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Chapter 4: Problem 122

Many oxidation鈥搑eduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(i)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

Short Answer

Expert verified
a. Balanced equation: \(2\mathrm{Al}(s)+6\mathrm{HCl}(a q) \rightarrow 2\mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2}(g)\). Oxidized: \(\mathrm{Al}\), Reduced: \(\mathrm{H}\) from \(\mathrm{HCl}\) b. Balanced equation: \(\mathrm{CH}_{4}(g)+2\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+2\mathrm{H}_{2}\mathrm{S}(g)\). Oxidized: \(\mathrm{C}\) from \(\mathrm{CH}_{4}\), Reduced: \(\mathrm{S}\) c. Balanced equation: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(l)\). Oxidized: \(\mathrm{C}\) and \(\mathrm{H}\) from \(\mathrm{C}_{3}\mathrm{H}_{8}\), Reduced: \(\mathrm{O}_{2}\) d. Balanced equation: \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\). Oxidized: \(\mathrm{Cu}\), Reduced: \(\mathrm{Ag}^{+}\)

Step by step solution

01

Recognize the Redox Reaction

In this reaction, Aluminum (\(\mathrm{Al}\)) reacts with Hydrochloric acid (\(\mathrm{HCl}\)) to produce Aluminum chloride (\(\mathrm{AlCl}_{3}\)) and Hydrogen gas (\(\mathrm{H}_{2}\)). Aluminum loses electrons in the process, while Hydrogen gains electrons. Thus, this is a redox reaction.
02

Balance the Equation

To balance the equation, we need to ensure that the number of atoms for each element is equal on both sides of the equation: \(\underline{2}\mathrm{Al}(s)+\underline{6}\mathrm{HCl}(a q) \rightarrow \underline{2}\mathrm{AlCl}_{3}(a q)+\underline{3}\mathrm{H}_{2}(g)\) Now, the equation is balanced.
03

Identify the Oxidized and Reduced Species

In this reaction, Aluminum (\(\mathrm{Al}\)) is oxidized as it loses electrons, and Hydrogen, from Hydrochloric acid (\(\mathrm{HCl}\)), is reduced as it gains electrons. b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2}\mathrm{S}(g)\)
04

Recognize the Redox Reaction

In this reaction, Methane (\(\mathrm{CH}_{4}\)) reacts with Sulfur (\(\mathrm{S}\)) to produce Carbon disulfide (\(\mathrm{CS}_{2}\)) and Hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)). Carbon loses electrons, while Sulfur gains electrons. Thus, this is a redox reaction.
05

Balance the Equation

To balance the equation, we need to ensure that the number of atoms for each element is equal on both sides of the equation: \(\mathrm{CH}_{4}(g)+\underline{2}\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\underline{2}\mathrm{H}_{2}\mathrm{S}(g)\) Now, the equation is balanced.
06

Identify the Oxidized and Reduced Species

In this reaction, Carbon, from Methane (\(\mathrm{CH}_{4}\)), is oxidized as it loses electrons, and Sulfur (\(\mathrm{S}\)) is reduced as it gains electrons. c. \(\mathrm{C}_{3}\mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(l)\)
07

Recognize the Redox Reaction

In this reaction, Propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) reacts with Oxygen (\(\mathrm{O}_{2}\)) to produce Carbon dioxide (\(\mathrm{CO}_{2}\)) and Water (\(\mathrm{H}_{2}\mathrm{O}\)). Carbon and Hydrogen lose electrons, while Oxygen gains electrons. Thus, this is a redox reaction.
08

Balance the Equation

To balance the equation, we need to ensure that the number of atoms for each element is equal on both sides of the equation: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\underline{5}\mathrm{O}_{2}(g) \rightarrow \underline{3}\mathrm{CO}_{2}(g)+\underline{4}\mathrm{H}_{2} \mathrm{O}(l)\) Now, the equation is balanced.
09

Identify the Oxidized and Reduced Species

In this reaction, Carbon and Hydrogen, from Propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)), are oxidized as they lose electrons, and Oxygen (\(\mathrm{O}_{2}\)) is reduced as it gains electrons. d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)
10

Recognize the Redox Reaction

In this reaction, Copper (\(\mathrm{Cu}\)) reacts with Silver ion (\(\mathrm{Ag}^{+}\)) to produce Silver (\(\mathrm{Ag}\)) and Copper ion (\(\mathrm{Cu}^{2+}\)). Copper loses electrons, while Silver gains electrons. Thus, this is a redox reaction.
11

Balance the Equation

To balance the equation, we need to ensure that the number of atoms for each element is equal on both sides of the equation: The equation is already balanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)
12

Identify the Oxidized and Reduced Species

In this reaction, Copper (\(\mathrm{Cu}\)) is oxidized as it loses electrons, and Silver ion (\(\mathrm{Ag}^{+}\)) is reduced as it gains electrons.

