Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 17, Problem 29E

Balance the following oxidation–reduction reactions that occur in acidic solution using the half–reaction method.

a. I ( a q ) + ClO ( a q ) I 3 ( a q ) + Cl ( a q )

b. As 2 O 3 ( s ) + NO 3 ( a q ) H 3 AsO 4 ( a q ) + NO ( g )

c. Br ( a q ) + MnO 4 ( a q ) Br 2 ( l ) + Mn 2+ ( a q )

d. CH 3 OH ( a q ) + Cr 2 O 7 2 ( a q ) CH 2 O ( a q ) + Cr 3 + ( a q )

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (a).

Answer to Problem 29E

The balanced equations are as follows,

ClO-(aq)+2H+(aq)+3I-(aq)Cl-(aq)+H2O(l)+I3-(aq)

Explanation of Solution

The balanced equation is defined as follows,

ClO-(aq)+2H+(aq)+3I-(aq)Cl-(aq)+H2O(l)+I3-(aq)

The reduction half cell reaction is,

ClO-(aq)Cl-(aq)

The change in the oxidation number of chlorine is from +1 to 1 .

The oxidation half reaction is,

I-(aq)I3-(aq)

The change in oxidation number of iodine is from 1 to (13) .

As the atoms other than hydrogen and oxygen are already balanced, so directly balance the oxygen atom in the reduction half reaction by adding water to right hand side,

ClO-(aq)Cl-(aq)+H2O(l)

Balance the hydrogen atoms in the reduction half reaction by adding H+ ions to the left hand side,

ClO-(aq)+2H+(aq)Cl-(aq)+H2O(l)

Balance the charge by adding appropriate number of electrons to the left hand side,

ClO-(aq)+2H+(aq)+2e-Cl-(aq)+H2O(l) (1)

Balance the oxidation half reaction as,

3I-(aq)I3-(aq)

Balance the charge by adding electrons at the appropriate side,

3I-(aq)I3-(aq)+2e- (2)

Add the oxidation and reduction half reaction to get the final equation,

ClO-(aq)+2H+(aq)+2e-Cl-(aq)+H2O(l)3I-(aq)I3-(aq)+2e-

Cancel similar terms on both the sides. The final equation is,

ClO-(aq)+2H+(aq)+3I-(aq)Cl-(aq)+H2O(l)+I3-(aq)

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (b).

Answer to Problem 29E

The balanced equations are as follows,

3As2O3(s)+4H+(aq)+4NO3-(aq)+7H2O(l)6H3AsO4(aq)+4NO(g)

Explanation of Solution

The balanced equation is defined as follows,

3As2O3(s)+4H+(aq)+4NO3-(aq)+7H2O(l)6H3AsO4(aq)+4NO(g)

The reduction half cell reaction is,

NO3-(aq)NO(g)

The change in the oxidation number of nitrogen is from +5 to +2 .

The oxidation half reaction is,

As2O3(s)H3AsO4(aq)

The change in oxidation number of arsenic is from +3 to +5 .

Elements except hydrogen and oxygen are already balanced, therefore, directly balance the oxygen in the reduction half reaction by adding water to right hand side,

NO3-(aq)NO(g)+2H2O(l)

Balance the hydrogen atoms in the reduction half reaction by adding H+ ions to the left hand side,

NO3-(aq)+4H+(aq)NO(g)+2H2O(l)

Balance the charge by adding appropriate number of electrons to the left hand side,

NO3-(aq)+4H+(aq)+3e-NO(g)+2H2O(l) (3)

Balance the oxidation half reaction as,

As2O3(s)2H3AsO4(aq)

Balance the oxygen in the reduction half reaction by adding water to left hand side,

As2O3(s)+5H2O(l)2H3AsO4(aq)

Balance the hydrogen atoms in the reduction half reaction by adding H+ ions to the right hand side,

As2O3(s)+5H2O(l)2H3AsO4(aq)+4H+

Balance the charge by adding electrons at the appropriate side,

As2O3(s)+5H2O(l)2H3AsO4(aq)+4H++4e- (4)

Multiply equation (3) by 4 and equation (4) by 3 to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

3As2O3(s)+15H2O(l)6H3AsO4(aq)+12H++12e-4NO3-(aq)+16H+(aq)+12e-4NO(g)+8H2O(l)

Cancel similar terms on both the sides to get the final equation as,

3As2O3(s)+4H+(aq)+4NO3-(aq)+7H2O(l)6H3AsO4(aq)+4NO(g)

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (c).

Answer to Problem 29E

The balanced equations are as follows,

2MnO4-(aq)+10Br-(aq)+16H+(aq)2Mn2+(aq)+5Br2(l)+8H2O(l)

Explanation of Solution

The balanced equation is defined as follows,

2MnO4-(aq)+10Br-(aq)+16H+(aq)2Mn2+(aq)+5Br2(l)+8H2O(l)

The reduction half cell reaction is,

MnO4-(aq)Mn2+(aq)

The change in the oxidation number of manganese is from +7 to +2 .

