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Write balanced equations for the following half-reactions. Specify whether each is an oxidation or reduction. (a) \(\operatorname{Cr}(\mathrm{s}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq}) \quad\) (in acid) (b) \(\mathrm{AsH}_{3}(\mathrm{g}) \longrightarrow \mathrm{As}(\mathrm{s}) \quad\) (in acid) (c) \(\mathrm{VO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{V}^{2+}(\mathrm{aq}) \quad\) (in acid) (d) \(\mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) (in base)

Short Answer

Expert verified
(a) Oxidation: \( \operatorname{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^- \); (b) Oxidation: \( \mathrm{AsH}_3 + 3H^+ + 3e^- \rightarrow \mathrm{As} + 3H_2O \); (c) Reduction: \( \mathrm{VO}_3^- + 4H^+ + e^- \rightarrow \mathrm{V}^{2+} + 3H_2O \); (d) Oxidation: \( 2\mathrm{Ag} + 2OH^- \rightarrow \mathrm{Ag}_2\mathrm{O} + H_2O + 2e^- \).

Step by step solution

01

Understand the Concept of Half-Reactions

A half-reaction is either the oxidation or reduction reaction component of a redox reaction. In a redox process, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. In acidic or basic solutions, half-reactions are balanced for both charge and mass, considering the presence of H鈦 for acidic and OH鈦 for basic solutions.
02

Write Half-Reaction for (a)

The unbalanced half-reaction is: \( \operatorname{Cr} \rightarrow \mathrm{Cr}^{3+} \). In acidic solution, to balance the charges, we add 3 electrons to the right: \( \operatorname{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^- \). This is an oxidation reaction as Cr loses electrons.
03

Write Half-Reaction for (b)

The unbalanced half-reaction is: \( \mathrm{AsH}_3 \rightarrow \mathrm{As} \). Balance elements first, and in acidic solution, add 3 electrons and 3 protons to get: \( \mathrm{AsH}_3 \rightarrow \mathrm{As} + 3H^+ + 3e^- \). Then balance the hydrogen with water: \( \mathrm{AsH}_3 + 3H^+ + 3e^- \rightarrow \mathrm{As} + 3H_2O \). This is an oxidation reaction.
04

Write Half-Reaction for (c)

The unbalanced half-reaction is: \(\mathrm{VO}_3^- \rightarrow \mathrm{V}^{2+} \). Balance V and O by adding water, and then balance the charge with H鈦 and electrons: \( \mathrm{VO}_3^- + 4H^+ + e^- \rightarrow \mathrm{V}^{2+} + 3H_2O \). This is a reduction reaction as electrons are gained.
05

Write Half-Reaction for (d)

The unbalanced half-reaction is: \( \mathrm{Ag} \rightarrow \mathrm{Ag}_2\mathrm{O} \). In basic solution, use OH鈦 to balance O and H: \( 2\mathrm{Ag} + 2OH^- \rightarrow \mathrm{Ag}_2\mathrm{O}+ H_2O + 2e^- \). This reaction is an oxidation reaction due to the loss of electrons by Ag.

