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What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at \(\mathrm{STP}\) 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 gases produced from the electrolysis of water by a current of 2.50 A in 15.0 minutes at STP are approximately \(0.327 \: L\) of H鈧(g) and \(0.164 \: L\) of O鈧(g).

Step by step solution

01

Calculate the total charge passed through the electrolyte

To calculate the total charge passed through the electrolyte, we'll use the formula: Total charge (Coulombs, C) = Current (Amperes, A) * Time (seconds, s) We are given the current as 2.50 A, and the time as 15.0 minutes. First, we need to convert the time to seconds: 15.0 minutes * 60 seconds/minute = 900 seconds Now, we can find the total charge: Total charge = 2.50 A * 900 s = 2250 C
02

Apply Faraday's Law of electrolysis

According to Faraday's Law of electrolysis, the amount of substance (in moles) produced is directly proportional to the charge passed through the electrolyte: Moles of substance = (Charge in Coulombs) / (Faraday's constant * moles of electrons required for the reaction) For water electrolysis, the balanced chemical equation is: 2 H鈧侽(l) 鈫 2 H鈧(g) + O鈧(g) From the balanced equation, we know that 2 moles of H鈧 are produced for every 1 mole of O鈧. Also, 4 moles of electrons are required to produce 2 moles of H鈧, and 4 moles of electrons are required to produce 1 mole of O鈧. Now, we can calculate the moles of H鈧 produced: Moles of H鈧 = (2250 C) / (96485 C/mol * 4 mol e鈦/2 mol H鈧) = 0.0146 mol H鈧 Next, we'll find the moles of O鈧 produced using a similar approach, noting that it is half the moles of H鈧 produced: Moles of O鈧 = 0.5 * moles of H鈧 = 0.5 * 0.0146 mol = 0.0073 mol O鈧
03

Convert moles to volumes at STP

At standard temperature and pressure (STP, 0掳C, and 1 atm), the molar volume of any ideal gas is 22.4 liters/mol. We can use this information to find the volumes of H鈧 and O鈧 gases produced: Volume of H鈧 = moles of H鈧 * molar volume at STP = 0.0146 mol * 22.4 L/mol = 0.327 L Volume of O鈧 = moles of O鈧 * molar volume at STP = 0.0073 mol * 22.4 L/mol = 0.164 L
04

Present the final answer

The volumes of hydrogen and oxygen gases produced from the electrolysis of water by a current of 2.50 A in 15.0 minutes at STP are approximately 0.327 L of H鈧(g) and 0.164 L of O鈧(g).

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Key Concepts

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

Faraday's Law
Faraday's Law of electrolysis is one of the foundational principles in electrochemistry. It quantitatively relates electric charge to the amount of substance liberated at an electrode during electrolysis. According to this law, the amount of chemical change is directly proportional to the charge passed through an electrolyte.

Let's break down what this means in a real-world scenario like our example exercise. When a current flows through an electrolytic cell, ions are reduced or oxidized at the electrodes. The amount of substance produced (in moles) depends on the total charge that has flown through the solution. But even more specifically, it also requires knowing the number of electrons needed for the specific chemical reaction occurring at the electrode.

For water electrolysis, generating hydrogen and oxygen gas, the stoichiometry of the reaction and the quantity of electrons involved are crucial. Knowing these variables allows us to use Faraday's Law to dictate the precise production of gases, tying the world of physics (electric charge) to the realms of chemistry (moles of compounds).
Electrolysis Calculations
Electrolysis calculations are all about translating electrical energy into chemical change. In the example exercise, we performed calculations to determine the volumes of hydrogen and oxygen gas produced during the electrolysis of water.

Navigating through these calculations involves several steps. First, we determine the total charge that has passed through the electrolyte using the product of current and time. Once we have this total charge, we can utilize Faraday's Law to calculate the moles of substance produced, taking into account the number of electrons transferred per mole of substance in the reaction.

To illustrate:
  • Total electric charge is current multiplied by time.
  • Using Faraday's Law, we link this charge to the chemical quantity, moles of substances produced.
  • The stoichiometry of the reaction provides the ratio between the moles of different substances, like how one mole of O鈧 is produced per two moles of H鈧 in water electrolysis.

With a clear understanding of these steps, the student can approach any electrolysis problem with confidence and solve for the amounts of substances produced during the process.
Molar Volume at STP
The molar volume at STP (Standard Temperature and Pressure) is a concept that simplifies the work with gases in chemical calculations. At STP, which is defined as 0掳C (273.15K) and 1 atm pressure, one mole of any ideal gas occupies 22.4 liters. This is a central assumption that we use to convert between moles and volume of gases.

In our exercise, we applied this concept to transform moles of hydrogen and oxygen into the corresponding volumes they would occupy at STP. It's important to remember that this is an approximation, best suited for ideal gases, which assumes no interactions between the gas molecules and that the volume of the gas particles themselves can be ignored. While real gases do not perfectly adhere to these assumptions, the molar volume at STP provides a close enough approximation for many practical purposes in chemistry, especially in educational contexts.

Understanding this concept allows students to easily toggle between the microscale (moles) and the macroscale (volume), enhancing their grasp of theoretical principles and their practical applications.
Stoichiometry
Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It's the calculation of the quantities of substances consumed and produced in a chemical reaction. Stoichiometry is governed by the laws of conservation of mass and conservation of charge, which dictate that atoms and charge must be balanced on both sides of a chemical equation.

In the context of our electrolysis example, understanding stoichiometry is crucial for determining the ratio of hydrogen to oxygen gas produced. The balanced chemical equation for the electrolysis of water shows us that two moles of hydrogen gas are produced for every mole of oxygen gas. This 2:1 ratio is an outcome of stoichiometry and is used as an intermediate step in our calculations, allowing us to link the moles of hydrogen and oxygen through their stoichiometric relationship.

Strengthening one's stoichiometry skills is essential not just for solving textbook exercises, but also for anticipating the outcomes of chemical reactions, whether it be in an academic setting or a professional laboratory.

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Most popular questions from this chapter

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 A. How long will it take to plate out \(99 \%\) of the platinum in 0.50 L of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{c} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\). \(\mathrm{Ox}\) ide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C}\) ). \(\Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is -380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between \(\Delta G\) and cell potential to do this.)

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

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