The oxidation half reaction is,

Br-(aq)Br2(l)

The change in oxidation number of bromine is from 1 to 0 .

All the elements except hydrogen and oxygen are already balanced so directly balance the oxygen in the reduction half reaction by adding water to right hand side,

MnO4-(aq)Mn2+(aq)+4H2O(l)

Balance the hydrogen atoms in the reduction half reaction by adding H+ ions to the left hand side,

MnO4-(aq)+8H+(aq)Mn2+(aq)+4H2O(l)

Balance the charge by adding appropriate number of electrons to the left hand side,

MnO4-(aq)+8H+(aq)+5e-Mn2+(aq)+4H2O(l) (5)

Balance the oxidation half reaction as,

2Br-(aq)Br2(l)

As there are no atoms of hydrogen or oxygen, so directly balance the charge by adding electrons at the appropriate side,

2Br-(aq)Br2(l)+2e- (6)

Multiply equation (5) by 2 and equation (6) by 5 to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

2MnO4-(aq)+16H+(aq)+10e-2Mn2+(aq)+8H2O(l)10Br-(aq)5Br2(l)+10e-

Cancel similar terms on both the sides. The final equation is,

2MnO4-(aq)+10Br-(aq)+16H+(aq)2Mn2+(aq)+5Br2(l)+8H2O(l)

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (d).

Answer to Problem 29E

The balanced equations are as follows,

Cr2O72-(aq)+8H+(aq)+3CH3OH(aq)2Cr3+(aq)+3CH2O(aq)+7H2O(l)

Explanation of Solution

The balanced equation is defined as follows,

Cr2O72-(aq)+8H+(aq)+3CH3OH(aq)2Cr3+(aq)+3CH2O(aq)+7H2O(l)

The reduction half cell reaction is,

Cr2O72-(aq)Cr3+(aq)

The change in the oxidation number of chromium is from +7 to +2 .

The oxidation half reaction is,

CH3OH(aq)CH2O(aq)

The change in oxidation number of carbon is from 2 to 0 .

Balance all the atoms except hydrogen and oxygen in the reduction half cell,

Cr2O72-(aq)2Cr3+(aq)

Balance the oxygen in the reduction half reaction by adding water to right hand side,

Cr2O72-(aq)2Cr3+(aq)+7H2O(l)

Balance the hydrogen atoms in the reduction half reaction by adding H+ ions to the left hand side,

Cr2O72-(aq)+14H+(aq)2Cr3+(aq)+7H2O(l)

Balance the charge by adding appropriate number of electrons to the left hand side,

Cr2O72-(aq)+14H+(aq)+6e-2Cr3+(aq)+7H2O(l) (7)

As all the atoms except hydrogen are already balanced so directly balance the oxidation half reaction as,

CH3OH(aq)CH2O(aq)+2H+(aq)

Balance the charge by adding electrons at the appropriate side,

CH3OH(aq)CH2O(aq)+2H+(aq)+2e- (8)

Multiply equation (8) by 3 to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

Cr2O72-(aq)+14H+(aq)+6e-2Cr3+(aq)+7H2O(l)3CH3OH(aq)3CH2O(aq)+6H+(aq)+6e-

Cancel similar terms on both the sides. The final equation is,

Cr2O72-(aq)+8H+(aq)+3CH3OH(aq)2Cr3+(aq)+3CH2O(aq)+7H2O(l)

