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Write complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, (b) oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution.

Short Answer

Expert verified
The balanced half-reactions for the given processes are: (a) Oxidation of nitrous acid to nitrate ion in an acidic solution: \(HNO_2 + H_2O \rightarrow NO_3^- + 2H^+ + e^-\) (b) Oxidation of N2 to N2O in an acidic solution: \(N_2 + H_2O \rightarrow N_2O + 2H^+ + 2e^-\)

Step by step solution

01

Write Down the Unbalanced Half-reaction

Identify the reactants and products: \(HNO_2 \rightarrow NO_3^-\)
02

Balance Atoms Except H and O

The nitrogen atoms are already balanced in the half-reaction.
03

Balance Oxygen Atoms With Water

Add 1 water molecule to the left side to balance the oxygen atoms: \(HNO_2 + H_2O \rightarrow NO_3^-\)
04

Balance Hydrogen Atoms With H+ Ions

Add 2 H+ ions to the right side to balance the hydrogen atoms: \(HNO_2 + H_2O \rightarrow NO_3^- + 2H^+\)
05

Balance Charge With Electrons

Add 1 electron to the right side to balance the charges: \(HNO_2 + H_2O \rightarrow NO_3^- + 2H^+ + e^-\)
06

Check the Balance

Atoms: 1 N, 3 O, and 3 H on each side. Charge: 0 on both sides. Thus, the balanced half-reaction for oxidation of nitrous acid to nitrate ion in an acidic solution is: \(HNO_2 + H_2O \rightarrow NO_3^- + 2H^+ + e^-\) (b) Oxidation of N2 to N2O in an acidic solution
07

Write Down the Unbalanced Half-reaction

Identify the reactants and products: \(N_2 \rightarrow N_2O\)
08

Balance Atoms Except H and O

The nitrogen atoms are already balanced in the half-reaction.
09

Balance Oxygen Atoms With Water

Add 1 water molecule to the left side to balance the oxygen atoms: \(N_2 + H_2O \rightarrow N_2O\)
10

Balance Hydrogen Atoms With H+ Ions

Add 2 H+ ions to the right side to balance the hydrogen atoms: \(N_2 + H_2O \rightarrow N_2O + 2H^+\)
11

Balance Charge With Electrons

Add 2 electrons to the right side to balance the charges: \(N_2 + H_2O \rightarrow N_2O + 2H^+ + 2e^-\)
12

Check the Balance

Atoms: 2 N, 1 O, and 2 H on each side. Charge: 0 on both sides. Thus, the balanced half-reaction for oxidation of N2 to N2O in an acidic solution is: \(N_2 + H_2O \rightarrow N_2O + 2H^+ + 2e^-\)

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Key Concepts

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

Oxidation-Reduction Reactions
Oxidation-reduction reactions, often called redox reactions, are chemical reactions where the oxidation state of atoms is changed. Essentially, these reactions involve the transfer of electrons between chemical species. In simple terms, one reactant is oxidized, meaning it loses electrons, while another is reduced, gaining those electrons. This transfer of electrons is what drives the chemical process.

In redox reactions:
  • Oxidation refers to loss of electrons or increase in oxidation state.
  • Reduction involves gain of electrons or decrease in oxidation state.
For example, in the reaction where nitrous acid ( HNO_2 ) is oxidized to nitrate ion ( NO_3^- ), the nitrous acid loses electrons, indicating it is being oxidized. The overall balance of gained and lost electrons is crucial, as it helps maintain the reaction's charge balance.
Acidic Solution Balancing
Balancing reactions in acidic solutions requires specific steps to ensure the reaction is chemically correct. When balancing redox reactions in acidic solutions, it is important to add water, hydrogen ions ( H^+ ), and sometimes electrons to balance out the atoms and charges.

Here’s how you balance equations in acidic solutions:
  • Balance all elements other than H and O first.
  • Add H_2O molecules to balance the oxygen atoms.
  • Use H^+ ions to balance the hydrogen atoms, which are plentiful in acidic solutions.
  • Finally, add electrons to one side of the equation to ensure that the charge is balanced across the reaction.
These steps ensure that both mass and charge are conserved, resulting in a balanced chemical equation suitable for acidic environments.
Chemical Equation Balancing
Balancing chemical equations is a fundamental chemical skill, crucial for following the law of conservation of mass. This process ensures that the number of each type of atom, as well as the overall electric charge, is the same on both sides of the equation. In a chemical reaction, atoms are neither created nor destroyed, so a properly balanced equation reflects this principle.

Steps for balancing equations include:
  • Clearly identify all reactants and products in their correct formulas.
  • Start by balancing atoms that appear in only one reactant and one product.
  • Leave hydrogen and oxygen to be balanced last, as they often appear in multiple compounds.
  • When necessary, adjust coefficients to keep the equation balanced.
Using coefficients instead of changing subscripts maintains the integrity of the original compounds. This technique ensures the accuracy of chemical equations in representing real chemical processes, making it essential for studying reactions like the oxidation of nitrous acid or nitrogen molecules.

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Most popular questions from this chapter

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\).

Complete and balance the following equations: (a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (f) \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow\)

Borazine, \((\mathrm{BH})_{3}(\mathrm{NH})_{3},\) is an analog of \(\mathrm{C}_{6} \mathrm{H}_{6},\) benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from \(2.00 \mathrm{~L}\) of ammonia at STP, assuming diborane is in excess?

Explain each of the following observations: (a) At room temperature \(\mathrm{I}_{2}\) is a solid, \(\mathrm{Br}_{2}\) is a liquid, and \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are both gases. (b) \(\mathrm{F}_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. \((\mathbf{c})\) The boiling point of \(\mathrm{HF}\) is much higher than those of the other hydrogen halides. (d) The halugems decrease in uxidizing puwer in the urder \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\).

Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

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