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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. \(\mathrm{Ag}\) 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
The assigned oxidation numbers for the given compounds are: a. HNO鈧�: H=+1, N=+5, O=-2 b. CuCl鈧�: Cu=+2, Cl=-1 c. O鈧�: O=0 d. H鈧侽鈧�: H=+1, O=-1 e. C鈧咹鈧佲倐O鈧�: C=-1, H=+1, O=-2 f. Ag: Ag=0 g. PbSO鈧�: Pb=+2, S=+6, O=-2 h. PbO鈧�: Pb=+4, O=-2 i. Na鈧侰鈧侽鈧�: Na=+1, C=+3, O=-2 j. CO鈧�: C=+4, O=-2 k. (NH鈧�)鈧侰e(SO鈧�)鈧�: N=-3, H=+1, Ce=+3, S=+6, O=-2

Step by step solution

01

Assign oxidation numbers to each atom

For H, the oxidation number is +1. For N, let x be the oxidation number. For O, the oxidation number is -2. Since the sum of the oxidation numbers must be equal to the overall charge of the compound, which is 0, we have: \[+1+x+(-2 \times 3) = 0\] Solve for x: \[x-5=0\] \[x=+5\] Thus, the oxidation numbers for H, N, and O are +1, +5, and -2, respectively. b. \(\mathrm{CuCl}_{2}\)
02

Assign oxidation numbers to each atom

For Cu, let x be the oxidation number. For Cl, the oxidation number is -1. Since there are 2 Cl atoms, the sum of the oxidation numbers for this compound is equal to 0: \[x+(-1 \times 2) = 0\] Solve for x: \[x-2=0\] \[x=+2\] Thus, the oxidation numbers for Cu and Cl are +2 and -1, respectively. c. \(\mathrm{O}_{2}\)
03

Assign oxidation number to O

Oxygen in its elemental state (uncombined with other elements) has an oxidation number of 0. d. \(\mathrm{H}_{2} \mathrm{O}_{2}\)
04

Assign oxidation numbers to each atom

For H, the oxidation number is +1. For O, since it is a peroxide, the oxidation number is -1. e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)
05

Assign oxidation numbers to each atom

For H, oxidation number is +1. For O, oxidation number is -2. For C, let x be the oxidation number, and since there are 6 C atoms, we have: \[6x+12(+1)+6(-2)=0\] Solve for x: \[6x = -6\] \[x = -1\] Thus, the oxidation numbers for C, H, and O are -1, +1, and -2, respectively. f. \(\mathrm{Ag}\)
06

Assign oxidation number to Ag

Silver (Ag) in its elemental state (uncombined with other elements) has an oxidation number of 0. g. \(\mathrm{PbSO}_{4}\)
07

Assign oxidation numbers to each atom

For S, let x be the oxidation number. For O, the oxidation number is -2. For Pb, let y be the oxidation number: \[y+x+(-2 \times 4)=0\] Since sulfate ion (SO4) has an overall charge of -2: \[x+(-2 \times 4)=-2\] Solve for x: \[x-8=-2\] \[x=6\] As the overall charge of the compound is 0: \[y+6+(-2 \times 4)=0\] Solve for y: \[y=-2\] Thus, the oxidation numbers for Pb, S, and O are +2, +6, and -2, respectively. h. \(\mathrm{PbO}_{2}\)
08

Assign oxidation numbers to each atom

For Pb, let x be the oxidation number. For O, the oxidation number is -2: \[x+(-2 \times 2)=0\] Solve for x: \[x-4=0\] \[x=+4\] Thus, the oxidation numbers for Pb and O are +4 and -2, respectively. i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)
09

Assign oxidation numbers to each atom

For Na, the oxidation number is +1. For C, let x be the oxidation number. For O, the oxidation number is -2. Since there are 2 Na atoms and 2 C atoms: \[2(+1)+2x+(-2 \times 4)=0\] Solve for x: \[2x-6=0\] \[x=+3\] Thus, the oxidation numbers for Na, C, and O are +1, +3, and -2, respectively. j. \(\mathrm{CO}_{2}\)
10

Assign oxidation numbers to each atom

For C, let x be the oxidation number. For O, the oxidation number is -2: \[x+(-2 \times 2)=0\] Solve for x: \[x-4=0\] \[x=+4\] Thus, the oxidation numbers for C and O are +4 and -2, respectively. k. \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\)
11

Assign oxidation numbers to each atom

For N, let x be the oxidation number. For H, the oxidation number is +1. For Ce, let y be the oxidation number. For S, let z be the oxidation number. For O, the oxidation number is -2. \[2x+2(1)+(-3)+3z+3(-8)=0\] Solve for x: \[x=-3\] Therefore, the oxidation numbers for N, H, Ce, S, and O are -3, +1, +3, +6, and -2, respectively.

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Key Concepts

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

Chemical Compounds
A chemical compound is a substance composed of two or more different elements that are chemically bonded together in fixed proportions. Compounds exhibit unique physical and chemical properties that are distinct from the elements they are made of. These properties include boiling and melting points, reactivity, and electrical conductivity.

For example, water (H2O) is a compound made from hydrogen and oxygen. Despite being composed of two gases, in compound form, water is a liquid at room temperature. Another compound, sodium chloride (NaCl), known as table salt, is made from sodium, a metal that reacts violently with water, and chlorine, a poisonous gas. Together, they make a safe, stable substance that's essential for life. Understanding the composition and properties of chemical compounds is foundational in chemistry and is a critical step when learning to assign oxidation states and analyze redox reactions.
Assigning Oxidation States
The oxidation state, or oxidation number, is a concept used in chemistry to help keep track of electrons in chemical reactions, especially redox reactions. It is a theoretical charge on an atom if all bonds were ionic and all electrons were assigned to the most electronegative element. Assigning oxidation states is based on a set of rules:
  • Elements in their elemental form (O2, H2, Na, Ag) have an oxidation number of 0.
  • The oxidation number of a monatomic ion equals its charge (Na+ has an oxidation number of +1).
  • Hydrogen is usually assigned +1; oxygen is usually -2.
  • Fluorine is always -1; other halogens are generally -1 unless they're bonded to oxygen or fluorine.
  • The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion's charge.

By following these rules, you can determine the oxidation states for any compound, such as HNO3, where hydrogen is +1, nitrogen is determined to be +5, and oxygen is -2. Understanding how to assign these states correctly is crucial for analyzing redox reactions and balancing chemical equations.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two species. It's a process where one substance gets oxidized (loses electrons) while the other gets reduced (gains electrons). The substance that loses electrons is known as the reducing agent, and the substance that gains electrons is the oxidizing agent.

For instance, in the reaction of hydrogen and oxygen to form water, hydrogen (the reducing agent) donates electrons to oxygen (the oxidizing agent). By understanding oxidation numbers, we can identify which species are oxidized and which are reduced. For example, in the combustion of glucose (C6H12O6) into carbon dioxide (CO2) and water, glucose is oxidized and oxygen is reduced. Recognizing redox processes is essential in various fields, including energy production, corrosion, and biological systems, as they are fundamental to understanding electron transfer and chemical reactivity.

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Most popular questions from this chapter

Consider a cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ} &=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{E}^{\circ} &=0.77 \mathrm{~V} \end{aligned} $$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0\) min. If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what reactions occur at the anode and the cathode (assuming standard conditions)?

An electrochemical cell consists of a silver metal electrode immersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.00 M\) separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of \(10.00 \mathrm{M} \mathrm{NH}_{3}\) that also contains \(2.4 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) The equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\) is: \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{13}\) and the two cell half-reactions are: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C}\) ?
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