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Chapter 17 Solutions

Chemistry: An Atoms First Approach

Ch. 17 - Prob. 2ALQCh. 17 - Prob. 3ALQCh. 17 - Prob. 4ALQCh. 17 - Sketch a cell that forms iron metal from iron(II)...Ch. 17 - Which of the following is the best reducing agent:...Ch. 17 - Prob. 7ALQCh. 17 - Prob. 8ALQCh. 17 - Explain why cell potentials are not multiplied by...Ch. 17 - What is the difference between and ? When is equal...Ch. 17 - Prob. 11ALQCh. 17 - Look up the reduction potential for Fe3+ to Fe2+....Ch. 17 - Prob. 13ALQCh. 17 - Is the following statement true or false?...Ch. 17 - Prob. 15RORRCh. 17 - Assign oxidation numbers to all the atoms in each...Ch. 17 - Specify which of the following equations represent...Ch. 17 - The Ostwald process for the commercial production...Ch. 17 - Prob. 19QCh. 17 - Prob. 20QCh. 17 - When magnesium metal is added to a beaker of...Ch. 17 - How can one construct a galvanic cell from two...Ch. 17 - The free energy change for a reaction, G, is an...Ch. 17 - What is wrong with the following statement: The...Ch. 17 - When jump-starting a car with a dead battery, the...Ch. 17 - Prob. 26QCh. 17 - Prob. 27QCh. 17 - Consider the following electrochemical cell: a. If...Ch. 17 - Balance the following oxidationreduction reactions...Ch. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Chlorine gas was first prepared in 1774 by C. W....Ch. 17 - Gold metal will not dissolve in either...Ch. 17 - Prob. 35ECh. 17 - Consider the following galvanic cell: a. Label the...Ch. 17 - Prob. 37ECh. 17 - Sketch the galvanic cells based on the following...Ch. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Prob. 41ECh. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Give the standard line notation for each cell in...Ch. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - The amount of manganese in steel is determined by...Ch. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Estimate for the half-reaction 2H2O+2eH2+2OH given...Ch. 17 - Prob. 54ECh. 17 - Glucose is the major fuel for most living cells....Ch. 17 - Direct methanol fuel cells (DMFCs) have shown some...Ch. 17 - Prob. 57ECh. 17 - Using data from Table 17-1, place the following in...Ch. 17 - Answer the following questions using data from...Ch. 17 - Prob. 60ECh. 17 - Consider only the species (at standard conditions)...Ch. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 68ECh. 17 - Consider the concentration cell shown below....Ch. 17 - Prob. 70ECh. 17 - The overall reaction in the lead storage battery...Ch. 17 - Prob. 72ECh. 17 - Consider the cell described below:...Ch. 17 - Consider the cell described below:...Ch. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - An electrochemical cell consists of a nickel metal...Ch. 17 - An electrochemical cell consists of a standard...Ch. 17 - Prob. 82ECh. 17 - Consider a concentration cell that has both...Ch. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Consider the following galvanic cell at 25C:...Ch. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - The solubility product for CuI(s) is 1.1 102...Ch. 17 - How long will it take to plate out each of the...Ch. 17 - The electrolysis of BiO+ produces pure bismuth....Ch. 17 - What mass of each of the following substances can...Ch. 17 - Prob. 96ECh. 17 - An unknown metal M is electrolyzed. It took 74.1 s...Ch. 17 - Electrolysis of an alkaline earth metal chloride...Ch. 17 - What volume of F2 gas, at 25C and 1.00 atm, is...Ch. 17 - What volumes of H2(g) and O2(g) at STP are...Ch. 17 - Prob. 101ECh. 17 - A factory wants to produce 1.00 103 kg barium...Ch. 17 - It took 2.30 min using a current of 2.00 A to...Ch. 17 - A solution containing Pt4+ is electrolyzed with a...Ch. 17 - A solution at 25C contains 1.0 M Cd2+, 1.0 M Ag+,...Ch. 17 - Consider the following half-reactions: A...Ch. 17 - In the electrolysis of an aqueous solution of...Ch. 17 - Copper can be plated onto a spoon by placing the...Ch. 17 - Prob. 109ECh. 17 - Prob. 110ECh. 17 - Prob. 111ECh. 17 - What reaction will take place at the Cathode and...Ch. 17 - Gold is produced electrochemically from an aqueous...Ch. 17 - Prob. 114AECh. 17 - The saturated calomel electrode. abbreviated SCE....Ch. 17 - Consider the following half-reactions: Explain why...Ch. 17 - Consider the standard galvanic cell based on the...Ch. 17 - Prob. 118AECh. 17 - The black silver sulfide discoloration of...Ch. 17 - Prob. 120AECh. 17 - When aluminum foil is placed in hydrochloric acid,...Ch. 17 - Prob. 122AECh. 17 - Prob. 123AECh. 17 - The overall reaction and equilibrium constant...Ch. 17 - What is the maximum work that can be obtained from...Ch. 17 - The overall reaction and standard cell potential...Ch. 17 - Prob. 127AECh. 17 - Prob. 128AECh. 17 - Prob. 129AECh. 17 - Prob. 130AECh. 17 - Prob. 131AECh. 17 - Prob. 132AECh. 17 - Prob. 133AECh. 17 - Prob. 134CWPCh. 17 - Consider a galvanic cell based on the following...Ch. 17 - Consider a galvanic cell based on the following...Ch. 17 - Consider a galvanic cell based on the following...Ch. 17 - An electrochemical cell consists of a silver metal...Ch. 17 - An aqueous solution of PdCl2 is electrolyzed for...Ch. 17 - Prob. 140CPCh. 17 - Prob. 141CPCh. 17 - The overall reaction in the lead storage battery...Ch. 17 - Consider the following galvanic cell: Calculate...Ch. 17 - Prob. 144CPCh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 146CPCh. 17 - The measurement of pH using a glass electrode...Ch. 17 - Prob. 148CPCh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 150CPCh. 17 - Prob. 151CPCh. 17 - Prob. 152CPCh. 17 - Consider the following galvanic cell: A 15 0-mole...Ch. 17 - When copper reacts with nitric acid, a mixture of...Ch. 17 - The following standard reduction potentials have...Ch. 17 - An electrochemical cell is set up using the...Ch. 17 - Three electrochemical cells were connected in...Ch. 17 - A silver concentration cell is set up at 25C as...Ch. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 160MP
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