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Key Concepts

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

Oxidation
Oxidation is a fundamental concept in chemistry, especially notable in redox reactions. It involves the loss of electrons by an atom, ion, or molecule. When an element undergoes oxidation, its oxidation state increases.
Consider the reaction of Aluminum (Al) with Hydrochloric acid (HCl) as an example. Here, Aluminum is oxidized as it loses electrons to form Aluminum chloride (AlCl鈧�). As Aluminum loses electrons, it transforms from a neutral metal into a positively charged Aluminum ion (Al鲁鈦�).
This electron loss is vital as it exemplifies one half of the redox reaction process.
  • The element being oxidized loses electrons.
  • Its oxidation state becomes more positive.
  • In chemical equations, such changes help indicate electron flow and energy transformation.
Understanding oxidation is crucial to predict how substances will react, especially when working with metals and their compounds.
Reduction
Reduction is the counterpart of oxidation in redox reactions. It involves the gain of electrons by a substance. This gain results in a decrease in the oxidation state of the element being reduced.
To illustrate, let's look at the reaction involving Copper (Cu) and Silver ions (Ag鈦�). Here, Silver ions are reduced to form elemental Silver (Ag). As Silver ions gain electrons, they transition from a positive charge to a neutral state.
Reduction is essential in balancing redox reactions and understanding chemical reactivity.
  • Reduction includes the gain of electrons.
  • The oxidation state of the reduced substance decreases.
  • It pairs with oxidation to form a complete redox cycle.
Mastering reduction allows chemists to harness electron transfer in processes ranging from electrolysis to energy storage.
Balancing Chemical Equations
Balancing chemical equations is a crucial skill in chemistry. It ensures that the law of conservation of mass is upheld by having the same number of each type of atom on both sides of the equation.
For instance, when balancing the reaction between Propane ( C鈧僅鈧�) and Oxygen ( O鈧�), one must ensure that each element's atoms are equal on both sides. This reaction produces Carbon dioxide (CO鈧�) and Water (H鈧侽) and needs a specific stoichiometric coefficient to be balanced.
The balanced equation reads: C鈧僅鈧� + 5O鈧� 鈫� 3CO鈧� + 4H鈧侽.
  • Each element must be balanced separately.
  • Changing coefficients, not subscripts, achieves balance.
  • Balancing helps predict how much of each reactant is needed or how much product is formed.
Balancing equations is foundational for quantitative analyses in chemistry.
Electron Transfer in Reactions
Electron transfer is at the heart of redox reactions, driving the transformation of substances. Every redox reaction involves a transfer of electrons between chemical species, altering their chemical identities.
In the case of Copper reacting with Silver ions, Copper loses electrons (oxidized) becoming Cu虏鈦�, while Silver ions gain electrons (reduced) to form Silver metal. This exchange maintains charge balance and enables the reaction to proceed.
Electron transfer is crucial for several applications:
  • It enables processes like corrosion, battery operation, and biological energy conversion.
  • Understanding electron flow helps design chemical processes and materials.
  • Electron transfers underpin energy-efficient reactions in industry and nature.
Grasping electron transfer chemistry opens the door to innovations in materials science and renewable energy.

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Most popular questions from this chapter

A 1.42 -g sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\) was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.

The iron content of iron ore can be determined by titration with a standard \(\mathrm{KMnO}_{4}\) solution. The iron ore is dissolved in \(\mathrm{HCl},\) and all the iron is reduced to \(\mathrm{Fe}^{2+}\) ions. This solution is then titrated with \(\mathrm{KMnO}_{4}\), solution, producing \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) ions in acidic solution. If it required 38.37 \(\mathrm{mL}\) of 0.0198 \(\mathrm{M}\) \(\mathrm{KMnO}_{4}\) to titrate a solution made from 0.6128 \(\mathrm{g}\) of iron ore, what is the mass percent of iron in the iron ore?

One of the classic methods for determining the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing the periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

What volume of 0.100\(M \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from 150.0 \(\mathrm{mL}\) of 0.250\(M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO\(_{4}\) obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?
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