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Key Concepts

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

Half-Reaction Balancing
In redox chemistry, balancing half-reactions is a crucial skill. Half-reactions represent the processes of either oxidation or reduction that occur during a redox reaction. To balance these reactions, we must focus on both mass and charge balance.
To start, identify the species being oxidized or reduced, and write the chemical equation. Then, to achieve mass balance, ensure that atoms other than oxygen and hydrogen are balanced first.
For oxygen, add water molecules (Step 4's VO3- example adds water to balance oxygen). To balance hydrogen, add protons (H鈦) in acidic solutions or hydroxide ions (OH鈦) in basic solutions. In step 5's Ag to Ag2O, OH鈦 was used to balance hydroxides.
Then, balance the charge by adding electrons鈥攑lace them on the side that requires charge balance. In acidic conditions, electrons are typically added to the reactant side, as shown in Step 2 with Cr to Cr鲁鈦.
  • Ensure both sides of the equation have equal numbers of each type of atom and equal overall charge.
  • Check charge and mass balance after finishing to confirm correctness.
  • Balance half-reactions first, and then combine them to get the total redox equation if necessary.
Particularly useful in understanding the detailed arrangements of redox reactions, balancing half-reactions provides a cleaner, organized view.
Oxidation and Reduction
Oxidation and reduction are two sides of a redox reaction, a core concept in chemistry.
Oxidation involves the loss of electrons by a substance, while reduction happens when a substance gains electrons.
In the steps you've seen:
  • Chromium (Cr) in Step 2 is oxidized to Cr鲁鈦, as it loses three electrons.
  • Arsine (AsH鈧) in Step 3 oxidizes to arsenic (As), with the release of three electrons and the subsequent balancing with protons.
  • The reduction of VO3鈦 to V2+ in Step 4 shows the gain of electrons.
The mnemonic "OIL RIG" can help: Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).
By understanding this concept, you determine the redox behavior of substances in a reactive environment. Always remember鈥攊f something is oxidized, another thing is being reduced to maintain balance. This interplay ensures that charge conservation is maintained.
Acidic and Basic Solutions
In chemistry, the conditions of the solution (acidic or basic) notably influence the balancing of redox reactions.
Acidic solutions are characterized by a high concentration of H鈦 ions. When balancing in an acidic medium, you use H鈦 ions to balance hydrogen atoms.
For example, in Step 2 and 3 involving Cr and AsH鈧 respectively, protons (H鈦) are added to balance the hydrogen.
  • Balancing oxygen often requires adding H鈧侽 molecules.
  • For charges, electrons are added to counterbalance the net positive or negative charge created by adding H鈦.
Basic solutions, on the other hand, have a high concentration of OH鈦 ions. In these solutions, OH鈦 ions are used instead.
For example, in Step 5 when balancing Ag to Ag鈧侽, OH鈦 ions are used to balance the reaction. Subsequently, H鈧侽 is often added to clear any excess hydrogen atoms introduced.
  • The final neutralization involves ensuring all charges and atoms (especially hydrogen and oxygen) are properly accounted for.
Understanding how these different conditions affect your approach ensures success in reaction balancing across varied chemical environments.

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Most popular questions from this chapter

A voltaic cell is constructed using the reaction of chromium metal and iron(II) ion. $$2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Fe}(\mathrm{s})$$ Complete the following sentences: Electrons in the external circuit flow from the _____ electrode to the _____ electrode. Negative ions move in the salt bridge from the _____ half-cell to the half-cell. The half-reaction at the anode is _____ and that at the cathode is ____.

A "silver coulometer" (Study Question 72) was used in the past to measure the current flowing in an electrochemical cell. Suppose you found that the current flowing through an electrolysis cell deposited \(0.089 \mathrm{g}\) of \(\mathrm{Ag}\) metal at the cathode after exactly 10 min. If this same current then passed through a cell containing gold(III) ion in the form of \(\left(\mathrm{AuCl}_{4}\right)^{-}\), how much gold was deposited at the cathode in that electrolysis cell?

An aqueous solution of KBr is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external source of electrical energy, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the half-reaction that occurs at this electrode. (b) Bromine is the primary product at the anode. Write an equation for its formation.

The reaction occurring in the cell in which \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and aluminum salts are electrolyzed is \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow\) \(\mathrm{Al}(\mathrm{s}) .\) If the electrolysis cell operates at \(5.0 \mathrm{V}\) and \(1.0 \mathrm{X}\) \(10^{5} \mathrm{A},\) what mass of aluminum metal can be produced in a 24-h day?

A voltaic cell can be built using the reaction between Al metal and \(\mathrm{O}_{2}\) from the air. If the Al anode of this cell consists of \(84 \mathrm{g}\) of aluminum, how many hours can the cell produce 1.0 A of electricity, assuming an unlimited supply of \(\mathrm{O}_{2} ?\)

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