Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 16

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

Short Answer

Expert verified
The oxidation numbers for the atoms in each compound are as follows: a. HNO3: H(+1), N(+5), O(-2) b. CuCl2: Cu(+2), Cl(-1) c. O2: O(0) d. H2O2: H(+1), O(-1) e. C6H12O6: C(0), H(+1), O(-2) f. Ag: Ag(0) g. PbSO4: Pb(+2), S(+6), O(-2) h. PbO2: Pb(+4), O(-2) i. Na2C2O4: Na(+1), C(+3), O(-2) j. CO2: C(+4), O(-2) k. (NH4)2Ce(SO4)3: N(-3), H(+1), Ce(+3), S(+6), O(-2) l. Cr2O3: Cr(+3), O(-2)

Step by step solution

01

a. Assign oxidation numbers in \(\mathrm{HNO}_{3}\)

We have a hydrogen, a nitrogen, and three oxygen atoms. 1. The oxidation number of hydrogen is +1. 2. The oxidation number of oxygen is -2. 3. We know that the sum of oxidation numbers must be equal to 0 (for a neutral molecule). Let's represent the oxidation number of nitrogen as x: +1 + x + 3(-2) = 0 Now, solve for x: x = +5 Thus, the oxidation numbers are: H: +1 N: +5 O: -2
02

b. Assign oxidation numbers in \(\mathrm{CuCl}_{2}\)

We have a copper (Cu) atom and two chlorine (Cl) atoms. 1. The oxidation number of chlorine in a compound is usually -1. 2. As there are two Cl atoms, their combined oxidation number is -2. 3. Let x be the oxidation number of Cu: x + 2(-1) = 0 Now, solve for x: x = +2 Thus, the oxidation numbers are: Cu: +2 Cl: -1
03

c. Assign oxidation numbers in \(\mathrm{O}_{2}\)

As this is an elemental form of oxygen, the oxidation number is 0.
04

d. Assign oxidation numbers in \(\mathrm{H}_{2}\mathrm{O}_{2}\)

We have two hydrogen atoms and two oxygen atoms. 1. The oxidation number of hydrogen is +1. 2. Unlike most oxygen-containing compounds, in \(\mathrm{H}_{2}\mathrm{O}_{2}\), the oxidation number of oxygen is -1. 3. Let's check the sum of oxidation numbers: 2(+1) + 2(-1) = 0, which is correct for a neutral molecule. Thus, the oxidation numbers are: H: +1 O: -1
05

e. Assign oxidation numbers in \(C_{6}H_{12}O_{6}\)

We have six carbon atoms, twelve hydrogen atoms, and six oxygen atoms. 1. The oxidation number of hydrogen is +1. 2. The oxidation number of oxygen is -2. 3. Let x be the oxidation number of carbon: 6x + 12(+1) + 6(-2) = 0 Now, solve for x: x = 0 Thus, the oxidation numbers are: C: 0 H: +1 O: -2
06

f. Assign oxidation numbers in \(\mathrm{Ag}\)

As this is an elemental form of silver, the oxidation number is 0.
07

g. Assign oxidation numbers in \(\mathrm{PbSO}_{4}\)

We have a lead (Pb) atom, a sulfur (S) atom, and four oxygen atoms. 1. The oxidation number of sulfur in a sulfate ion is +6. 2. The oxidation number of oxygen is -2. 3. Let x be the oxidation number of Pb: x + (+6) + 4(-2) = 0 Now, solve for x: x = +2 Thus, the oxidation numbers are: Pb: +2 S: +6 O: -2
08

h. Assign oxidation numbers in \(\mathrm{PbO}_{2}\)

We have a lead (Pb) atom and two oxygen atoms. 1. The oxidation number of oxygen is -2. 2. Let x be the oxidation number of Pb: x + 2(-2) = 0 Now, solve for x: x = +4 Thus, the oxidation numbers are: Pb: +4 O: -2
09

i. Assign oxidation numbers in \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)

We have two sodium atoms, two carbon atoms, and four oxygen atoms. 1. The oxidation number of sodium is +1. 2. The oxidation number of oxygen is -2. 3. Let x be the oxidation number of carbon: 2(+1) + 2x + 4(-2) = 0 Now, solve for x: x = +3 Thus, the oxidation numbers are: Na: +1 C: +3 O: -2
10

j. Assign oxidation numbers in \(\mathrm{CO}_{2}\)

We have a carbon atom and two oxygen atoms. 1. The oxidation number of oxygen is -2. 2. Let x be the oxidation number of carbon: x + 2(-2) = 0 Now, solve for x: x = +4 Thus, the oxidation numbers are: C: +4 O: -2
11

k. Assign oxidation numbers in \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\)

We have N, H, Ce, S, and O atoms 1. The oxidation number of hydrogen is +1. 2. The oxidation number of sulfur in a sulfate ion is +6. 3. The oxidation number of oxygen is -2. 4. Let x and y be the oxidation numbers of nitrogen (N) and cerium (Ce) respectively: +1 + x + 2y + 6(+6) + 12(-2) = 0 Now, solve for x and y From the ammonium ions (NH4): x+(1) * 4 = +1 (ammonium charge is +1); x = -3 From the sulfate ions (SO4): -2+(6) * 3 = -2 (sulfate charge is -2); y = +3 Thus, the oxidation numbers are: N: -3 H: +1 Ce: +3 S: +6 O: -2
12

l. Assign oxidation numbers in \(\mathrm{Cr}_{2}\mathrm{O}_{3}\)

We have two chromium (Cr) atoms and three oxygen atoms. 1. The oxidation number of oxygen is -2. 2. Let x be the oxidation number of Cr: 2x + 3(-2) = 0 Now, solve for x: x = +3 Thus, the oxidation numbers are: 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12} .\) Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

Electrolysis of an alkaline earth metal chloride using a current of 5.00 A for 748 s deposits 0.471 g of metal at the cathode. What is the identity of the alkaline earth metal chloride?

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at \(\mathrm{STP}\) are produced from the electrolysis of water by a current of \(2.50 \mathrm{A}\) in \(15.0 \mathrm{min} ?\)

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 \(M\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

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.

Sign-up for free