.1cm} \text{L/mol} = 0.262 \hspace{0.1cm} \text{L}\)
  • Volume of \(\text{O}_2\) = \(0.00583 \hspace{0.1cm} \text{moles} \times 22.4 \hspace{0.1cm} \text{L/mol} = 0.131 \hspace{0.1cm} \text{L}\)
This calculation reveals that under standard conditions, 0.262 liters of hydrogen gas and 0.131 liters of oxygen gas are produced from the electrolysis process.

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