/*! 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 16 Assign oxidation numbers to all ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 16

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. Ag j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Short Answer

Expert verified
a. H: +1, N: +5, O: -2 e. C: 0, H: +1, O: -2 i. Na: +1, C: +3, O: -2 b. Cu: +2, Cl: -1 f. Ag: 0 j. C: +4, O: -2 c. O: 0 g. Pb: +2, S: +6, O: -2 k. N: -3, H: +1, Ce: +3, S: +6, O: -2 d. H: +1, O: -1 h. Pb: +4, O: -2 1. Cr: +3, O: -2

Step by step solution

01

a. HNO3

(+1) + (x) + 3(-2) = 0 (Since the total charge of the compound is 0) x = +5 Oxidation numbers: H: +1, N: +5, O: -2
02

e. C6H12O6

6(x) + 12(+1) + 6(-2) = 0 x = 0 Oxidation numbers: C: 0, H: +1, O: -2
03

i. Na2C2O4

2(+1) + 2(x) + 4(-2) = 0 x = +3 Oxidation numbers: Na: +1, C: +3, O: -2
04

b. CuCl2

(x) + 2(-1) = 0 x = +2 Oxidation numbers: Cu: +2, Cl: -1
05

f. Ag

This is a free element, so its oxidation number is 0.
06

j. CO2

(x) + 2(-2) = 0 x = +4 Oxidation numbers: C: +4, O: -2
07

c. O2

This is a free element, so its oxidation number is 0.
08

g. PbSO4

(x) + (+6) + 4(-2) = 0 x = +2 Oxidation numbers: Pb: +2, S: +6, O: -2
09

k. (NH4)2Ce(SO4)3

2(+1) * 2 + (x) + 3(+6) + 12(-2) = 0 x = +3 Oxidation numbers: N: -3, H: +1, Ce: +3, S: +6, O: -2
10

d. H2O2

2(+1) + 2(-1) = 0 Oxidation numbers: H: +1, O: -1
11

h. PbO2

(x) + 2(-2) = 0 x = +4 Oxidation numbers: Pb: +4, O: -2
12

1. Cr2O3

2(x) + 3(-2) = 0 x = +3 Oxidation numbers: Cr: +3, O: -2

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.

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between substances. These reactions involve two simultaneous changes: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons.

In a redox reaction, the substance that gives away electrons is known as the reducing agent, while the one that takes electrons is the oxidizing agent. By tracking these electron transfers, you can determine which elements are oxidized and which are reduced. This is often clarified using oxidation numbers, a key concept in balancing redox reactions and understanding electron flow.
  • **Oxidation**: Increase in oxidation number
  • **Reduction**: Decrease in oxidation number
  • **Reducing Agent**: Donates electrons
  • **Oxidizing Agent**: Accepts electrons
These reactions are significant in various applications, such as metabolism in biological systems, energy production in batteries, and many industrial processes.
Oxidation States
Oxidation states, or oxidation numbers, are useful for understanding the electron balance in chemical reactions. They indicate the degree of oxidation of an atom in a compound.

Assigning oxidation numbers helps identify the changes occurring during redox reactions and involves specific rules:
  • The oxidation number of any free element is zero. Examples: O2, Ag.
  • For ions, the oxidation number equals the charge of the ion. Example: Na+ is +1.
  • Oxygen usually has an oxidation number of -2, except in peroxides where it's -1.
  • Hydrogen is generally +1, but in hydrides, it's -1.
By following these rules, one can calculate the oxidation state of each element in a compound, helping to determine the nature of redox reactions.
Chemical Compounds
Chemical compounds are substances composed of two or more different elements bonded together. The properties of each compound are determined by the types and ratios of atoms involved.

To better understand the distribution of electrons in compounds, oxidation numbers are employed, revealing how electrons are shared or transferred among elements. For example:
  • **HNO3**: This compound has nitrogen with an oxidation state of +5.
  • **C6H12O6**: Carbon in this organic compound has a balanced oxidation state of 0.
  • **Na2C2O4**: Carbon here carries an oxidation state of +3.
Understanding these properties is essential in predicting the behavior of compounds during chemical reactions. Each compound's unique oxidation state configuration allows for the interpretation of their reactivity and function in various chemical processes.

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

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 \mathrm{M}\) KF solution b. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{M} \mathrm{MgI}_{2}\) solution

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q)\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of HCl? How can you harness this reaction to do useful work?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation
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.