On electrolysis of acidified water, if volume of hydrogen liberated is \(5.6 \mathrm{~cm}^{3}\), then the volume of oxygen liberated equal to: (a) \(1.4 \mathrm{~cm}^{3}\) (b) \(2.8 \mathrm{~cm}^{3}\) (c) \(8.2 \mathrm{~cm}^{3}\) (d) \(5.6 \mathrm{~cm}^{3}\)

Short Answer

Expert verified
The volume of oxygen liberated is \(2.8 \mathrm{~cm}^{3}\).

Step by step solution

01

Understanding the Electrolysis of Water

Recognize that during the electrolysis of water, hydrogen and oxygen gases are formed at cathode and anode respectively. The reaction is represented by the equation: \(2H_2O (l) \rightarrow 2H_2 (g) + O_2 (g)\). From the balanced chemical equation, it can be seen that two volumes of hydrogen gas are formed for every one volume of oxygen gas.
02

Calculate the Volume of Oxygen

Using the stoichiometry of the reaction, the volume of oxygen is half the volume of hydrogen produced because the ratio of volume of hydrogen to oxygen is 2:1. Therefore, to find the volume of the oxygen produced, divide the volume of hydrogen by 2: \(\frac{5.6 \mathrm{~cm}^3}{2} = 2.8 \mathrm{~cm}^3\).

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Key Concepts

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

Stoichiometry in Electrolysis
Stoichiometry is akin to a recipe for chemists. It involves using balanced chemical equations to calculate the quantities of reactants and products involved in chemical reactions. When solving problems related to the electrolysis of water, stoichiometry allows us to predict the amount of hydrogen and oxygen gas that will be produced.

In the electrolysis of water, the stoichiometric coefficients in the balanced chemical equation indicate the proportions of water (reactant) that decompose into hydrogen and oxygen gases (products). The balanced equation shows a 2:1 ratio, meaning two hydrogen molecules (or two volumes) are produced for every one oxygen molecule (or one volume).

This relationship can be applied whether we are measuring in moles, molecules, or—as in this exercise—volumes. Knowing how to use the stoichiometric ratios from the balanced equation is crucial in accurately determining the quantities of each substance produced or required.
Understanding Chemical Equations
Chemical equations represent the shorthand of chemistry, conveying the transformation of reactants to products during a chemical process. A well-balanced equation is essential because it follows the law of conservation of mass—matter is neither created nor destroyed.

The equation for the electrolysis of water, \(2H_2O (l) \rightarrow 2H_2 (g) + O_2 (g)\),is balanced. This means that the number of atoms for each element is the same on both the reactant and product sides. Such equations also inform us about the phase of matter, with '(l)' denoting liquid for water and '(g)' denoting gas for hydrogen and oxygen. By mastering the interpretation of chemical equations, students can contextualize the changes occurring during a reaction and predict the outcomes effectively.
Volume Ratio of Gases
When gases are produced or consumed in a chemical reaction, they occupy volumes that can be related through stoichiometric principles. In ideal conditions, at a constant temperature and pressure, Avogadro's law states that equal volumes of gases contain an equal number of molecules. This allows us to relate the volumes of gases directly to the coefficients in the balanced chemical equation.

In our exercise, the 2:1 volume ratio of hydrogen to oxygen reflects this stoichiometric relationship. For every two parts of hydrogen, one part of oxygen is produced during the electrolysis. It's a direct and essential relation that comes from the stoichiometry of the chemical equation involved—practical knowledge for handling problems about the gaseous products in electrolysis.

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Most popular questions from this chapter

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The reaction potential values of \(\mathrm{M}, \mathrm{N}\) and \(\mathrm{O}\) are \(+2.46\), \(-1.13\) and \(-3.13 \mathrm{~V}\) respectively. Which of the following order is correct, regarding their reducing property? (a) \(\mathrm{O}>\mathrm{N}>\mathrm{M}\) (b) \(\mathrm{O}>\mathrm{M}>\mathrm{N}\) (c) \(\mathrm{M}>\mathrm{N}>\mathrm{O}\) (d) \(\mathrm{M}>\mathrm{O}>\mathrm{N}\)

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