/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Write complete balanced half-rea... [FREE SOLUTION] | 91影视

91影视

Write complete balanced half-reactions for (a) reduction of nitrate ion to NO in acidic solution, (b) oxidation of HNO \(_{2}\) to \(\mathrm{NO}_{2}\) in acidic solution.

Short Answer

Expert verified
(a) Reduction of nitrate ion to NO in acidic solution: \(NO_3^- + 4 H^+ + 3 e^- \rightarrow NO + 2 H_2O \) (b) Oxidation of HNO鈧 to NO鈧 in acidic solution: \( HNO_2 + e^- \rightarrow NO_2 + H_2O + 2 H^+ \)

Step by step solution

01

Write the Unbalanced Half Reaction

First, let's write down the unbalanced half-reactions for each process. (a) Reduction of nitrate ion (NO鈧冣伝) to NO: NO鈧冣伝 鈫 NO (b) Oxidation of HNO鈧 to NO鈧: HNO鈧 鈫 NO鈧
02

Balance Atoms besides Oxygen and Hydrogen

In this case, nitrogen is already balanced in both half-reactions. (a) NO鈧冣伝 鈫 NO (b) HNO鈧 鈫 NO鈧
03

Balance Oxygen Atoms with H鈧侽

Add the appropriate number of H鈧侽 molecules to balance the oxygen atoms in the reaction. (a) NO鈧冣伝 鈫 NO + 2 H鈧侽 (b) HNO鈧 鈫 NO鈧 + H鈧侽
04

Balance Hydrogen Atoms with H鈦 ions

Add the appropriate number of H鈦 ions to balance the hydrogen atoms. (a) NO鈧冣伝 + 4 H鈦 鈫 NO + 2 H鈧侽 (b) HNO鈧 鈫 NO鈧 + H鈧侽 + 2 H鈦
05

Balance the Charge with e鈦 (electrons)

Add electrons to the half-reaction with a lower overall charge to ensure the charges on both sides of the reaction are equal. (a) NO鈧冣伝 + 4 H鈦 + 3 e鈦 鈫 NO + 2 H鈧侽 (b) HNO鈧 + e鈦 鈫 NO鈧 + H鈧侽 + 2 H鈦 Now, we have the complete balanced half-reactions for each process: (a) Reduction of nitrate ion to NO in acidic solution: NO鈧冣伝 + 4 H鈦 + 3 e鈦 鈫 NO + 2 H鈧侽 (b) Oxidation of HNO鈧 to NO鈧 in acidic solution: HNO鈧 + e鈦 鈫 NO鈧 + H鈧侽 + 2 H鈦

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Half-Reaction Method
The half-reaction method is a systematic approach for balancing redox reactions. Redox reactions involve the transfer of electrons between substances, and these reactions are split into two parts: reduction and oxidation. The half-reaction method separates these parts so that they can be balanced individually, making the overall process more manageable.

To balance a half-reaction, one must ensure that the number of atoms of each element and the total charge is the same on both sides of the reaction. This involves adding appropriate coefficients and, where necessary, water molecules, hydrogen ions (H鈦), and electrons (e鈦). Once both half-reactions are balanced, they can be combined to give the balanced overall redox reaction.

This process often requires an understanding of the conditions of the reaction, such as whether it occurs in an acidic or basic solution, as this affects the balancing agents used (H鈦 in acidic conditions and OH鈦 in basic conditions).
Reduction of Nitrate Ion
In the context of balancing redox reactions, understanding the process for the reduction of nitrate ion is crucial. The nitrate ion (NO鈧冣伝) is a common oxidizing agent and can be reduced to a variety of products depending on the conditions. In this case, we are focusing on the reduction of nitrate to nitrogen monoxide (NO) in an acidic solution.

The balancing of this reaction begins by writing the unbalanced half-reaction: NO鈧冣伝 鈫 NO. Next, we balance the oxygen atoms by adding water molecules. To balance the hydrogen atoms in an acidic solution, we use H鈦 ions. Finally, we add electrons to balance the charge, as the nitrate ion starts with a negative charge, and the resulting products must have a neutral charge. The balanced half-reaction retains the conservation of mass and charge, which states that matter is neither created nor destroyed, and the total charge must be the same on both sides of the equation.
Oxidation of Nitrous Acid
The oxidation state of an atom within a compound can change through chemical reactions. When we discuss the oxidation of nitrous acid (HNO鈧), we are referring to a process where the nitrogen in nitrous acid is increasing its oxidation state. In the reaction that was outlined, nitrous acid is oxidized to form nitrogen dioxide (NO鈧), which has a higher oxidation state of nitrogen.

The initial step is to write the skeleton reaction, then balance the atoms other than oxygen and hydrogen. In an acidic environment, water and hydrogen ions are used to balance the oxygen and hydrogen atoms, respectively. Lastly, electrons are added to balance the charge because oxidation involves the loss of electrons. A key aspect of this process is ensuring that the electrons lost in the oxidation half-reaction are the same as those gained in the reduction half-reaction, allowing them to cancel each other out when combined to give the overall balanced redox equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Complete and balance the following equations:$$ \begin{array}{l}{\text { (a) } \mathrm{ZnCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}} \\ {\text { (b) } \mathrm{BaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (c) } \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow}\end{array} $$ $$\begin{array}{l}{\text { (c) ~ } \mathrm{CS}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow} \\ {\text { (e) } \mathrm{Ca}(\mathrm{CN})_{2}(s)+\mathrm{HBr}(a q) \longrightarrow}\end{array}$$

Give a reason why hydrogen might be placed along with the group 1 A elements of the periodic table.

Consider the elements Li, \(\mathrm{K}, \mathrm{Cl}, \mathrm{C}, \mathrm{Ne},\) and Ar. From this list, select the element that (a) is most electronegative, (b) has the greatest metallic character, (c) most readily forms a positive ion, (d) has the smallest atomic radius, (e) forms \(\pi\) bonds most readily, (f) has multiple allotropes.

When aluminum replaces up to half of the silicon atoms in \(\mathrm{SiO}_{2},\) a mineral class called feldspars result. The feldspars are the most abundant rock-forming minerals, comprising about 50\(\%\) of the minerals in Earth's crust. Orthoclase is a feldspar in which Al replaces one-fourth of the Si atoms of \(\mathrm{SiO}_{2},\) and charge balance is completed by \(\mathrm{K}^{+}\) ions. Determine the chemical formula for orthoclase.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.